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just olya [345]
3 years ago
12

When you throw a ball, the work you do to accelerate it equals the kinetic energy the ball gains. If you do twice as much work w

hen throwing the ball, does it go twice as fast? Explain. Yes. Twice as much work will give the ball twice as much kinetic energy. Since KE is proportional to the speed, the speed will double as well. Yes. Twice as much work will give the ball four times as much kinetic energy. Since KE is proportional to the speed squared, the speed will be the square root of 4, or twice as fast. No. Twice as much work will give the ball four times as much kinetic energy. Since KE is proportional to the speed, the speed will be four times larger. No. Twice as much work will give the ball twice as much kinetic energy. But since KE is proportional to the speed squared, the speed will be 2 times larger.
Physics
1 answer:
aniked [119]3 years ago
5 0

Answer:

No. Twice as much work will give the ball twice as much kinetic energy. But since KE is proportional to the speed squared, the speed will be sqrt{2} times larger.

Explanation:

The work done on the ball is equal to the kinetic energy gained by the ball:

W=K

So when the work done doubles, the kinetic energy doubles as well:

2W = 2 K

However, the kinetic energy is given by

K=\frac{1}{2}mv^2

where

m is the mass of the ball

v is its speed

We see that the kinetic energy is proportional to the square of the speed, v^2. We can rewrite the last equation as

v=\sqrt{\frac{2K}{m}}

which also means

v=\sqrt{\frac{2W}{m}}

If the work is doubled,

W'=2W

So the new speed is

v'=\sqrt{\frac{2(2W)}{m}}=\sqrt{2}\sqrt{\frac{2W}{m}}=\sqrt{2} v

So, the speed is \sqrt{2} times larger.

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A circular loop of wire is in the plane of the paper. The south pole of a bar magnet is being moved from a position in front of
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3 years ago
A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio
natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}

For the maximum velocity of the ball at the top of the vertical circular motion,

v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}

where g is the acceleration due to gravity,

Substituting the known values,

\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

8 0
1 year ago
How does this relate to Newton's second law?
taurus [48]

Answer:

In ideal case, when no resistive forces are present then both the balls will reach the ground simultaneously. This is because acceleration due to gravity is independent of mass of the falling object. i.e. g = GM/R² where G = 6.67×10²³ Nm²/kg², M = mass of earth and R is radius of earth.

Let us assume that both are metallic balls. In such case, we have to take into account the magnetic field of earth (which will give rise to eddy currents, and these eddy currents will be more, if surface area will be more) and viscous drag of air ( viscous drag is proportional to radius of falling ball), then bigger ball will take slightly more time than the smaller ball.

Explanation:

In ideal case, when no resistive forces are present then both the balls will reach the ground simultaneously. This is because acceleration due to gravity is independent of mass of the falling object. i.e. g = GM/R² where G = 6.67×10²³ Nm²/kg², M = mass of earth and R is radius of earth.

Let us assume that both are metallic balls. In such case, we have to take into account the magnetic field of earth (which will give rise to eddy currents, and these eddy currents will be more, if surface area will be more) and viscous drag of air ( viscous drag is proportional to radius of falling ball), then bigger ball will take slightly more time than the smaller ball.

4 0
3 years ago
In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1
lara31 [8.8K]

Answer:

Part a)

f = \frac{8}{9}

Part b)

f = \frac{120}{169}

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 500 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

2v_o = 2v_{1f} + v_o + v_{1f}

now we have

v_{1f} = \frac{v_o}{3}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{8}{9}

Part b)

Let say the collision between Camel and the car is elastic collision

So here we can use momentum conservation

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 300 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

10v_o = 10v_{1f} + 3(v_o + v_{1f})

now we have

v_{1f} = \frac{7v_o}{13}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{120}{169}

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

8 0
3 years ago
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