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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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The cylinder with piston locked in place is immersed in a mixture of ice and water and allowed to come to thermal equilibrium wi
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Answer:

a. volume of gas:  (decreases)

b. temperature of gas:  (same)

c. internal energy of gas: (same)

d. pressure of gas: (increases)

Explanation:

We have a gas (let's suppose that is ideal) in a piston with a fixed volume V.

Then we put in a reservoir at 0°C (the mixture of water and ice)

remember that the state equation for an ideal gas is:

P*V = n*R*T

and:

U = c*n*R*T

where:

P = pressure

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n = number of mols

R = constant

c = constant

T = temperature.

Now, we have equilibrium at T = 0°C, then we can assume that T is also a constant.

Then in the equation:

P*V = n*R*T

all the terms in the left side are constants.

P*V = constant

And knowing that:

U = c*n*R*T

then:

n*R*T = U/c

We can replace it in the other equation to get:

P*V = U/c = constant.

Now, the piston is (slowly) moving inwards, then:

a) Volume of the gas: as the piston moves inwards, the volume where the gas can be is smaller, then the volume of the gas decreases.

b) temperature of the gas: we know that the gas is a thermal equilibrium with the mixture (this happens because we are in a slow process) then the temperature of the gas does not change.

c) Internal energy of the gas:

we have:

P*V = n*R*T = constant

and:

P*V = U/c = constant.

Then:

U = c*Constant

This means that the internal energy does not change.

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Here we can use the relation:

P*V = constant

then:

P = (constant)/V

Now, if V decreases, the denominator in that equation will be smaller. We know that if we decrease the value of the denominator, the value of the quotient increases.

And the quotient is equal to P.

Then if the volume decreases, we will see that the pressure increases.

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Suppose an object is launched from a point 320 feet above the earth with an initial velocity of 128 ft/sec upward, and the only
Ne4ueva [31]

Answer:

(a)Therefore the highest altitude attained by the object is =576 ft .

(b)Therefore the object takes 6 sec to fall to the ground.

Explanation:

Initial velocity: Initial velocity is a velocity from which an object starts to move.

u is usually used for notation of initial notation.

Final velocity: Final velocity is a velocity of an object after certain second from starting.

The final velocity is denoted by v.

Acceleration: The difference of final velocity and initial velocity per unit time

The S.I unit of acceleration is m/s².

(a)

Given that u= 128 ft\sec and g = 32 ft/sec².

At highest point the velocity of the object is 0 i.e v=0

Since the displacement is opposite to the gravity.

Therefore acceleration( a)= -g = -32 ft/sec².

To find the time this to happen we use the following formula

v=u+at

Here v=0

⇒0=128+(-32) t

⇒32t=128

⇒t = 4 sec

To determine the height we use the following formula

s=ut+\frac{1}{2} at^2

\Rightarrow s= (128\times4)+\frac{1}{2}\times (-32) \times4^2

⇒s= 256 ft

Therefore the highest altitude attained by the object is =(320+256)ft=576 ft .

(b)

At the highest point the velocity of the object is 0.

so u=0. a=g= 32 ft/sec²  [ since the direction of gravity and the displacement are same] s= 576 ft

To determine the time to fall we use the following formula

s=ut+\frac{1}{2} at^2

\Rightarrow 576 = (0\times t)+\frac{1}{2} \times 32 \times t^2

\Rightarrow 16\times t^2=576

\Rightarrow t^2=\frac{576}{16}

\Rightarrow t^2=36

⇒t=6 sec

Therefore the object takes 6 sec to fall to the ground.

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