Answers
ANSWER:
The Moon's gravity is less than Earth's.
STEP-BY-STEP EXPLANATION:
When you jump, you fall back to the ground. Apples or leaves also fall: we are all attracted to the Earth. It is the terrestrial attraction due to the force of gravity.
The force of gravity also exists on the Moon. But since the Moon is smaller than the Earth, the attraction felt on the Moon is smaller than the Earth's attraction.
As the gravity is less, you can do things such as the one shown in the image.
As the force of attraction is less, the weight is less on the Moon, which can cause things that would be impossible on Earth.
Related Questions
Assume a water strider has a roughly circular foot of radius 0.0203 mm. The surface tension of water is 0.0700 N/m.A. What is the maximum possible upward force on the foot due to surface tension of the water? NB. What is the maximum mass of this water strider so that it can keep from breaking through the water surface? The strider has six legs. mg
Answers
Part (A)
The maximum possible upward force acting on the foot is,
[tex]F=2\pi r\sigma[/tex]Substitute the known values,
[tex]\begin{gathered} F=2(3.14)(0.0203\text{ mm)(}\frac{10^{-3}\text{ m}}{1\text{ mm}})(0.0700\text{ N/m)} \\ =8.9\times10^{-6}\text{ N} \end{gathered}[/tex]Thus, the maximum possible upward force on the foot is
[tex]8.9\times10^{-6}\text{ N}[/tex]Part (B)
The maximum force due to six legs can be expressed as,
[tex]6F=mg[/tex]Substitute the known values,
[tex]\begin{gathered} 6(8.9\times10^{-6}N)=m(9.8m/s^2) \\ m=\frac{6(8.9\times10^{-6}\text{ N)}}{9.8m/s^2}(\frac{1kgm/s^2}{1\text{ N}}) \\ =(5.45\times10^{-6}\text{ kg)(}\frac{1\text{ mg}}{10^{-6}\text{ kg}}) \\ =5.45\text{ mg} \end{gathered}[/tex]Thus, the maximum mass of water strider is 5.45 mg.
Which shape fits a position vs. time graph of an object that is slowing down? Which shape fits a position vs. time graph of an object that is speeding up?
Answers
1.
In a position x time graph, if the velocity is constant, so the position increases at the same rate over the time, so we have a linear relation between the position and the time.
Therefore the shape that represents this relation is C.
2.
In a velocity x time graph, if the velocity is constant, its value is always the same over the time, it doesn't change. That is represented graphically by a horizontal line, therefore the shape that represents this relation is B.
.
Point charges 88μC,-55μC and 70 μC are placed in a straight line. The central one is 0.75m from each of the others. Calculate the net force on each due to the other two.
Answers
The net force on the charges is 139.04 N.
What is the net force between the charges?
The net force between the charges is calculated by applying Coulomb's law as follows;
F = kq₁q₂/r²
where;
k is Coulomb's constantq₁ is the first chargeq₂ is the second charger is the distance between the chargesF(12) = (9 x 10⁹ x 88 x 10⁻⁶ x 55 x 10⁻⁶) / (0.75²)
F(12) = 77.44 N
F(23) = (9 x 10⁹ x 55 x 10⁻⁶ x 70 x 10⁻⁶) / (0.75²)
F(23) = 61.6 N
The net force on the charges is calculated as follows;
F(net) = 77.44 N + 61.6 N
F(net) = 139.04 N
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There are three point charges placed in a straight line, so the net force on each due to the other two is 139.04 N.
What is a charge?Charged matter experiences a force when it comes into contact with an electromagnetic field because electric charge is a property of matter. An electric field can have either a positive or negative charge. Charges that are similar to one another and dissimilar to one another are drawn to one another.
Given information in the question,
Charge, q₁ = 88 μC
Charge, q₂ = -55 μC
Charge, q₃ = 70 μC
Distance, r = 0.75 meters.
F = kq₁q₂/r²
Put the values in the above formula,
F₁₂ = (9 x 10⁹ x 88 x 10⁻⁶ x 55 x 10⁻⁶) / (0.75²)
F₁₂ = 77.44 N
F₂₃ = (9 x 10⁹ x 55 x 10⁻⁶ x 70 x 10⁻⁶) / (0.75²)
F₂₃ = 61.6 N
Now, calculate the net force :
F(net) = 77.44 N + 61.6 N
F(net) = 139.04 N
Hence, the net force due to the other two is 139.04 N.
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a projectile starting from ground hits a target on the ground located at the distance of 1000m after 40 sec
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A projectile starting from ground hits a target on the ground located at a distance of 1000 meters after 40 seconds.
a) The size of the angle θ is = 83°
b)The initial velocity was the projectile launched is (u)=205.13 m/s.
What is velocity?The velocity is a physical term that is refer to how much the object has covered the distance in a given time. It can be measured in m/s and cm/s.
How can we calculate the velocity?a) To calculate the angle we are using two formulas, they are
r= u cos(θ)T
T= 2u sin (θ)/g
Here we are given,
r= The distance covered in the motion = 1000m.
T = The time covered in the motion = 40 Seconds.
g= The acceleration due to gravity = 9.8 m/s²
We have calculate the values of angle = θ
Now we put the values in above equation we get,
1000= u cosθ*40...(1)
40*g= 2u sinθ.....(2)
Equation(2) divided by equation(1) we get,
2tanθ=40*40*g/1000
Or, tanθ=7.84
Or, θ= 82.7°≈83°
From the above calculation we can say that, The size of the angle θ is = 83°
b) The initial velocity was the projectile launched is = u
As we know, r= u cos(θ)T
Now we put the values in the equation we get,
1000= u cos(83)*40
Or, u=205.13 m/s
From the above calculation we can say that, The initial velocity was the projectile launched is (u)=205.13 m/s.
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A projectile starting from ground hits a target on the ground located at a distance of 1000 meters after 40 seconds.
a) What is the size of the angle θ?
b) At what initial velocity was the projectile launched?
A sailboat starts from rest and accelerates at a rate of 0.13 m/s? over a distance of 344 m.(a) Find the magnitude of the boat's final velocitym/s (b)find the time it takes the boat to travel this distance
Answers
We know that
• It starts from rest. (The initial velocity is zero).
,• The acceleration rate is 0.13 m/s^2.
,• The distance covered is 344 m.
To find the magnitude of the boat's final velocity, we have to use the following formula.
[tex]v^2_f=v^2_0+2ad[/tex]Let's use the given magnitudes.
[tex]\begin{gathered} v^2_f=0^2+2(0.13(\frac{m}{s^2}))(344m) \\ v^2_f=89.44(\frac{m^2}{s^2}) \\ v_f=\sqrt[]{89.44(\frac{m^2}{s^2})} \\ v_f\approx9.46(\frac{m}{s}) \end{gathered}[/tex](a) Therefore, the final velocity is 9.46 meters per second.Now, to find the time it takes the boat to travel this distance, we use the following formula.
[tex]d=v_0\cdot t+\frac{1}{2}at^2[/tex]Using the given magnitudes, we have the following.
[tex]344m=\frac{1}{2}(0.13(\frac{m}{s^2}))t^2[/tex]Let's solve for t.
[tex]\begin{gathered} t=\sqrt[]{\frac{2\cdot344m}{0.13(\frac{m}{s^2})}} \\ t=\sqrt[]{\frac{688m}{0.13}}\sec \\ t\approx72.75\sec \end{gathered}[/tex](b) Therefore, it takes 72.75 seconds.Apassenger in a drops a and it freely from rest to the groundWhat is the speed of the ball a distance of 20 massuming the acceleration due to gravity equal to 9.01 m/s ^ 2
Answers
Given data
*The given initial speed is u = 0 m/s
*The given distance is s = 20 m
*The value of the acceleration due to gravity is g = 9.01 m/s^2
The formula for the final speed of the ball at a distance of 20 m is given by the kinematic equation of motion as
[tex]v^2=u^2+2gs[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v^2=(0)^2+2(9.01)(20) \\ v^2=\text{36}0.04 \\ v=18.98\text{ m/s} \end{gathered}[/tex]Hence, the speed of the ball at a distance of 20 m is v = 18.98 m/s
Two children are riding on a merry-go-round. Child A is at a greater distance from the axis of rotation than child B. Which child hasthe larger linear displacement?Select one:a. There is not enough information given to answer the question.b. Child Ac. They have the same non-zero linear displacementd. Child Be. They have the same zero linear displacement.
Answers
The linear displacement in a rotation is given by:
[tex]s=r\theta[/tex]where r is the distance from the axis of rotation and theta is the angular displacement.
Since the linear displacement is proportional to the radius we conclude that child A has a larger linear displacement.
Find the force (N) on a 7.36cm radius piston on the other end of a hydraulic system that is driven by 23.37N of force from a 2.98cm radius piston.
Answers
The force (N) on a 7.36cm radius piston would be 142.55 Newtons, if on the other end of a hydraulic system that is driven by 23.37 N of force from a 2.98cm radius piston.
What is pressure?The total applied force per unit of area is known as the pressure.
The pressure depends both on externally applied force as well the area on which it is applied.
Pressure = Force / Area
As given in the problem we have to find out the force (N) on a 7.36cm radius piston on the other end of a hydraulic system that is driven by 23.37N of force from a 2.98cm radius piston,
By using the Pascal Law,
F₁ / A ₁ = F₂ / A₂
23.37 / 2.98² = F₂ / 7.36²
F₂ = 23.37 × 7.36² / 2.98 ²
= 142.55 Newtons
Thus, the force (N) on a 7.36cm radius piston would be 142.55 Newtons ,
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40) Climbing the Empire State Building A new record for running the stairs of the Empire State Building was set on February 4, 2003. The 86 flights, with a total of 1576 steps, was run in 9 minutes and 33 seconds. If the height gain of each step was 0.20 m, and the mass of the runner was 70.0 kg, what was his average power output during the climb? Give your answer in both watts and horsepower.
Answers
The power is given by:
[tex]P=\frac{W}{t}[/tex]where W is the work and t is the time.
We know that the work done is related to the gravitational potential energy by:
[tex]W=mg(y_f-y_0)[/tex]where yf is the final height of the object and y0 is the initial height.
Now, in this case we have a total of 1576 steps, each of them with a height of 0.2 meters that means that the total height gained is:
[tex](1576m)\cdot0.2=315.2\text{ m}[/tex]This in turns means that the work done is:
[tex]W=(70\operatorname{kg})(9.8\frac{m}{s^2})(315.2m)=216227.2\text{ J}[/tex]Now, the time it takes to achieve this is 9 minutes 33 seconds, this is the same as:
[tex]9(60)+33=573\text{ s}[/tex]Finally we use the power formula:
[tex]P=\frac{216227.2\text{ J}}{573\text{ s}}=377.36\text{ W}[/tex]Now we need to remember that 1 Hp is equal to 745.7 W, then we have that this is the same as:
[tex]377.36\text{ W}\cdot\frac{1\text{ Hp}}{745.7\text{ W}}=0.506[/tex]Therefore the power to make that climb is 377.36 W or 0.506 Hp
A block of mass 0.500 kg slides on a flat smooth surface with a speed of 2.80 m/s. It then slides over a rough surface with μk and slows to a halt. While the block is slowing, (a) What is the frictional force on the block? (b) What is the magnitude of the block’s acceleration? (c) How far does the block slide on the rough part before it comes to a halt?
Answers
a ) The frictional force on the block = 1.47 N
b ) The magnitude of the block’s acceleration = - 2.94 m / s²
c ) Distance travelled on rough part before it comes to a halt = 1.33 m
m = 0.5 kg
v = 2.8 m / s
Since there is no vertical motion,
∑ [tex]F_{y}[/tex] = 0
N - mg = 0
N = 0.5 * 9.8
N = 4.9 N
μ = 0.3
[tex]f_{k}[/tex] = μ N
[tex]f_{k}[/tex] = 0.3 * 4.9
[tex]f_{k}[/tex] = 1.47 N
The net force acting on the block is due to friction,
F = [tex]f_{k}[/tex] = 1.47 N
F = m a
1.47 = 0.5 * a
a = 2.94 m / s²
Since, acceleration is towards the opposite of motion,
a = - 2.94 m / s²
v² = u² + 2 a s
0 = 2.8² + ( 2 * - 2.94 * s )
s = 1.33 m
Therefore,
a ) The frictional force on the block = 1.47 N
b ) The magnitude of the block’s acceleration = - 2.94 m / s²
c ) Distance travelled on rough part before it comes to a halt = 1.33 m
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Block 1 has a mass of 12 kg is moving to the right on a levelsurface at a speed of 2 m/s. Block 2 has a mass of 2.5 kg andis at rest on the surface. Block 1 collides with block 2, causingblock 2 to move to the right with a speed of 4 m/s. How fast,and in what direction, is block 1 moving after the collision?
Answers
Given:
The mass of block 1, m₁=12 kg
The mass of block 2, m₂=2.5 kg
The velocity of block 1 before the collision, u=2 m/s
The velocity of block 2 after the collision, v₂=4 m/s
To find:
The velocity of block 1 after the collision.
Explanation:
From the law of conservation of momentum, the total momentum of the blocks before the collision will be equal to the total momentum of the blocks after the collision.
Thus,
[tex]m_1u=m_1v_1+m_2v_2[/tex]Where v₁ is the velocity of block 1 after the collision.
On rearranging the above equation,
[tex]v_1=\frac{m_1u-m_2v_2}{m_1}[/tex]On substituting the known values,
[tex]\begin{gathered} v_1=\frac{12\times2-2.5\times4}{12} \\ =1.17\text{ m/s} \end{gathered}[/tex]The positive sign of the velocity indicates that block 1 will continue to move to the right.
Final answer:
The velocity of block 1 after the collision will be 1.17 m/s and its direction is to the right.
2. A 14000 kg air jet accelerates from rest to 70 m/s before it takes off. What is the changein momentum of the jet?
Answers
Answer:
980,000 kg m/s
Explanation:
The change in momentum can be calculated as
[tex]\begin{gathered} \Delta p=m\Delta v \\ \Delta p=m(v_f-v_i) \end{gathered}[/tex]Where m is the mass, vf is the final velocity and vi is the initial velocity. Replacing m = 14000 kg, vf = 70 m/s and vi = 0 m/s, we get
[tex]\begin{gathered} \Delta p=14000\text{ kg \lparen}70\text{ m/s - 0 m/s\rparen} \\ \Delta p=14000\text{ kg \lparen70 m/s\rparen} \\ \Delta p=980000\text{ kg m/s} \end{gathered}[/tex]Therefore, the change in momentum of the jet is 980,000 kg m/s
If a car has a momentum of 2315Ns and its mass is 382kg, how fast is it moving?
Answers
ANSWER:
6.06 m/s
STEP-BY-STEP EXPLANATION:
Given:
Momentum = 2315 N*s
Mass = 382 kg
We can calculate the speed as follows:
[tex]\begin{gathered} p=m\cdot v \\ v=\frac{p}{m} \\ \text{ we replacing} \\ v=\frac{2315}{382} \\ v=6.06\text{ m/s} \end{gathered}[/tex]It moves at a speed of 6.06 m/s
A student fires a cannonball vertically upwards. The cannonball returns to the ground after a 3.80s flight. Determine all unknowns and answer the following questions. Neglect drag and the initial height and horizontal motion of the cannonball. Use regular metric units (ie. meters).How long did the cannonball rise? What was the cannonball's initial speed? What was the cannonball's maximum height?
Answers
The time taken by ball to rise is given as,
[tex]t=3.80\text{ s}[/tex]Therefore, the total time taken to move the ball in the upward direction is calculated as,
[tex]\begin{gathered} t_0=\frac{3.80\text{ s}}{2} \\ =1.90\text{ s} \end{gathered}[/tex]Therefore, the ball rise for 1.90 s.
The final speed of the ball can be given as,
[tex]v=u+gt[/tex]At the maximum height the final speed of ball is zero.
Substitute the known values,
[tex]\begin{gathered} 0m/s=u+(9.8m/s^2)(1.90\text{ s)} \\ u=-(9.8m/s^2)(1.90\text{ s)} \\ =-18.62\text{ m/s} \end{gathered}[/tex]Therefore, the initial speed of the ball is -18.62 m/s where negative sign indicates the direction of ball.
The final speed of the ball can be given as,
[tex]v^2=u^2-2gh_m[/tex]At the maximum height the final speed is zero. Substitute the known values,
[tex]\begin{gathered} (0m/s)^2=(-18.62m/s)^2-2(9.8m/s^2)h_m \\ h_m=\frac{(-18.62m/s)^2}{2(9.8m/s)^2} \\ =17.7\text{ m} \end{gathered}[/tex]Therefore, the maximum height of the canon is 17.7 m.
Two traveling pulses on a rope move toward each other at a speed of 1.0 m/s. The waves have the same amplitude. The drawing shows the position of the waves at time t = 0 s. Which one of the following drawings depicts the waves on the rope at t = 4.0 s?
Answers
ANSWER:
STEP-BY-STEP EXPLANATION:
We know the graph when t = 0. Therefore, at t = 4 pulse at "A" will reach to "B" and pulse at "B" will reach to "A".
This means that the graph is contrary, the only one that fulfills this is the graph (e).
After reading your textbook, you are able to maintain the bold, key words in coded representations in a network of neurons in your brain. In memory, this process is called
Answers
In memory, the process in which after reading your textbook, you are able to maintain the bold, key words in coded representations in a network of neurons in your brain is called storage.
In memory there are three phases. They are:
Encoding StorageRetrievalIn the given scenario, the process in which converting the bold, key words to coded representations is encoding process. The process which maintains the coded representations in a network of neurons in your brain is storage. The process of remembering the information stored when needed is retrieval.
Therefore, in memory, this process is called Storage
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What is the wavelength of the sound produced by a bat if the frequency of the sound is 90 kHz on a night when the air temperature is 22°C?
Answers
We know that the wavelength is related to the speed of the wave by:
[tex]v=\lambda f[/tex]where f is the frequency.
The speed of sound on air at a given temperature is given by:
[tex]v=331\sqrt[]{1+\frac{T}{273}}[/tex]so in this case the speed is:
[tex]v=331\sqrt[]{1+\frac{22}{273}}=344.08[/tex]Plugging this and the frequency in the first expression above we have:
[tex]\begin{gathered} 344.08=90\times10^3\lambda \\ \lambda=\frac{344.08}{90\times10^3} \\ \lambda=3.82\times10^{-3} \end{gathered}[/tex]Therefore the wavelength is:
[tex]3.82\times10^{-3}\text{ m}[/tex]For each letter, write a word to describe its role:A + B = C D x E = F
Answers
A is an addend
B is an addend
C is a sum
D is a factor
E is a factor
F is a product
Explanations:Note:Numbers(or characters) that are added together with the addition operator are called addends
The result of an addition operation is called sum
Numbers that are multiplied together are called factors
The result of a multiplication operator is called product
Considering the definitions above:In A + B = C
A is an addend
B is an addend
C is a sum
In D x E = F
D is a factor
E is a factor
F is a product
What are the units for buoyant force?
Answers
Answer Unit of Buoyant Force The unit of the buoyant force is the Newton (N). = Here, F= buoyant force
Explanation:
i hope this helpes
carts, bricks, and bands
8. What acceleration results when 2 rubber bands stretched to 20 cm are used to pull a cart with three bricks?
a. About 0.25 m/s2
b. About 0.33 m/s2
c. About 0.50 m/s2
d. About 1.00 m/s2
Answers
A. The acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with three bricks is 0.25 m/s².
What is the applied force on an object?The force applied on object is obtained by multiplying the mass and acceleration of the object.
According to Newton's second law of motion, the force exerted on an object is directly proportional to the product of mass and acceleration of the object.
Also, the applied force is directly proportional to the change in the momentum of the object.
Mathematically, the force acting on object is given as;
F = ma
a = F/m
where;
a is the acceleration of the objectm is the mass of the objectF is the applied forceFrom the trials, the acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with three bricks is 0.25 m/s².
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A 61 −kg ice skater coasts with no effort for 50 m until she stops. If the coefficient of kinetic friction between her skates and the ice is μk=0.10 , how fast was she moving at the start of her coast?
Answers
The 61 Kg ice skater was moving with a velocity of 9.9 m/s at the start of her coast.
How do I determine the initial velocity ?We'll begin by calculating the force. This can be obatined as follow:
Mass (m) = 61 KgAcceleration due to gravity (g) = 9.81 m/s²Normal reaction (N) = mg = 61 × 9.8 = 597.8 NCoefficient of kinetic friction (μK) = 0.10Force (F) = ?F = μKN
F = 0.1 × 597.8
F = 59.78 N
Next, we shall obtain the deceleration. This can be obtained as follow:
Force (F) = 59.78 NMass (m) = 61Deceleration (a) = ?a = -F / m (since the skater is coming to rest)
a = -59.78 / 61
a = -0.98 m/s²
Finally, we shall determine the initial velocity. This can be obtained as follow:
Distance (s) = 50 mFinal velocity (v) = 0 m/sDeceleration (a) = -0.98 m/s²Initial velocity (u) = ?v² = u² + 2as
0² = u² + (2 × -0.98 × 50)
0 = u² - 98
Collect like terms
u² = 0 + 98
u² = 98
Take the square root of both sides
u = √98
u = 9.9 m/s
Thus, the skater was moving with a velocity of 9.9 m/s
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Popeye the Sailor man, who has a mass of 85 kg, ran ( at a constant rate ) up a flight of stairs that are 3.55 m high in 6 seconds. How many watts of power did he generate during his run ?
Answers
Given data
*The given mass of the Sailorman is m = 85 kg
*The given height is h = 3.55 m
*The given time is t = 6 s
*The value of the acceleration due to gravity is
[tex]g=9.8m/s^2[/tex]The formula for the power generated by the Sailorman during his run is given as
[tex]\begin{gathered} P=\frac{W}{t} \\ =\frac{\text{mgh}}{t} \end{gathered}[/tex]*Here W is the work done
[tex]\begin{gathered} P=\frac{85\times9.8\times3.55}{6} \\ =492.85\text{ W} \end{gathered}[/tex]When x = –4, what is the value of g(x)? g(x)=-5x+3 help please
Answers
Replace the value x=-4 into the expression for g(x) and simplify, as follow:
[tex]\begin{gathered} g(-4)=-5(-4)+3 \\ g(-4)=20+3 \\ g(-4)=23 \end{gathered}[/tex]Hence, g(-4) = 23
A fuzzy die that has a weight of 1.70N hangs from the ceiling of a car by a massless string. The car travels on a horizontal road and has an acceleration of 2.70m/s^2 to the left. The string makes an angle theta with respect to the vertical, as shown in the figure below. 1) What is the angle theta?
Answers
First we calculate the mass of the fuzzy die
[tex]m=\frac{W}{g}[/tex]m is the mass, W is the weight, and g is the gravity
m=?
W=1.70N
g=9.8 m/s^2
we susbtitute
[tex]m=\frac{1.70}{9.80}=0.17\text{ kg}[/tex]Then we calculate the force of x
[tex]Fx=0.17(2.70)=0.468\text{ N}[/tex][tex]Fy=1.70N[/tex]Then we have the next diagram
Therefore for the angle
[tex]\theta=\arctan (\frac{F_x}{F_y})=arctan(\frac{0.468}{1.70})=15.4\text{ \degree}[/tex]ANSWER
The angle is 15.4°
Which optical instrument produces a magnified, virtual, and inverted image of small objects?1) a refracting telescope2) a single lens reflex camera3) a microscope4) a pair of binoculars
Answers
Microscope
Explanations:Microscopes are known for magnifying tiny objects
The images produced by a microscope are:
virtual (formed behind the screen)
Inverted
Magnified or enlarged
Therefore, the optical instrument which produces a magnified, virtual, and inverted image is the microscope
Jacob Grena raises a spoon 0.210 m above a table . If the spoon and its contents have a mass of 30.0 g, what is the gravitational potential energy associated with the spoon at that height relative to the table's surface?
Answers
Explanation:
The spoon is raised so it gains Gravitational potential energy. Formula to find Gravitational potential energy;
Gravitational potential energy = mass × Gravitational field strength × height of the body from the surface (table in this scenario)
In symbols; E = m×g×h
Substitute values:
m = 30g = 0.03kg (don't forget to convert grams to kg)
g = 10N/kg
h = 0.210 m
So it's;
0.03kg × 10N/kg × 0.210m = 0.063 Joules
SI Unit of energy is joules
I want to know if I’m doing this correctly, and if not, then what would be the best process to do solve it?
Answers
Given
[tex]3\times10^{-10}m^{}[/tex]
Factor Name Symbol
10-1 deci d
10-2 centi c
10-3 milli m
10-6 micro µ
10-9 nano n
10-12 pico p
10-15 femto f
10-18 atto a
10-21 zepto z
10-24 yocto y
a. nanometers
0.3 nm
[tex]3\times10^{-1}nm^{}[/tex]b. picometers
300 pm
[tex]3\times10^2pm[/tex]Megan kicks a soccer ball with a mass of 2 kg. The ball leaves the ground moving 50 meters per second. What is the kinetic energy of the ball?
Answers
ANSWER
[tex]2500J[/tex]EXPLANATION
The kinetic energy of a body is the energy it possesses due to its motion and it can be found by applying the formula:
[tex]E=\frac{1}{2}mv^2[/tex]where m = mass, v = velocity
From the question:
[tex]\begin{gathered} m=2\text{ kg} \\ v=50\text{ m/s} \end{gathered}[/tex]Therefore, the kinetic energy of the ball is:
[tex]\begin{gathered} E=\frac{1}{2}\cdot2\cdot50^2 \\ E=2500J \end{gathered}[/tex]A penguin runs 29,000 m/s how far will it travel in 10 seconds
Answers
Speed is measured as the ratio of distance to the time in which the distance was covered. Speed is a scalar quantity as it has only direction and no magnitude.
[tex]s=\frac{d}{t}[/tex]s = speed
d = distance
t = time
[tex]\begin{gathered} d=st \\ d=29000\text{ m/s}\cdot10s \\ d=290000\text{ m} \end{gathered}[/tex]The distance would be 290,000 m
name the quatity whose SI unit is j/kg/°c
Answers
Answer:
Explanation:
Specific heat:
[ c ] = 1 J / (kg·°C)