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finlep [7]
3 years ago
11

the kinetic energy of an object with mass m moving with a velocity of 5 m/s is 25 j what will be its kinetic energy when its vel

ocity is doubled what will its kinetic energy when it is velocity increased 3 time
Physics
1 answer:
Dmitriy789 [7]3 years ago
4 0

Answer:

100 J, 225 J

Explanation:

The kinetic energy of an object is given by:

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

where

m is the mass of the object

v is the velocity of the object

In this problem, the initial kinetic energy of the object is

K = 25 J

Then, the velocity is doubled, which means

v' = 2v

Therefore, the new kinetic energy will be

K'=\frac{1}{2}m(2v)^2 = 4(\frac{1}{2}mv^2)=4K

Therefore, the kinetic energy has quadrupled:

K' = 4(25)=100 J

Later, the velocity is tripled, which means

v'' = 3v

Therefore, the new kinetic energy will be

K''=\frac{1}{2}m(3v)^2 = 9(\frac{1}{2}mv^2)=9K

Therefore, the kinetic energy has increased by a factor of 9:

K' = 9(25)=225 J

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Answer:

a_{c} =  13.46\ m/s^2

Explanation:

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a_{c} = \frac{v^2}{r}

where,

ac = centripetal acceleration = ?

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r = distance of ball from center of rotation = 0.91 m

Using these values in equation, we get:

a_{c} = \frac{(3.5\ m/s)^2}{0.91\ m}\\\\a_{c} =  13.46\ m/s^2

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3 years ago
Suppose you are at the center of a large freely-rotating horizontal turntable in a carnival funhouse. As you crawl toward the ed
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Answer:

c. remains the same, but the RPMs decrease.

Explanation:

Because there aren't external torques on the system composed by the person and the turntable it follows that total angular momentum (I) is conserved, that means the total angular momentum is a constant:

\overrightarrow{L}=constant

The total angular momentum is the sum of the individual angular momenta, in our case we should sum the angular momentum of the turntable and the angular momentum of a point mass respect the center of the turntable (the person)

\overrightarrow{L_{turnatble}}+\overrightarrow{L_{person}}=constant (1)

The angular momentum of the turntable is:

\overrightarrow{L_{turnatble}}=I\overrightarrow{\omega} (2)

with I the moment of inertia and ω the angular velocity.

The angular momentum of the person respects the center of the turntable is:

\overrightarrow{L_{person}}=\overrightarrow{r}\times m\overrightarrow{v} (3)

with r the position of the person respects the center of the turntable, m the mass of the person and v the linear velocity

Using the fact v=\omega r:

\overrightarrow{L_{person}}=\overrightarrow{r}\times rm\overrightarrow{\omega}(3)

By (3) and (2) on (1) and working only the magnitudes (it's all that we need for this problem):

I\omega+r^{2}m\omega=constant

\omega(I+r^{2}m)=constant

Because the equality should be maintained, if we increase the distance between the person and the center of the turntable (r), the angular velocity should decrease to maintain the same constant value because I and m are constants, so the RPM's (unit of angular velocity) are going to decrease.

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A car starting from rest accelerates at a constant 2.0 m/s2 for 10 s. It then travels with constant speed it has achieved for an
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Answer:

The total distance will be 400 m.

Explanation:

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For portion BC:

V= u + at

V=0 + 2 x 10

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In this portion car moves with constant velocity  20 m/s for 10 s.

So distance S= V x t

S=20 x 10 =200 m.

For portion CD:

The velocity at point C will be 20 m/s

In this portion the final speed of car will be zero because given that at final car come to rest.

So the acceleration will be in the negative direction to stop the car.

We know that

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The total distance will be 400 m.

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