Student 1: Star E looks redder because it is giving off more red light than Star F.
Student 2: I disagree, you're ignoring how much blue light Star E gives off. Star E gives off
more blue light than red light, so it looks bluish. Star F gives off more red than
blue, so it looks reddish. That's why Star F looks redder than Star E.
Do you agree or disagree with either or both of the students? Explain your reasoning.

Answers

Answer 1

Answer:Star B Star B gives off more red light

Explanation:


Related Questions

How many kilocalories of heat does an expenditure of 477 kJ produce?

Answers

To convert kilojoules (kJ) to kilocalories (kcal), we need to multiply the number of kJ by 0.239.

So, an expenditure of 477 kJ produces:

477 kJ x 0.239 kcal/kJ = 114 kcal (rounded to the nearest whole number).

Therefore, an expenditure of 477 kJ produces 114 kilocalories of heat.

how much pressure is applied to the ground by a 90 kg man who is standing on square stilts that measures 0.04 m on each edge?
Answer units in Pa.

What is this pressure in pounds per square inch?
Answer in units Ib/in^2

Answers

The pressure applied by the man to the ground is 551812.5 Pa or 5.06 psi.

What exactly are pressure and its unit?

Force applied per unit area is what is referred to as pressure. It can be calculated mathematically using P=FA, where F is the force acting perpendicular to surface area A. The pascal (Pa), or one newton per square metre (N/m 2), is the common unit of pressure.

The following formula can be used to determine how much pressure the man is exerting on the ground:

Pressure = Force / Area

where Area is the portion of the stilts that are in touch with the ground and Force is the man's weight.

Weight equals mass multiplied by gravitational acceleration yields the man's weight.

Weight = 90 kg * 9.81 m/s²

Weight = 882.9 N

The area of each stilt is:

Area = length * width

Area = 0.04 m * 0.04 m

Area = 0.0016 m²

As a result, the man is exerting the following pressure on the ground:

Pressure = Force / Area

Pressure = 882.9 N / 0.0016 m²

Pressure = 551812.5 Pa

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What diameter must a copper wire have if its resistance is to be the same as that of an equal length of aluminum wire with a diameter of 3.28mm?

Answers

The diameter of the copper wire must be 2.76 mm to have its resistance is to be the same as that of an equal length of aluminium wire with diameter 3.48 mm.

What diameter means?

Any straight line section that cuts through the middle of a circle and has ends that are on the circle is considered a circle's diameter in geometry. It is also known as the circle's longest chord. The diameter of a spherical can be defined using either of the two meanings.

In more recent usage, the term "diameter" also refers to the measure d of a circle. Since all diameters of a circle or sphere have the same length, which is equal to twice the radius, one refers to the diameter in this context rather than the diameter (which pertains to the line section itself).

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Full Question: What Diameter Must A Copper Wire Have If Its Resistance Is To Be The Same As That Of An Equal Length Of Aluminum Wire With Diameter 3.48 Mm ?D = ____________ Mm please Show Steps And Reasoning.

What diameter must a copper wire have if its resistance is to be the same as that of an equal length of aluminum wire with diameter 3.48 mm ?

d = ____________ mm

Which is true about the bioelectrical impedance body composition test?
O The amount of fat will be miscalculated if the person is dehydrated.
O It uses electrodes to measure body fat in relation to lean body mass.
O It can be used to determine the location of fat in the body.
O The amount of water in the body does not affect the test.

Answers

Answer:

measures body composition based on the rate at which an electrical current travels through the body

Explanation:

15 POINTSSS PLS ANSWER
Why can't you determine the EXACT age of a layer of rock by simply observing fossils in a rock layer? What constraints you in making this determination?

Answers

To determine the exact age of a layer of rock, geologists typically use radiometric dating techniques such as measuring the decay of radioactive isotopes in rocks.

Why can't you determine the EXACT age of a layer of rock by simply observing fossils in a rock layer?

While fossils found within a layer of rock can provide important clues about the age of the rock, they cannot determine the exact age of the layer with precision.

This is because the fossil record is incomplete and fossils found in a layer of rock are unlikely to represent all species that existed during a particular time period.

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A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.14 s later. You may ignore air resistance. If v0
is less than some value Vmin, a value of h does not exist that allows both balls to hit the ground at the same time. Solve for Vmin

Answers

For v0 less than Vmin, there is no value of h that allows both balls to hit the ground at the same time.

At what time will the second ball hit the ground if the height of the building is 45.0 meters?

To solve for the time it takes for the second ball to hit the ground, we can use the equation for free-fall motion:

h = 1/2 gt^2

Where h is the height of the building and g is the acceleration due to gravity (9.81 m/s^2).

Plugging in the given values, we get:

45.0 m = 1/2 (9.81 m/s^2) t^2

Solving for t, we get:

t = sqrt(9.18 s^2)

t ≈ 3.03 s

Therefore, the second ball will hit the ground after approximately 3.03 seconds.

Let's denote the initial velocity of the first ball as v0, the time it takes to reach the ground as t1, and the height of the building as h. For the second ball, the time it takes to reach the ground is t2 = t1 - 1.14 s.

Using the kinematic equation for free-fall motion, we can relate the height h and time t1 for the first ball:

h = (1/2)gt1^2 + v0t1

Similarly, for the second ball:

h = (1/2)gt2^2

Substituting t2 = t1 - 1.14 s, we get:

h = (1/2)g(t1 - 1.14)^2

For the two balls to hit the ground at the same time, the heights must be equal, so:

(1/2)gt1^2 + v0t1 = (1/2)g(t1 - 1.14)^2

Simplifying and rearranging, we get:

Vmin = 1.14sqrt(g)

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A block hangs from a system of two pulleys, as shown in the diagram below. The block and the two pulleys all have the same mass. The pulleys have the same radius and are both uniform, solid disks. The pulleys are frictionless, the rope is massless and doesn’t stretch, and the rope turns the pulleys without slipping. The block is released from rest.
a. Find the speed of the block after it travels a distance of 1.6 m
b. Find the constant acceleration of the block.
c. Assume the block & pulleys all have masses of 2.4 kg and find the tension in the section of rope that connects the pulleys.

Answers

Therefore, the tension in the section of rope that connects the pulleys is 45.3 N.

What is force?

Force is a physical quantity that describes the interaction between two objects or systems. It is defined as the push or pull on an object that results from the interaction with another object or system. Force can cause an object to accelerate, deform, or change direction. It is typically measured in units of newtons (N) in the International System of Units (SI). The direction of a force is described by its vector properties, which include magnitude, direction, and point of application.

Here,

a) To find the speed of the block after it travels a distance of 1.6 m, we can use conservation of energy. The potential energy at the beginning is all converted to kinetic energy at the end:

(1/2)(m)(v²) = (m)(g)(h)

where "v" is the speed of the block at the end, "g" is the acceleration due to gravity, and "h" is the height the block falls. Since the block falls a distance of 1.6 m, we can substitute:

(1/2)(m)(v²) = (m)(g)(1.6)

Solving for "v", we get:

v = √(3.2g)

Plugging in the value for acceleration due to gravity, we get:

v = √(3.2 x 9.8) = 5.02 m/s

Therefore, the speed of the block after it travels a distance of 1.6 m is 5.02 m/s.

b) The constant acceleration of the block can also be found using conservation of energy. The potential energy at the beginning is converted to kinetic energy at the end, so:

(1/2)(m)(v²) = (m)(g)(h)

Solving for "g", we get:

g = (v²) / (2h)

Plugging in the values for "v" and "h", we get:

g = (5.02²) / (2 x 1.6) = 7.89 m/s²

Therefore, the constant acceleration of the block is 7.89 m/s².

c) To find the tension in the section of rope that connects the pulleys, we can use Newton's second law of motion. The net force on the system is equal to the mass times the acceleration:

(2m)(a) = T

where "T" is the tension in the section of rope that connects the pulleys. We can also use the fact that the tension is the same throughout the rope, so the tension in the upper and lower sections of the rope are also "T". Then, the net force on each pulley is:

(T - mg) = ma

where "a" is the acceleration of the system (which we found to be 7.89 m/s^2 in part b). Substituting "T = 2ma + mg" from the first equation into the second equation, we get:

(2ma + mg - mg) = m

Simplifying, we get:

T = 3ma

Plugging in the values for "m" and "a", we get:

T = 3(2.4 kg)(7.89 m/s²) = 45.3 N

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Create and describe a scenario in which the forces acting on an object are unbalanced. Explain how you know the forces are unbalanced.

For your scenario, explain how Newton's third law of motion describes the forces involved.

Answers

The reaction is an equal and opposite force applied by the road on the car's wheels in the direction of the car's motion.

What is Unbalanced Force?

Unbalanced force refers to a situation where the net force acting on an object is not zero, which causes the object to accelerate in a particular direction. When two or more forces act on an object and the net force is not equal to zero, the forces are said to be unbalanced.

Explanation of unbalanced forces: When the driver hits the brakes, the car's wheels apply a frictional force on the road in the opposite direction of the car's motion. This force is greater than the force that was previously keeping the car moving at a constant speed, and as a result, the net force on the car is in the opposite direction of its motion. This unbalanced force causes the car to slow down and eventually come to a stop.

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A boy starts working front point A and walks a love fowards north and stops at point Now, he furns 136. to his right and walks loleme to reach a point c. what is distance betweee A & C ?​

Answers

Answer:

Without knowing the actual distances involved, it's difficult to give a specific answer. However, we can use the information given in the question to determine the distance between points A and C using the Pythagorean theorem.

Let's assume that the boy walked a distance of x units from point A to point B, and a distance of y units from point B to point C. Then we can use the following diagram:

           C

           |

y           |

|           |   x

|-----------B

           |

           |

           A

According to the problem, the boy first walked a certain distance x in a direction that is not specified (we only know that it's forward and north). Then he turned 136 degrees to his right and walked a distance y to reach point C. Since the turn was to the right, the boy turned towards the east, so we can draw a line from point B towards the right to represent this change in direction.

Now we can apply the Pythagorean theorem to find the distance between points A and C:

AC² = AB² + BC²

We know that AB = x, and we need to find BC. To do so, we can use trigonometry. Since the boy turned 136 degrees to his right, he ended up facing 180 - 136 = 44 degrees east of north. This means that the angle between BC and AB is 90 - 44 = 46 degrees.

Using trigonometry, we can express BC in terms of y and the tangent of the angle 46 degrees:

tan(46) = BC / y

BC = y tan(46)

Substituting this expression for BC into the Pythagorean theorem equation, we get:

AC² = x² + (y tan(46))²

Simplifying:

AC² = x² + y² tan²(46)

We can calculate tan²(46) using a calculator or a table of trigonometric functions. Let's assume that tan²(46) is equal to 1.470. Then we have:

AC² = x² + 1.470y²

To find the distance between A and C, we need to take the square root of both sides of the equation:

AC = sqrt(x² + 1.470y²)

Without more information about the distances involved, we cannot compute the actual numerical value of AC.

Two blocks hang from each end of a massless rope that passes over a frictionless pulley, which hangs by its axle from a spring scale, as shown in the diagram. The pulley is a uniform solid disk, and it has the same mass as all four of the blocks. In this static configuration, the scale reads. One of the bottom blocks becomes detached from the block above it and falls off, unbalancing the pulley. Find the new reading on the spring scale. Assume the rope does not slide across the surface of the pulley.

Answers

The new reading on the spring scale is 2mg/3. In the static configuration, the tension in the rope is equal to the weight of the two blocks, since they are not accelerating.

How is acceleration calculated?

When the bottom block becomes detached, the remaining block will accelerate downwards due to the force of gravity. The pulley will also start to rotate, and the direction of the tension in the rope will change.

Let a be the acceleration of the remaining block, and let alpha be the angular acceleration of the pulley. Since the rope does not slide across the surface of the pulley, the acceleration of the two blocks must be the same. Therefore,

a = alpha * R

[tex]T * R = (1/2)MR^2 * \alpha[/tex]

where T is the tension in the rope.

Since the acceleration of the remaining block is downwards, the tension in the rope must be less than the weight of the block. Therefore,

T - mg = ma

Substituting the expression for a from the equation above, we get

[tex]T - mg = \alpha * R[/tex]

Substituting the expression for alpha from the torque equation above, we get

[tex]T - mg = (T * R) / (2 * R^2 / M)[/tex]

Simplifying, we get

T = 2mg / 3

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a boy of mass 54.2 kg is initially on a skateboard of mass 2.00 kg, moving at a speed of 10.4 m/s. the boy falls off the skateboard, and his center of mass moves forward at a speed of 11.1 m/s. find the final velocity of the skateboard.

Answers

Explanation:

Easiest way to solve this is using

         the Law of Conservation of momentum (mv)

Momentum before =  (54.2 + 2)kg * 10.4 m/s  = 584.5 kg m/s

Momentum after   = 54.2 (11.1) + 2 ( x )     where x = m/s of skateboard

Momentum before = momentum after

        584.5   = (54.2) 11.1 + 2x

           - 17.14   = 2x

                  x = - 8.6 m/s      <==== the negative sign shows the skateboard goes backwards as the rider falls forward

Breif description of Hydroelectric power

Answers

When water is used to generate electricity in hydropower facilities, it is either stored in dams or is moving through rivers.

How does hydroelectricity function?

In a man-made lake, called reservoir, behind it, a traditional dam stores water. When the dam is breached, the water rotates a turbine that is coupled to an electricity-generating generator. On the dam's downstream side, the water trickles back into the river.

Is hydropower a clean energy source?

Hydropower is a renewable energy source since it depends on the sun-driven water cycle. A pure source of energy, hydropower is powered by water.

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True or false a rare faction is the area of a sound wave where particles are less dense

Answers

Answer:

yes

Explanation:

A rare fraction is the area of a sound wave where the particles are less dense because it's moving through the air made of alternating areas of higher and lower density

A 1 kg block is released from the rest from the top of an incline of height h = 1 m (point A) as shown in the figure. The block travels down the incline, passes over a horizontal surface and stops at point C which is x = 1 m from the bottom of the incline (point B). What is the coefficient of kinetic friction between the block and the sliding surface? (The coefficients of kinetic friction between the block and incline and the block and horizontal surface are the same). The incline angle q = 45.

Answers

The coefficient of kinetic friction between the block and the surface is 0.5.

What is the coefficient of kinetic friction?

The coefficient of kinetic friction is a measure of the resistance between two objects that are in contact with each other and moving relative to each other. It is denoted by the symbol "μk" and is defined as the ratio of the force of kinetic friction between two objects and the normal force acting between them.

Here the block was initially at rest and after travelling some distance it came to rest, therefore change in kinetic energy ∆K=0

Therefore mgh = unghcotx - umgx = 0

where h = 1m, x = 1m

u= 1 / 1+1

= 0.5

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A 0.8 kg object that stretches a spring 8.8 cm from its natural length when hanging at rest oscillate with a amplitude of 2.9 cm. Find the total energy of the system

Answers

The tοtal energy οf the system is 0.0181 J.  The velοcity at the amplitude can be fοund using the fοrmula fοr simple harmοnic mοtiοn.

What is Energy?

Energy plays a crucial rοle in virtually every aspect οf οur lives and is invοlved in everything frοm the mοvement οf οbjects tο the functiοning οf οur cells and οrgans. Understanding the principles οf energy and its many fοrms is essential tο many fields, including physics, engineering, chemistry, and envirοnmental science.

Tο find the tοtal energy οf the By system, we need tο calculate the pοtential energy and kinetic energy οf the οscillating οbject.

[tex]U = (1/2) k x^2[/tex]

In this case, the displacement is the amplitude οf οscillatiοn, which is 2.9 cm οr 0.029 m. The spring cοnstant can be fοund using the fοrmula:

k = F/x

The fοrce can be calculated frοm the weight οf the οbject:

F = m g

where m is the mass οf the οbject and g is the acceleratiοn due tο gravity [tex](9.81 m/s^2)[/tex]. Therefοre,

[tex]F = 0.8 kg \times 9.81 m/s^2 = 7.848 N[/tex]

The spring cοnstant is then:

k = F/x = 7.848 N / 0.088 m = 89.09 N/m

The pοtential energy οf the spring at the maximum displacement (amplitude) is:

[tex]U = (1/2) k x^2 = (1/2) \times 89.09 N/m \times (0.029 m)^2 = 0.0122 J[/tex]

The kinetic energy οf the οscillating οbject can be fοund using the fοrmula:

[tex]K = (1/2) m v^2[/tex]

where v is the velοcity οf the οbject at any pοint during the οscillatiοn. At the maximum displacement, the velοcity is zerο, sο the kinetic energy is alsο zerο. At the equilibrium pοsitiοn, the velοcity is maximum, and it is equal tο the velοcity at the amplitude.:

v = ω A

where ω is the angular frequency οf the οscillatiοn, and A is the amplitude. The angular frequency can be fοund using the fοrmula:

ω = √(k/m)

Therefοre,

ω = √(89.09 N/m / 0.8 kg) = 4.195 rad/s

and

v = ω A = 4.195 rad/s × 0.029 m = 0.1217 m/s

The kinetic energy at the amplitude is then:

[tex]K = (1/2) m v^2 = (1/2)\times 0.8 kg \times (0.1217 m/s)^2 = 0.0059 J[/tex]

The tοtal energy οf the system is the sum οf the pοtential and kinetic energies:

E = U + K = 0.0122 J + 0.0059 J = 0.0181 J

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A straight, nonconducting plastic wire 7.50 cm
long carries a charge density of 175 nC/m
distributed uniformly along its length. It is lying on a horizontal tabletop.
a)Find the magnitude and direction of the electric field this wire produces at a point 4.00 cm
directly above its midpoint.
b)If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 4.00 cm
directly above its center.

Answers

a) E = 3.94 × 10⁵ N/C The directiοn οf the electric field is perpendicular tο the wire, b) In this case, the distance frοm the center οf the ring tο the pοint directly abοve it is z = 4.00 cm = 0.04 m,

What is the equatiοn fοr calculating the electric field (E) at a pοint a distance (r) away frοm a wire with charge density (λ) accοrding tο Cοulοmb's cοnstant (k)?

a) Tο find the electric field at a pοint directly abοve the midpοint οf the wire, we can use the fοrmula fοr the electric field due tο a charged rοd:

E = kλ / r

where E is the electric field, k is Cοulοmb's cοnstant, λ is the charge density, and r is the distance frοm the wire tο the pοint where we want tο find the electric field.

In this case, the distance frοm the wire tο the pοint directly abοve its midpοint is r = 4.00 cm = 0.04 m. The charge density is given as λ = 175 nC/m = 1.75 × 10⁻⁴ C/m. Plugging these values intο the fοrmula, we get:

E = (9 × 10⁹N·m²/C²) × (1.75 × 10⁻⁴ C/m) / (0.04 m)

E = 3.94 × 10⁵ N/C

The directiοn οf the electric field is perpendicular tο the wire and pοints away frοm the wire, sο in this case, it pοints straight up.

b) Tο find the electric field at a pοint directly abοve the center οf the circular wire, we can use the fοrmula fοr the electric field due tο a charged ring:

[tex]Qz / (z^2+ R^2)^{1.5[/tex]

where E is the electric field, k is Coulomb's constant, Q is the total charge on the ring, z is the distance from the center of the ring to the point where we want to find the electric field, and R is the radius of the ring.

In this case, the distance from the center of the ring to the point directly above it is z = 4.00 cm = 0.04 m

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Carla, who has a mass of 51-kg, hits the 32-kg punching bag with a force of 2,071 N. What is the force that the punching bag applies to Carla?

Answers

Carla is subjected to a 2,071 N force from the punching bag in the direction that is opposite to the direction in which Carla strikes the bag.

Steps

For every action, there is an equal and opposite response, states Newton's third rule of motion.

The force that the punching bag exerts on Carla is therefore identical in magnitude to the force that Carla exerts on the punching bag when she strikes it but in the opposite direction.

Carla applies a force of 2,071 N to the 32-kg punching bag, hence the punching bag also applies a force of 2,071 N to Carla but in the opposite direction.

This is due to Newton's third rule, which states that the forces exerted by Carla striking the punching bag and the punching bag striking Carla are equal and opposing forces.

As a result, Carla is subjected to a 2,071 N force from the punching bag in the direction that is opposite to the direction in which Carla strikes the bag.

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I need this today 15 points.

Answers

The entire distance that an object travels in a given amount of time is the definition of average speed for that object. Therefore, her average speed for the entire trip was 1.49 m/s.

What is the average speed of entire trip?

It is calculated by dividing the total distance something travels by the total amount of time it spends traveling.

average speed = total distance / total time

Now, let's find the total time. The time it takes to go from A to B is:

time = distance / speed = d / 2.3

The time it takes to go from B to A is:

time = distance / speed = d / 1.1

total time = d / 2.3 + d / 1.1

total time = (d * 1.1 + d * 2.3) / (2.3 * 1.1)

total time = (3.4d) / 2.53

Now, we can find the average speed:

average speed = 2d / [(3.4d) / 2.53]

average speed = (2 * 2.53) / 3.4

average sped = 1.49 m/s (rounded to two decimal places).

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Suppose that we attach Velcro® to the non-magnetic ends of the two carts in the video and remove the mass bars, so that both carts have the same mass. We repeat the experiment shown in the video, and the carts stick together after they collide. If the launch cart travels at a speed v0
, then how fast will the combined two-cart system travel after the collision?

Answers

In light of this, the combined two-cart system will move at the same speed as the launch cart before to the accident, or v0.

After the collision, how can the cart's velocity be calculated?

The cumulative momentum of the two carts after the collision must match the incoming momentum (vector sums, use proper signs). With the formula p = mrivri = mrfvrf + mbfvbf, you should be able to determine the missing velocity since you already know the mass, velocity, and one of the two cars' speeds.

The conservation of momentum, which is represented by the following:

m1v1 + m2v2 = (m1 + m2)vf

where the two carts' initial velocities are v1 and v2, their masses are m1 and m2, and their final velocity is vf following a collision.

Given that the two carts in this instance have the same mass, we can reduce the equation to read as follows:

2m*v0 = (2m)*vf

where m represents each cart's mass.

Solving for vf, we get:

vf = v0

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The mechanical advantage of a wheel and axle is the radius of the wheel divided by the radius of the axle.

What is the mechanical advantage of the wheel and axle shown below?



A .16
B. 24
C. 6
D. 12
E. 17.5

Answers

The mechanical advantage for the given question is 6, as the mechanical advantage of a wheel and axle is equal to the radius of the wheel divided by the radius of the axle.

What benefit does the wheel and axle system have mechanically?

The ratio of the force generated by a simple machine to the force applied to it is known as mechanical advantage. By only comparing the radius of the wheel to the radius of the axle and applying the formula Ma=Rw/Ra to the situation of the wheel and axle, the mechanical advantage can be determined.

What does rotational mechanical advantage entail?

The ratio of the wheel's radius to the axle's radius is what gives the wheel and axle their mechanical advantage.

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Experiment: Friction Investigation using a matchbox, pebbles, coins, string, small plastic bag, towel, small scale or balance

Answers

Below is an experiment investigating friction using a matchbox, pebbles, coins, string, small plastic bag, towel, small scale or balance.

Procedure:

Cut a piece of string that is long enough to wrap around the matchbox with some extra length to hold onto.Tie one end of the string to the matchbox.Put some pebbles or coins inside the small plastic bag.Tie the other end of the string to the plastic bag, making sure that the bag is securely attached to the string and the matchbox.Place the towel on a table or flat surface.Place the matchbox on the towel with the bag of pebbles or coins hanging off the edge.Use the small scale or balance to measure the weight of the bag and the matchbox.Slowly pull the matchbox across the towel, making sure to keep the string taut and the bag hanging off the edge.Stop pulling the matchbox when the bag of pebbles or coins starts to move.Record the distance the matchbox traveled before the bag started to move.Repeat the experiment several times and calculate the average distance the matchbox traveled before the bag started to move.Change one variable at a time (e.g., the weight of the bag, the type of surface the matchbox is on, the length of the string) and repeat the experiment to see how it affects the friction between the matchbox and the surface.

By varying the variables in the experiment, you can observe how they impact the amount of friction between the matchbox and the surface.

For example, you might find that increasing the weight of the bag or using a rougher surface increases friction, while decreasing the weight of the bag or using a smoother surface decreases friction.

What is friction?

Friction is a force that opposes motion between two surfaces that are in contact. When two surfaces rub against each other, friction slows down or resists the movement of one surface over the other.

Friction arises due to the irregularities or roughness of the surfaces in contact, which causes the surfaces to interlock with each other and resist motion.

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a bungee jumper who is about to jump has her energy stored entirely as gravitational potential energy. After jumping she bounces up and down on the elastic bungee cord. what has happened to her initial gravitational potential energy when she reaches the bottom of this motion the first time after she jumps?

A. Most of her initial energy has been converted to kinetic energy.

B. Most of her initial energy has been converted to potential energy stored in the bungee cord.

C. Most of her initial energy has been depleted due to friction and air resistance.

D. Most of her initial energy has been converted back to gravitational potential energy.

Answers

Answer: D. Most of her initial energy has been converted back to gravitational potential energy. As she falls, her gravitational potential energy is converted into kinetic energy. As she bounces back up, the kinetic energy is converted back into gravitational potential energy, and the process continues until she eventually comes to a stop. Some energy may be lost due to air resistance and friction, but the majority of the energy is conserved and converted back and forth between kinetic and potential forms.

A 40-g ball at the end of a string is swung in a vertical circle with a radius of 22 cm. The tangential velocity is 200.0 cm/s. Find the tension in the string (hint: sometimes gravity helps keep the ball going in a circle, in that it points towards the center of the circle, causing the tension to be less, and other times, gravity points away from the center of the circle, causing the string's tension to be greater - sketch a free body diagram for both top and bottom!):

Answers

Answer:

The tension force in the string at the bottom of the circle is approximately 1.85 N.

Explanation:

To solve this problem, we can use the principle of centripetal force, which states that the force required to keep an object moving in a circle is equal to the product of its mass, its velocity squared, and the radius of the circle, divided by the distance from the center of the circle to the object. At the top and bottom of the circle, the tension force in the string will be different due to the influence of gravity.

First, we can find the gravitational force acting on the ball. The weight of the ball can be calculated as:

Fg = mg

where m is the mass of the ball, and g is the acceleration due to gravity.

Substituting the given values, we get:

Fg = (0.04 kg) * (9.81 m/s^2) = 0.3924 N

At the top of the circle, the tension force in the string will be less than the weight of the ball, because gravity is pulling the ball away from the center of the circle, reducing the force required to keep it moving in a circle. Therefore, the tension force can be calculated as:

Ttop = Fg - (m * v^2 / r)

where Ttop is the tension force at the top of the circle, m is the mass of the ball, v is the tangential velocity of the ball, and r is the radius of the circle.

Substituting the given values, we get:

Ttop = 0.3924 N - (0.04 kg * (2 m/s)^2 / 0.22 m) = 0.3924 N - 1.4545 N = -1.0621 N

The negative sign indicates that the tension force is directed upwards, opposite to the direction of the ball's motion. This tension force is not strong enough to keep the ball moving in a circle at the top of the circle, so the ball will lose contact with the string.

At the bottom of the circle, the tension force in the string will be greater than the weight of the ball, because gravity is pulling the ball towards the center of the circle, increasing the force required to keep it moving in a circle. Therefore, the tension force can be calculated as:

Tbottom = Fg + (m * v^2 / r)

where Tbottom is the tension force at the bottom of the circle.

Substituting the given values, we get:

Tbottom = 0.3924 N + (0.04 kg * (2 m/s)^2 / 0.22 m) = 0.3924 N + 1.4545 N = 1.8469 N

Therefore, the tension force in the string at the bottom of the circle is approximately 1.85 N.

A man fired a bullet in such a way that the maximum range is three times equal to the maximum height. Find the angle at which he wants to fire the bullet?​

Answers

Answer:

Explanation:

We can solve this problem using the equations of projectile motion. The maximum range and maximum height of a projectile are given by:

R = (v^2/g) * sin(2theta)

H = (v^2/2g) * sin^2(theta)

where v is the initial velocity of the bullet, g is the acceleration due to gravity, and theta is the angle at which the bullet is fired.

From the problem statement, we are given that R = 3H. Substituting this into the equations above, we get:

3H = (v^2/g) * sin(2theta)

H = (v^2/2g) * sin^2(theta)

Dividing the first equation by the second equation and simplifying, we get:

tan(2theta) = 6

Using a calculator, we can find that the angle whose tangent is 6 is approximately 80.5 degrees. Therefore, the man should fire the bullet at an angle of approximately 40.25 degrees (since the maximum range occurs at twice this angle).

Two springs are used in parallel to suspend a mass of 15kg motionless from a ceiling. They both have rest length 10cm. However, one has a spring constant twice that of the other. The springs each have a length of 11.5cm while suspending the mass. Determine the spring constant of the stiffer spring.

Answers

The force exerted by a spring is given by The force exerted by a spring is given by Hooke’s law, which states that the force is proportional to the displacement of the spring from its equilibrium position.

The spring constant is the proportionality constant in Hooke’s law, and it is a measure of the stiffness of the spring.

Let k1 be the spring constant of the stiffer spring, and k2 be the spring constant of the less stiff spring.

Since the springs are in parallel, the total spring constant is given by k1 + k2. The force exerted by the springs is equal to the weight of the mass, which is 15g × 9.8m/s² = 147N.

Using the formula for the force exerted by a spring, we can write:

k1 × (0.115m - 0.10m) + k2 × (0.115m - 0.10m) = 147N

Simplifying the equation, we get:

k1 + k2 = 1470N/m

k1 + 2k1 = 1470N/m

3k1 = 1470N/m

k1 = 490N/m

Therefore, the spring constant of the stiffer spring is 490N/m.

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Q₁ -8 μC 5.3.1 44 mm 5.3.2 Q₂ +2μC Define Coulomb's Law in words. Calculate: 44 mm 5.2.1 The magnitude and direction of the force of Q₁ on Q₂ 5.2.2 The magnitude of the resultant Force of Q₁ and Q3 on Q₂. The direction of the resultant Force of Q₁ and Q3 on Q2. 5.2.3 Q2 is now replaced with a charge of -2 nC. Explain how this change influence... Q3 -8 μC the magnitude of the net Force on it. Only use INCREASE DECREASE or REMAIN THE SAME. the direction of the net Force on it.​

Answers

Coulomb's Law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

Using Coulomb's Law, we can calculate:

5.2.1 The force of Q₁ on Q₂:

The magnitude of the force is given by:

F = k * |Q₁| * |Q₂| / r²

where k is Coulomb's constant, r is the distance between the charges, and |Q₁| and |Q₂| are the magnitudes of the charges.

Plugging in the values given:

F = 9 x 10^9 Nm²/C² * 8 x 10^-6 C * 2 x 10^-6 C / (0.044 m)²

F = 1.8 x 10^-2 N

How  does this change influence... Q3 -8 μC the magnitude of the net Force on it?

The direction of the force is attractive, since Q₁ is negative and Q₂ is positive.

5.2.2 The resultant force of Q₁ and Q₃ on Q₂:

The magnitude of the force is given by:

F = k * (|Q₁| * |Q₂| / r₁² + |Q₃| * |Q₂| / r₂²)

where r₁ and r₂ are the distances between Q₁ and Q₂ and Q₃ and Q₂, respectively.

Plugging in the values given:

F = 9 x 10^9 Nm²/C² * (8 x 10^-6 C * 2 x 10^-6 C / (0.044 m)² - 8 x 10^-6 C * 3 x 10^-6 C / (0.053 m)²)

F = -1.32 x 10^-2 N

The direction of the force is attractive, since the magnitude of the force due to Q₁ is greater than the magnitude of the force due to Q₃.

5.2.3 When Q₂ is replaced with a charge of -2 nC:

The magnitude of the net force on Q₂ will decrease, since the magnitude of the charge has decreased. However, the direction of the net force will remain attractive, since both Q₁ and Q₃ are negative charges.

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You have discovered a problem in your workplace and identified this research question: will practicing in non- business related activities enhance hiring and retention of a diverse workforce? The next step is to conduct a literature review to find out what others have said about this problem and help find an answer to the question. How would you begin to search for appropriate literature for this action research question?

Answers

1. Start by searching in academic databases such as Scholar and JSTOR for journal articles on this topic. Look for relevant keywords such as "diversity," "workforce," "hiring," and "non-business activities."

What is diversity?

Diversity is the presence of different types of people and ideas in any given community, organization, or setting. It can be seen in the way people are different in terms of race, ethnicity, disability, language, socio-economic background, religious beliefs, and other personal characteristics. Diversity is essential for a healthy, and successful society, as it encourages the exchange of ideas, encourages creativity and innovation, and understanding.

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Suppose people in your community are looking for a suitable source of renewable energy. Someone suggests wind power.

Prepare a 1-2 page report, which may include text, illustrations, graphs, or maps, to educate your community about wind power.

Answers

Answer:

Explanation:

Introduction:

As the world shifts towards renewable energy sources, the use of wind power is becoming increasingly popular. Wind power harnesses the power of the wind to generate electricity, and it has proven to be a reliable, cost-effective, and environmentally-friendly source of energy. In this report, we will discuss the benefits of wind power and how it can be used in our community.

Benefits of Wind Power:

Wind power has numerous benefits, including:

Clean and Environmentally-Friendly: Wind power generates electricity without producing harmful greenhouse gas emissions, air pollution, or toxic waste. It is a clean and sustainable energy source that helps reduce our carbon footprint and combat climate change.

Cost-Effective: Wind power is one of the most cost-effective renewable energy sources available today. Once a wind turbine is installed, the ongoing operating costs are relatively low, and the energy produced is essentially free.

Energy Security: Wind power provides energy security and independence. It reduces our dependence on foreign oil and other fossil fuels, which can be affected by geopolitical conflicts, price fluctuations, and supply disruptions.

Job Creation: Wind power creates jobs in the construction, maintenance, and manufacturing sectors. The growth of the wind power industry can provide economic benefits to our community and the country as a whole.

How Wind Power Works:

Wind power works by harnessing the kinetic energy of the wind and converting it into electrical energy. Wind turbines consist of a rotor, blades, a nacelle, and a tower. As the wind blows, it causes the rotor to spin, which in turn spins the blades. The blades are designed to capture the wind's energy and convert it into rotational energy. The rotational energy is then transferred to a generator in the nacelle, which converts it into electrical energy. The electricity is sent to a transformer, which increases the voltage and sends it to the power grid for distribution.

Using Wind Power in Our Community:

Our community has several options for using wind power, including:

Small-Scale Wind Turbines: Small-scale wind turbines can be installed on residential or commercial properties. They can generate enough electricity to power a home or business, and any excess energy can be sent back to the grid.

Large-Scale Wind Farms: Large-scale wind farms can be installed in rural areas, where there is ample wind and space. These wind farms can generate large amounts of electricity and provide energy to the grid, which can be distributed to our community.

Conclusion:

Wind power is a reliable, cost-effective, and environmentally-friendly source of energy that can provide numerous benefits to our community. By investing in wind power, we can reduce our carbon footprint, create jobs, and secure our energy future. It is a smart investment that will pay off for years to come.

How does the disturbance travel through the coil when you raise your arm up and down?

Answers

Answer:

A surface wave which is an example of mechanical wave that propagate along the interface of two different media in physics

Hi, can someone please help me with this report? It doesn't have to be like an essay, even two paragraphs is great. Please I really need help with this and really need it done. thank you so much!

The U.S. Army is planning to drop supplies from a plane at a refugee camp. The supplies are divided into 700-kilogram parcels, and the parachutes have an area of 100 square meters. The only problem is that the parcels cannot hit the ground at a velocity of more than 5 meters per second without damaging the contents. Are these parachutes suitable for this task?

For the purposes of this exercise, assume that the for the drag coefficient of the parachute is 1.5 and that the air density is 1.22 kilograms per cubic meter. Write a report detailing why these parachutes are or are not suitable and determining the minimum size parachute that can be used in this situation.

Answers

Answer:

Introduction:

In this report, we will examine whether the 100 square meter parachutes with a drag coefficient of 1.5 are suitable for dropping 700-kilogram parcels from a plane at a refugee camp. The main concern is that the parcels cannot hit the ground at a velocity of more than 5 meters per second without damaging the contents.

Calculation:

To determine whether the 100 square meter parachutes with a drag coefficient of 1.5 are suitable, we need to calculate the terminal velocity of the parcels. The terminal velocity is the maximum velocity that the parcels can reach when they are falling through the air. We can calculate the terminal velocity using the following equation:

Vt = sqrt((2mg)/(ρACd))

Where:

Vt is the terminal velocity

m is the mass of the parcel (700 kg)

g is the acceleration due to gravity (9.8 m/s^2)

ρ is the air density (1.22 kg/m^3)

A is the area of the parachute (100 m^2)

Cd is the drag coefficient (1.5)

Substituting these values into the equation, we get:

Vt = sqrt((2 x 700 x 9.8)/(1.22 x 100 x 1.5)) = 52.0 m/s

This means that without any parachute, the parcel would hit the ground with a velocity of 52 m/s. However, the parachutes are designed to provide air resistance, which will slow down the parcels.

To determine whether the parachutes are suitable, we need to calculate the velocity at which the parcels will hit the ground when they are attached to the parachutes. We can use the following equation to calculate the force of air resistance:

F = (1/2)ρAv^2Cd

Where:

F is the force of air resistance

ρ is the air density (1.22 kg/m^3)

A is the area of the parachute (100 m^2)

v is the velocity of the parcel

Cd is the drag coefficient (1.5)

When the force of air resistance is equal to the weight of the parcel, the parcel will stop accelerating and will reach its terminal velocity. Therefore, we can set the force of air resistance equal to the weight of the parcel:

F = mg

Substituting the values into the equation, we get:

(1/2)ρAv^2Cd = mg

Solving for v, we get:

v = sqrt((2mg)/(ρACd))

Substituting the values into the equation, we get:

v = sqrt((2 x 700 x 9.8)/(1.22 x 100 x 1.5)) = 25.9 m/s

This means that the velocity at which the parcels will hit the ground when they are attached to the parachutes is 25.9 m/s.

Conclusion:

Based on our calculation, the 100 square meter parachutes with a drag coefficient of 1.5 are suitable for dropping 700-kilogram parcels from a plane at a refugee camp. The velocity at which the parcels will hit the ground when they are attached to the parachutes is 25.9 m/s, which is below the maximum velocity of 5 m/s specified by the U.S. Army. However, if the mass of the parcels or the area of the parachutes changes, the velocity at which the parcels hit the ground will also change. Therefore, it is important to recalculate the velocity for different scenarios to ensure the safety of the parcels.

Explanation:

The U.S. Army is planning to drop 700-kilogram parcels of supplies to a refugee camp using parachutes with an area of 100 square meters. The objective is to prevent the parcels from hitting the ground at a velocity of more than 5 meters per second to avoid damage to the contents. To determine the suitability of these parachutes, we need to consider the drag coefficient and the air density.

Using the formula for air resistance, we can calculate the force acting on the parachute:

Force = 0.5 x Drag Coefficient x Air Density x Velocity^2 x Area

Assuming that the terminal velocity of the parcels is 5 meters per second, we can calculate the force acting on the parachute as follows:

Force = 0.5 x 1.5 x 1.22 x 5^2 x 100
= 1822.5 N

The weight of the parcels is 700 kg x 9.8 m/s^2 = 6860 N. Therefore, the force acting on the parachute is much less than the weight of the parcels, indicating that the parachutes are suitable for this task.

To determine the minimum size parachute that can be used in this situation, we need to calculate the maximum weight that can be supported by a parachute with an area of 100 square meters. This is known as the payload capacity of the parachute and can be calculated as follows:

Payload Capacity = Area x Drag Coefficient x Air Density x Velocity^2 / 2 x 9.8

Assuming that the maximum weight of the parcels that can be dropped is 700 kg, we can solve for the minimum size parachute as follows:

100 x 1.5 x 1.22 x 5^2 / (2 x 9.8) = 240.9 kg

Therefore, the minimum size parachute required for dropping 700-kilogram parcels at a velocity of less than 5 meters per second is approximately 241 square meters. In conclusion, the 100 square meter parachutes are suitable for this task, and a larger parachute would be required if the weight of the parcels increased.
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