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3 years ago

What happens to Roberto’s kinetic energy when he runs twice as fast?

Physics

1 answer:

Scilla [17]3 years ago

8 0

Well I would assume it would increase due to the increase in body movement creating more energy

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A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and c

Nady [450]

(a) +9.30 kg m/s

The impulse exerted on an object is equal to its change in momentum:

$I= \Delta p = m \Delta v = m (v-u)$

where

m is the mass of the object

$\Delta v$ is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

m = 0.272 kg

u = -12.6 m/s

v = +21.6 m/s

So the impulse is

$I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s$

(b) 155 N

The impulse can also be rewritten as

$I=F \Delta t$

where

F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)

$\Delta t$ is the duration of the collision

In this situation, we have

$\Delta t = 0.06 s$

So we can re-arrange the equation to find the magnitude of the average force:

$F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N$

6 0

3 years ago

A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from

Ulleksa [173]

Answer:

$1270.64\ \text{J}$

Explanation:

m = Mass of object = $\dfrac{mg}{g}$

mg = Weight of object = 20 N

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

$\theta$ = Angle of slope = $30^{\circ}$

f = Force of friction

fd = Work done against friction

The force balance of the system is

$\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}$

The work done against friction is $1270.64\ \text{J}$ .

8 0

3 years ago

Write your opinion about achievement made by during rana rule

Sedaia [141]

Answer:

Your opinion about achievement made by during rana rule

Explanation:

April Fools !

4 0

3 years ago

A CFL bulb has an efficiency of 8.9% and a power of 22 W. How much light energy does the lightbulb produce in 1 second

hjlf

The power that the light is able to utilize out of the supply is only 0.089 of the given.
Power utilized = (0.089)(22 W)
= 1.958 W
= 1.958 J/s
The energy required in this item is the product of the power utilized and the time. That is,
Energy = (1.958 J/s)(1 s) = 1.958 J
Thus, the light energy that the bulb is able to produce is approximately 1.958 J.

6 0

3 years ago

Read 2 more answers

What amount of heat is required to raise the temperature of 25 grams of copper to cause a 15ºC change? The specific heat of copp

Lina20 [59]

The amount of heat required is B) 150 J

Explanation:

The amount of heat energy required to increase the temperature of a substance is given by the equation:

$Q=mC\Delta T$