Answers
The force exerted to push the 25 Kg wooden box across the wooden floor at a constant speed of 1.0 m/s is 49 N (2nd option)
How do i determine the force exerted?Frictional force is given by the following formula
Frictional force = coefficient of friction (μ) × normal reaction (N)
Since the speed is constant,
Frictional force = Force exerted
Thus,
Force exerted = coefficient of friction (μ) × normal reaction (N)
F = μN
Now, we shall determine the force exerted to push the box. Details below:
Mass of box (m) = 25 KgCoefficient of friction (μ) = 0.2Acceleration due to gravity (g) = 9.8 m/s² Normal reaction (N) = mg = 25 × 9.8 = 245 NForce exerted (F) =?F = μN
F = 0.2× 245
F = 49 N
Thus, the force exerted to push the box is 49 N (2nd option)
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Related Questions
You have a concave mirror. A candle is located 22 cm from the mirror. The image of the candle appears at a distance 7.0 cm from the mirror. What is the focal length of the mirror?
Answers
I will give you an example to know how to do it, watch and learn.
Question:
The image of a candle flame placed at a distance of 30 cm from a mirror is formed on a screen placed in front of mirror at a distance of 60 cm from its pole. What is the nature of the mirror. Find its focal length. If the height of flame is 2.4 cm , find the height of image.
Given :-
u = - 30 cm
v = 60 cm
h = 2.4 cm
Solution :-
Using mirror formula,
1/f = 1/v - 1/u
⇒ 1/f = 1/60 - (1/- 30)
⇒ 1/f = 1/60 + 1/30
⇒ 1/f = 3/60
⇒ 1/f = 1/20
⇒ f = 20 cm
Magnification, M = v/u = Height of image/height of object
⇒ v/u = 60/-30
⇒ v/u = h'/2.4
⇒ h' = - 60/30 × 2.4
⇒ h' = - 4.8 cm
Hence, The required height of the image is - 4.8 cm
The image formed by the mirror is inverted.
Now thats how you do it. get the idea?
Hope it helps :)
If you have a concave mirror. A candle is located 22 cm from the mirror. The image of the candle appears at a distance of 7.0 cm from the mirror. Then the focal length of the mirror is 5.31 cm.
To find the focal length of the concave mirror, we can use the mirror formula:
1/f = 1/u + 1/v
where:
f = focal length of the mirror
u = object distance (distance of the candle from the mirror)
v = image distance (distance of the candle's image from the mirror)
We are given that:
u = -22 cm (negative sign indicates that the object is in front of the mirror)
v = -7.0 cm (negative sign indicates that the image is formed behind the mirror)
Plugging in the values, we get:
1/f = 1/-22 + 1/-7.0
Simplifying, we get:
1/f = (-7.0 + (-22)) / (-22 × -7.0)
1/f = 29 / 154
f = 5.31 cm (rounded to two decimal places)
Therefore, the focal length of the concave mirror is 5.31 cm.
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Rank the capacitors on the basis of the charge stored on the positive plate. Rank from largest to smallest: To rank items as equivalent; overlap them: A) A = 4 cm ; C=8nF B) A = 2 cm? =8 nFC) A = 2 cm? =4 nF D) A = 4 cm? C = 2 nF E) A =1 cm? = 1ln F
Answers
According to the charge that is stored on the positive plate, the order is B, A, C, D, and E, from biggest to smallest.
The charge stored on the positive plate of a capacitor is given by Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. Since the capacitors are connected in parallel, they all have the same voltage V. Therefore, the capacitor with the largest capacitance will have the largest charge stored on its positive plate.
Ranking the capacitors from largest to smallest based on their capacitances, we get:
B) A = 2 cm, C = 8 nF
A) A = 4 cm, C = 8 nF
C) A = 2 cm, C = 4 nF
D) A = 4 cm, C = 2 nF
E) A = 1 cm, C = 1 nF
Therefore, the ranking from largest to smallest based on the charge stored on the positive plate is: B, A, C, D, E.
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a water skier is pulled behied a boat by a rope the rope has a tension of 650 N and is at an angle of 27. what is the y compoent of the tension
Answers
Assuming it’s horizontal,
Answer is 650sin27
A battery of EMF 1. 5V has terminal pd of 1. 25V when a resistor of 25 ohm is joined to it. Calculate the current flowing, internal resistance and terminal pd when resistance of 25 ohm is replaced by 10 ohm
Answers
Using Ohm's law, we can find the current flowing in the circuit: I = V / R where I is the current, V is the voltage, and R is the resistance.
When a resistor of 25 ohms is connected to the battery, the terminal pd is 1.25 V. This means that the voltage drop across the internal resistance of the battery is: 1.5 V - 1.25 V = 0.25 V. Using Ohm's law, we can find the internal resistance of the battery: 0.25 V / I = R_internal. Substituting the given values, we get: R_internal = 0.25 V / (1.5 V / 25 ohm) = 4.17 ohm. The current flowing in the circuit is: I = 1.25 V / 25 ohm = 0.05 A. When the resistance is replaced by 10 ohms, the current flowing in the circuit is: I = 1.5 V / 35 ohm = 0.043 A. Using the same approach as before, the terminal pd can be calculated as: 1.5 V - 0.043 A * (R_internal + 10 ohm) = V_terminal. Substituting the value of R_internal, we get: V_terminal = 1.5 V - 0.043 A * (4.17 ohm + 10 ohm) = 1.34 V. Therefore, when the resistance is replaced by 10 ohms, the current flowing in the circuit is 0.043 A, the internal resistance of the battery is 4.17 ohm, and the terminal pd is 1.34 V.
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An object is moving to the right, and experiencing a net force that is directed to the right. The magnitude of the force is decreasing with time. The speed of the object is:A. increasingB. constant in timeC. decreasing
Answers
Answer:
The speed of the object is increasing.
F = M a as long as the magnitude of the acceleration is acting towards the right there will be an acceleration towards the right
1. Find FG between the earth and a football player 100 kg in mass. Fgrav =980 N
2. What is the radius of the moon when an astronaut of madd 70kg is having a gravitational force
117 N when standing in the surface of the moon
3. Determine the mass of Jupiter if a gravitational force on a scientist whose weight when in earth is 686 N, is Fgrav = 1823 N
Answers
Fgrav = G * (m1 * m2) / d^2
where G is the gravitational constant (6.67 x 10^-11 Nm^2/kg^2), m1 is the mass of the earth (5.97 x 10^24 kg), m2 is the mass of the football player (100 kg), and d is the distance between their centers of mass.
Since the football player is standing on the surface of the earth, we can assume that the distance between their centers of mass is equal to the radius of the earth (6.37 x 10^6 m).
Plugging in the values, we get:
Fgrav = (6.67 x 10^-11 Nm^2/kg^2) * (5.97 x 10^24 kg) * (100 kg) / (6.37 x 10^6 m)^2
Fgrav = 981 N
Therefore, the force of gravity between the earth and the 100 kg football player is approximately 980 N (rounded to the nearest whole number).
2. To find the radius of the moon when an astronaut with a mass of 70 kg is experiencing a gravitational force of 117 N while standing on the surface of the moon, we can use the same formula as above:
Fgrav = G * (m1 * m2) / d^2
where G is the gravitational constant (6.67 x 10^-11 Nm^2/kg^2), m1 is the mass of the moon (7.35 x 10^22 kg), m2 is the mass of the astronaut (70 kg), and d is the distance between their centers of mass, which is equal to the radius of the moon (r).
Plugging in the values, we get:
117 N = (6.67 x 10^-11 Nm^2/kg^2) * (7.35 x 10^22 kg) * (70 kg) / r^2
Solving for r, we get:
r = 1.74 x 10^6 m
Therefore, the radius of the moon when an astronaut with a mass of 70 kg is experiencing a gravitational force of 117 N while standing on the surface of the moon is approximately 1.74 x 10^6 m.
3. To determine the mass of Jupiter if a gravitational force on a scientist whose weight is 686 N on earth is Fgrav = 1823 N, we can use the same formula as above:
Fgrav = G * (m1 * m2) / d^2
where G is the gravitational constant (6.67 x 10^-11 Nm^2/kg^2), m1 is the mass of Jupiter, m2 is the mass of the scientist (weight on earth divided by gravitational acceleration on earth, which is 9.81 m/s^2), and d is the distance between their centers of mass, which is the radius of Jupiter (r).
Plugging in the values, we get:
1823 N = (6.67 x 10^-11 Nm^2/kg^2) * (m1) * (686 N / 9.81 m/s^2) / (r^2)
Solving for m1, we get:
m1 = 1.90 x 10^27 kg
Therefore, the mass of Jupiter is approximately 1.90 x 10^27 kg.
Princess Leia once fell out of the spaceship. She came back with the Force. From the farthest spot (zero speed) back to the spaceship was 10 meters, and it took 5 second. She weighed 50 kg. Assume constant force during this trip. Find the force needed. (It is not much, but to get a force while floating in space is magic.)
Answers
The force needed to propel Princess Leia from a stationary position 10 meters back to the spaceship in 5 seconds would be 100 Newtons. This force can be determined by using the formula F = m x a, where m is the mass of the object and a is the acceleration.
The force required to propel Princess Leia from a stationary position 10 meters back to the spaceship in 5 seconds would be equal to the work done divided by the time taken, or 50 kg x 10 m/5 s = 100 Newtons. This is the force that is needed to move her 10 meters in 5 seconds, taking into account her 50 kg mass.
To explain this in a little more detail, the force required is determined by the formula F = ma, where F is force, m is mass and a is acceleration. Since the acceleration is constant in this scenario, the force applied is equal to the mass multiplied by the acceleration, or F = m x a. This means that to move the 50 kg mass of Princess Leia from zero speed to 10 meters in 5 seconds, a force of 100 Newtons is required.
To illustrate this with an example, if you were to pick up a 50 kg weight and move it 10 meters in 5 seconds, you would need to exert 100 Newtons of force. This is the same force required to move Princess Leia in the same way in space, taking into account her 50 kg mass.
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Work is done only when..... _ causes an object to _
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Work is characterized as the resultant action when a force moves an object in the force's direction.
What does force mean in science?It is clear what the word "force" means. The terms "push" and "pull" are perfectly acceptable at this level to describe forces. An object does not have a force inside of it or within it. Another object applies a force to the first.
In what ways do forces differ?When two items come into contact with one another physically and interact, these kinds of forces are involved. Frictional, tensional, normal, air resistance, applied, and spring forces are examples of the several types of contact forces.
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A car traveling at 28. 5 m/s E skids to a stop in 2. 58 s. Determine the skidding displacement of the car (assume uniform acceleration). 27
Answers
The car's displacement when sliding is 73.66 metres.
We can use the kinematic equation for displacement with constant acceleration to solve for the skidding displacement of the car:
d = vi * t + 0.5 * a * t^2
where d is the displacement, vi is the initial velocity, t is the time, and a is the acceleration.
First, we need to find the acceleration of the car. We can use the equation for average acceleration:
a = (vf - vi) / t
where vf is the final velocity (which is 0 since the car comes to a stop), vi is the initial velocity (28.5 m/s to the east), and t is the time (2.58 s).
a = (0 - 28.5 m/s) / 2.58 s
a = -11.05 m/s^2
The negative sign indicates that the acceleration is in the opposite direction to the initial velocity.
Now we can substitute the values into the first equation:
d = vi * t + 0.5 * a * t^2
d = (28.5 m/s) * (2.58 s) + 0.5 * (-11.05 m/s^2) * (2.58 s)^2
d = 73.66 m
Therefore, the skidding displacement of the car is 73.66 meters.
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A cup of hot chocolate with a temperature of 40°c is placed inside a
refrigater at 5°c .An identical cup of hot chocolate at 90°c is placed on a table in a
room at 25°c. A third identical cup of hot chocolate at 80°c is placed on an
outdoor table, where the surrounding air has a temperature at 0°c, for which
of the three cups has the most energy been transferred as heat when
equilibrium has been reached?
Al The first cup has the largest energy transfer
B/ The second cup has the largest energy transfer
c/ The third cup has the largest energy transfer
O The same amount of energy is transferred as heat for all three cups
Answers
Answer:
The answer is c
Explanation:
The third cup has the largest energy transfer
if a lot of material (from the accretion disk) makes it onto the white dwarf (before it can all fuse away), what will happen? a. the winds from the white dwarf will blow all the material off. b. the white dwarf will have enough mass to collapse. c. the white dwarf will turn into a red giant. d. the white dwarf will transfer the mass back to the companion star e. the material will sink to the core of the white dwarf and the star will return to the main sequence.
Answers
Option b is correct. If enough material from the accretion disk makes it onto the white dwarf then it will cause the white dwarf to collapse and produce a Type Ia supernova.
A white dwarf is a remnant of low or medium-mass stars that have stopped producing energy through nuclear fusion. White dwarfs are not planets; instead, they are the small cores of stars that have collapsed under the weight of their own gravity. They are the densest form of matter apart from black holes and neutron stars.
If a lot of material from the accretion disk makes it onto the white dwarf (before it can all fuse away), the material will accumulate on the surface of the white dwarf, increasing its mass.
The white dwarf will have a mass greater than the Chandrasekhar limit (1.4 solar masses). This can lead to a catastrophic event known as a Type Ia supernova, in which the white dwarf collapses and explodes. hence correct option is b.
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Which of the following descriptions best describes all of the factors that need to be considered when determining an object’s terminal velocity?(1 point) Responses the object’s weight, length, and width the object’s weight, length, and width air density and the object’s drag coefficient air density and the object’s drag coefficient the object’s weight and area it presents, as well as air density the object’s weight and area it presents, as well as air density the object’s weight and area it presents
Answers
The object’s weight and area it presents, as well as air density.
The correct option is C.
What is velocity and example?Velocity may be defined as the rate that a thing travels in a certain direction. as the speed of a car driving north on a highway or the pace at which a rocket takes off. the proportion between the distance travelled in a simple machine by the point where the effort being applied to the distance moved in the same time by the place where the load is applied.
Who first defined velocity?Galileo Galilei, an Italian physicist, is recognized with being the initial person to calculate speed by dividing it by the distance travelled and the time required. Galileo defined speed as the amount of distance travelled in a given amount of time.
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The complete question is -
Which of the following descriptions best describes all of the factors that need to be considered when determining an object’s terminal velocity?
A) The object’s weight, length, and width
B) Air density and the object’s drag coefficient
C) The object’s weight and area it presents, as well as air density
D) The object’s weight, length, width, and area it presents, as well as air density and the object’s drag coefficient
Suppose you are driving through a valley whose bottom has a circular shape. If your mass is M, what is the magnitude of the normal force Fn exerted on you by the car seat as you drive past the bottom of the hill A) Fn < Mg B) Fn = Mg C) Fn > Mg
Answers
The right response is (c) Fn > Mg
When you are driving through a circular valley, the direction of the net force acting on you is constantly changing. At the bottom of the valley, your velocity vector is horizontal and your acceleration vector is pointing upwards.
According to Newton's second law, F = ma, where F is the net force, m is your mass, and a is your acceleration. Therefore, the net force acting on you is greater than your weight (Mg) and is equal to the sum of the normal force and the force of gravity.
Thus, the correct answer is (c) Fn > Mg. The normal force is greater than your weight to provide the additional force required to maintain the circular motion.
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how much work dose billy perform if he pushes the 800N stalled car a distence of 25m?
Answers
Answer:
20,000 J
Explanation:
Work = Force multiplied by Displacement
The work done on a object can only be calculated when the applied force is causing an object to move in the direction the force is being applied towards.
The given information:
Billy applies a form of 800N on a staller car and the total distance it was pushed was 25 meters.
To calculate the work done, substitute the given information into the formula for work.
Work = (800 N) × (25 m)
Work = 20,000 J
Note: Remember the SI unit for Work is Joules (J).
balance the chemical equation
N₂ + 02 → - N₂0
Answers
The balanced form of the given chemical equation is as follows: 2N₂ + 0₂ → 2N₂0.
What is a balanced chemical equation?A chemical equation in chemistry is a symbolic representation of a chemical reaction where reactants are represented on the left, and products on the right.
A chemical equation results in the formation of new substances called products from substances called reactants.
However, to obey the law of conservation of mass, the atoms of each element on both sides of the equation must be the same. This is referred to as a balanced chemical equation.
The balanced form of the reaction between nitrogen and oxygen gas is given above.
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An organism that breaks down dead plants and animals, returning matter back into the soil, is called a
A.
herbivore.
B.
producer.
C.
competitor.
D.
decomposer.
Answers
Answer:
d. decomposer
Explanation:
they play a critical role in the flow of energythrough an ecosystem. They break apart dead organisms into simpler inorganic materials, making nutrients available to primary producers
you are driving your car around a roundabout when you get a flat tire and you decelerate at a constant rate to a stop. the diameter of the roundabout is 70 m m . it takes you 32 s s to come to a complete stop. while slowing down, you continue to drive in a circle and you stop at the same spot that you popped your tire. what must have been your speed before the pop?
Answers
Speed before the tire popped must have been 35.8 m/s. To calculate your speed before the tire popped, we will use the equation vf = vi + at. In this equation, vf is the final velocity (0 m/s), vi is the initial velocity, a is the acceleration (deceleration in this case), and t is the time.
Since we know the diameter of the roundabout, the time it took to stop, and the acceleration is constant, we can calculate the initial velocity as:
vi = vf + at
vi = 0 m/s + (-1.125 m/s2) * 32 s
vi = -35.8 m/s
So, your speed before the tire popped must have been 35.8 m/s.
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a 47.0 kg stunt pilot who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane.if the plane's speed at the lowest point of the circle is 96.0 m/s , what is the minimum radius of the circle so that the acceleration at this point will not exceed 4.00 g ?what is the apparent weight of the pilot at the lowest point of the pullout?
Answers
The minimum radius of the circle so that the acceleration at this point will not exceed 4.00 is 2304 m, and the apparent weight of the pilot at the lowest point of the pullout is 272.6 N.
The minimum radius of the circle.
The equation for centripetal acceleration is:
a = (v²)/r, where a is the centripetal acceleration, v is the velocity, and r is the radius of the circle.
The equation can be rearranged to solve for the radius of the circle,
r = (v²)/a.
Plugging in the given values, we get:
r = (96)² / 4.00g
r = 2304 m.
The apparent weight of the pilot.
The equation for the apparent weight of the pilot is
[tex]w_{app}[/tex] = mg - m* [tex]a_{centripetal}[/tex], where [tex]w_{app}[/tex] is the apparent weight, m is the mass of the pilot, g is the acceleration due to gravity, and [tex]a_{centripetal}[/tex] is the centripetal acceleration.
Plugging in the given values, we get [tex]w_{app}[/tex] = (47.0)(9.8) - (47.0)(4) = 272.6 N.
Therefore, the minimum radius of the circle is 2304 m, and the apparent weight of the pilot is 272.6 N.
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A car needs to generate 75. 0 hp in order to maintain a constant velocity of 18. 2 m/s on a flat road. What is the magnitude of the total resistive force acting on the car (due to friction, air resistance, etc. )
Answers
Answer:
the magnitude of the car is
F=307.9N
Explanation:
The expression for the power in terms of force and velocity is as follows;
P=Fv
Here, P is the power, F is the force and v is the velocity.
Rearrange the above expression to get the value of the force.
F=\frac{P}{v}
Convert power from horsepower to watt.
P=75.0 hp
P= (75)(745.7) W
P= 55,927.5 W
Put P= 55,927.5 W and v= 18.2 meter per second in the above expression.
F=\frac{55,927.5}{18.2}
F= 3072.9 N
Therefore, the magnitude of the force is F= 3072.9 N.
Answer it quickly please
Answers
The coefficient of the kinetic friction of the sand paper would be = 0.5. That is option C.
How to calculate the coefficient of the kinetic friction of sand paper?To calculate the coefficient of the kinetic friction of the sand paper the slope of the sandpaper line is obtained. This is because slope of kinetic friction and normal force = coefficient of kinetic friction.
To calculate the slope = Y2 - Y1/ X2 - X1
Where Y2 = 3.3
Y 1 = 1.0
X2 = 6.9
X1 = 2
Slope = 3.3-1.0/6.9-2
= 2.3/4.9
= 0.5
Therefore the coefficient of the kinetic friction of the sand paper which is the slope of kinetic friction and normal force for the sandpaper line would be = 0.5.
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1. The speed of a bobsled is increasing because it has an acceleration of 2. 4 m/3. At a given instant in time, the forces resisting the motion, including kinetic friction and air resistance, total 450 N. The combined mass of the bobsled and its riders is 270 kg. (a) What is the magnitude of the force propelling the bobsled forward? (b) What is the magnitude of the net force that acts on the bobsled?
Answers
(a) The magnitude of the force propelling the bobsled forward is 648 N. (b) The magnitude of the net force acting on the bobsled is 198 N.
Use Newton's second law of motion, which states that force equals mass times acceleration (F = ma). Rearranging the formula, we get F = ma, where F is the force, m is the mass, and a is the acceleration.
So, the force propelling the bobsled forward is,
F = ma = (270 kg)(2.4 m/s^2) = 648 N
The net force acting on the bobsled can be calculated by subtracting the forces resisting the motion from the force propelling the bobsled forward. So, the net force acting is,
Net force = Force propelling forward - Forces resisting motion
Net force = 648 N - 450 N
Net force = 198 N
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A 46.4 kg wolf is running at 12.2 m/s what is the wolfs kinetic energy
Answers
The 46.4 kg wolf moving at a speed of 12.2 m/s has a kinetic energy of 3,456.98 J.
What establishes an object's entire amount of kinetic energy?The amount of work done on an item and its acceleration following a moment of inertia caused by external forces are two aspects that affect the overall kinetic energy of an object.
The formula: gives the kinetic energy (KE) of an item.
KE = (1/2) × mass × velocity²
where velocity is measured in meters per second (m/s) and mass is measured in kilograms (kg).
Plugging in the following numbers will allow us to determine the wolf's kinetic energy:
KE = (1/2) × 46.4 kg × (12.2 m/s)²
= (1/2) × 46.4 kg × 148.84 m²/s²
= 3,456.98 joules (J)
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During a tug of war game, both teams apply a force of 75 N. What is the net force?
Answers
The net force is 0 N. This is because the forces from each team are equal but opposite in direction, so the net force is the difference between the two forces, which is 0 N.
What is force ?Force is an interaction between two or more objects that results in a change in the motion of one or more of the objects. It is a vector quantity, meaning it has both magnitude and direction. Force can be divided into two categories: contact forces, which occur when two objects physically touch one another, and non-contact forces, which occur when objects interact without direct physical contact. Examples of contact forces include friction, tension, normal force, and air resistance. Examples of non-contact forces include gravitational force, electric force, and magnetic force.
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A 15-cm-long aluminum tank is filled with ethyl alcohol. A high-frequency ultrasound wave travels horizontally through one wall of the tank and then through the alcohol. There are 275 times more cycles of the wave in the alcohol than in the aluminum wall. How thick is the wall of the tank?
Answers
The aluminium wall measures about 0.273 centimetres in thickness.
Let's assume that the frequency of the ultrasound wave is denoted by f and the thickness of the aluminum wall is denoted by x. We know that the speed of the ultrasound wave in aluminum is greater than that in ethyl alcohol. Since the wavelength of the wave is constant, we can use the formula:
f1 x1 = f2 x2,
where f1 and f2 are the frequencies of the ultrasound wave in aluminum and ethyl alcohol, respectively, and x2 is the thickness of the alcohol.
We are given that the ratio of the number of cycles in the alcohol to that in the aluminum wall is 275. Therefore:
f2 = 275 f1.
We also know that the speed of the ultrasound wave in aluminum is approximately 5 times that in ethyl alcohol. Therefore:
x1 = 5 x2.
Substituting these values into the equation gives:
275 f1 x2 = f1 (5 x2),
which simplifies to:
x2 = (1/55) x1.
Finally, we substitute the length of the aluminum tank (15 cm) for x1 to get:
x2 = (15 cm) / 55 = 0.273 cm.
Therefore, the thickness of the aluminum wall is approximately 0.273 cm.
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Object X of mass m0 travels to the right with a velocity v0. Object Y of mass 2m0 moves to the left at 2v0, as shown in the figure. The objects collide and then stick together. What is the change in kinetic energy of the two-object system from immediately before the collision to immediately after the collision?
ATTACHMENT OF DIAGRAM BELOW. Ignore my calculations
Answers
The kinetic energy decreases by 3movo^2
Does the kinetic energy increase for an inelastic collision?In an inelastic collision, kinetic energy is not conserved. Some of the initial kinetic energy of the system is converted into other forms of energy, such as heat, sound, or deformation energy, and is not available to do work on the system.
In fact, the total kinetic energy of the system usually decreases in an inelastic collision. This is because the objects involved in the collision stick together or deform upon impact, resulting in a loss of kinetic energy. The amount of energy lost depends on the specific properties of the objects and the nature of the collision.
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Forces of 7.6N at 38 degrees and 11.8N at 143 degrees act at a point.Calculate the magnitude and direction of their resultant.
Answers
Answer:
Approximately [tex]12.3\; {\rm N}[/tex] at approximately [tex]106^{\circ}[/tex].
Explanation:
Assume the two given directions are measured with respect to the positive [tex]x[/tex] axis.
If a vector of magnitude [tex]\| a\|[/tex] is at an angle of [tex]\theta[/tex] from the positive [tex]x[/tex] axis, this vector can be written in the component form as:
[tex]\begin{aligned}\| a\|\begin{bmatrix}\cos(\theta) \\ \sin(\theta)\end{bmatrix}\end{aligned}[/tex];
Or equivalently:
[tex]\begin{aligned}\begin{bmatrix}\|a\|\, \cos(\theta) \\ \|a\|\, \sin(\theta)\end{bmatrix}\end{aligned}[/tex].
For example, the [tex]7.6\; {\rm N}[/tex] force is a vector with magnitude [tex]7.6\; {\rm N}[/tex] at a direction of [tex]38^{\circ}[/tex] from the positive [tex]x[/tex] axis. This vector can be represented as:
[tex]\begin{aligned} 7.6\, \begin{bmatrix}\cos(38^{\circ}) \\ \sin(38^{\circ})\end{bmatrix} &= \begin{bmatrix}7.6\, \cos(38^{\circ}) \\ 7.6\, \sin(38^{\circ})\end{bmatrix} \approx \begin{bmatrix}5.9889 \\ 4.6790 \end{bmatrix}\end{aligned}[/tex].
Similarly, the [tex]11.8\; {\rm N}[/tex] vector can be represented as:
[tex]\begin{aligned}11.8\, \begin{bmatrix}\cos(143^{\circ}) \\ \sin(143^{\circ})\end{bmatrix} &= \begin{bmatrix}11.8\, \cos(143^{\circ}) \\ 11.8\, \sin(143^{\circ})\end{bmatrix} \approx \begin{bmatrix}-9.4239 \\ 7.1014 \end{bmatrix}\end{aligned}[/tex].
To find the sum of the two vectors, take the sum of each component separately:
[tex]\begin{aligned}& \begin{bmatrix}5.9889 \\ 4.6790 \end{bmatrix} + \begin{bmatrix}-9.4239 \\ 7.1014\end{bmatrix} \\ =\; & \begin{bmatrix}5.9889 + (-9.4239)\\ 4.6790 + 7.1014\end{bmatrix} \\ \approx\; & \begin{bmatrix}-3.4350 \\ 11.780\end{bmatrix} \end{aligned}[/tex].
Apply the Pythagorean Theorem to find the magnitude of this vector sum:
[tex]\displaystyle \sqrt{(-3.4350)^{2} + (11.780)^{2}} \approx 12.271[/tex].
Note that the first component ([tex]x[/tex]-component) of this vector is negative, such that this vector would point to the left of the vertical axis. Since the second component ([tex]y[/tex]-component) of this vector is positive, this vector would point above the horizontal axis. Hence, the direction of this vector (relative to the positive [tex]x\![/tex]-axis) would be an angle between [tex]90^{\circ}[/tex] and [tex]180^{\circ}[/tex].
Divide the [tex]x[/tex]-component of this vector by its magnitude to find the cosine of the angle between this vector and the positive [tex]x\![/tex]-axis. Apply the inverse cosine function to find this angle:
[tex]\begin{aligned}\arccos \left(\frac{-3.4350}{12.271}\right) \approx 106^{\circ}\end{aligned}[/tex].
A spacecraft is orbiting a planet at a distance of 10,000 km from the planet's center. The spacecraft's speed is constant at 20,000 km/hr. Suddenly, the spacecraft's engines fail and it begins to free fall towards the planet's surface. Assuming that the planet has a uniform density and a radius of 5,000 km:
1. What is the gravitational acceleration experienced by the spacecraft as it falls towards the planet's surface?
2. What is the maximum velocity that the spacecraft can reach just before impacting the planet's surface?
3. What is the kinetic energy of the spacecraft just before impacting the planet's surface?
Answers
The gravitational acceleration experienced by the spacecraft would be 13.33 m/s^2.
The maximum velocity the spacecraft can reach would be 1633.5 m/s.
The kinetic energy of the spacecraft would be KE = 1/2mv^2
Gravitational acceleration problemThe gravitational acceleration experienced by the spacecraft as it falls toward the planet's surface can be calculated using the formula
g = G*M/r^2Thus:
g = (6.67 x 10^-11 N m^2/kg^2) * (4/3 * pi * (5000 km)^3 * 5500 kg/m^3) / (10000 km)^2 = 13.33 m/s^2.
The maximum velocity that the spacecraft can reach just before impacting the planet's surface can be calculated using the formula
v = sqrt(2gh)The height from which the spacecraft falls is equal to the distance from the planet's surface to the spacecraft's initial position, which is 10,000 km - 5000 km = 5000 km.
Thus, v = sqrt(2 * 13.33 m/s^2 * 5,000,000 m) = 1633.5 m/s.
The kinetic energy of the spacecraft just before impacting the planet's surface can be calculated using the formula
KE = 1/2mv^2More on gravitational acceleration can be found here: https://brainly.com/question/30429868
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can one entangled particle be measured for momentum while the other entangled particle is measured for position?
Answers
According to the principles of quantum mechanics, the answer to the question, "Can one entangled particle be measured for momentum while the other entangled particle is measured for the position?" is no.
If you measure the position of one particle in an entangled pair, you are creating a definite outcome for that particle. By measuring the other particle's momentum, you are causing it to have a definite outcome. That, however, is incompatible with the fact that the two particles are entangled. Their correlation will be broken if this is done.
The position and momentum measurements are incompatible in quantum mechanics. If you want to measure the position of a particle, the uncertainty in the momentum of the particle must be large. Conversely, if you want to measure the momentum of a particle, the uncertainty in the position of the particle must be large.
Furthermore, measuring the position of a particle has a detrimental effect on the uncertainty in its momentum. Similarly, measuring the momentum of a particle has a detrimental effect on the uncertainty in its position. The measurement outcomes of two entangled particles are determined by correlations between them.
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a rock is thrown upward with a velocity of 11 meters per second from the top of a 44 meter high cliff, and it misses the cliff on the way back down. when will the rock be 10 meters from ground level? round your answer to two decimal places.
Answers
The rock will be 10 meters from the ground level after 2.63 seconds when the rock is thrown upward with velocity of 11m/s .
When the rock is thrown upward with a velocity of 11 m/s from the top of the 44 m high cliff, it will take approximately 4.00 seconds for the rock to reach the top of its trajectory and start heading back down.
The velocity of the rock on its way back down will be 11 m/s in the opposite direction. After 2.63 seconds, the rock will have travelled a distance of 10 m from the ground level.
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2. A disc speeds up from rest at a constant rate of 2. 5 rad/s^2.
a. What is the final angular speed of the disc after 18 secs. ? (45 rad/s)
b. How many revolutions will the disc go through during this time period?
(64. 5 revs)
C. WHat is the linear speed of a point at the end of the disc after 18 secs. (The radius
of the disc is 50 cm. (22. 5 m/s)
I
Answers
a. The final angular speed of the disc after 18 seconds is 45 rad/s.
b. The disc goes through approximately 64.5 revolutions during this time period.
c. The linear speed of a point at the end of the disc after 18 seconds is 22.5 m/s.
a. The angular acceleration of the disc is constant at 2.5 [tex]rad/s^2[/tex].
Using the formula for angular velocity with constant angular acceleration, we have:
ωf = ωi + αt
Where:
ωi = initial angular velocity (0, since the disc starts from rest)
α = angular acceleration (2.5 [tex]rad/s^2[/tex])
t = time (18 s)
ωf = 0 + 2.5 × 18
ωf = 45 rad/s
Therefore, the final angular speed of the disc after 18 seconds is 45 rad/s.
b. To find the number of revolutions the disc goes through during this time period, we need to first find the total angle rotated by the disc. Using the formula for angular displacement with constant angular acceleration, we have:
θ = ωit + 1/2 α[tex]t^2[/tex]
Where:
ωi = initial angular velocity (0, since the disc starts from rest)
α = angular acceleration (2.5 [tex]rad/s^2[/tex])
t = time (18 s)
θ = 0 + 1/2 × 2.5 × [tex]18^2[/tex]
θ = 405 rad
One revolution is equal to 2π radians,
so the number of revolutions is given by:
N = θ / 2π
N = 405 / 2π
N ≈ 64.5 revolutions
Therefore, the disc goes through approximately 64.5 revolutions during this time period.
c. The linear speed of a point at the end of the disc is given by:
v = rω
Where:
r = radius of the disc (50 cm = 0.5 m)
ω = angular velocity (45 rad/s)
v = 0.5 × 45
v = 22.5 m/s
Therefore, the linear speed of a point at the end of the disc after 18 seconds is 22.5 m/s.
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