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ahrayia [7]
3 years ago
12

An amount of work W is done on an object of mass m initially at rest, and as result it winds up moving at speed v. Suppose inste

ad it were already moving at speed v and the same amount of work W was done on it. What would be its final speed
Physics
1 answer:
sesenic [268]3 years ago
8 0

Answer:

Explanation:

Given

W amount of work is done on the system such that it acquires v velocity after operation(initial velocity)

According to work energy theorem work done by all the forces is equal to change in kinetic energy of object

W=\frac{1}{2}mv^2---1

where m=mass of object

v=velocity of object

When the object is already have velocity v then the final speed is given by work energy theorem

W=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2-----2

From 1 and 2 we get

\frac{1}{2}mv^2=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2

2\times \frac{1}{2}mv^2=\frac{1}{2}mv_f^2

v_f^2=2v^2

v_f=\sqrt{2}v                

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Sand is pouring out of a pipe and is forming a conical pile on the ground. The radius of the pile is increasing at a rate of 2 f
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Answer:

a)\ V=3053.628059 \ ft^3

b)\ V'=1017.876\ ft^3/day

Explanation:

<u>Rate of Change</u>

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\displaystyle V=\frac{\pi r^2h}{3}

The height is said to be 1/2 of the radius, thus

\displaystyle V=\frac{\pi r^2\cdot r}{2\cdot 3}

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a) Knowing r=18 feet, the volume is

\displaystyle V=\frac{\pi 18^3}{6}

V=3053.628059 \ ft^3

b) The rate of change of the volume is computed by taking the derivative of both sides respect to the time

\displaystyle V'=3\frac{\pi r^2}{6}r'

\displaystyle V'=\frac{\pi r^2}{2}r'

Where r' is the given rate of change of the radius: 2 feet/day.

Now we compute

\displaystyle V'=\frac{\pi 18^2}{2}\cdot 2=1017.876

\boxed{V'=1017.876\ ft^3/day}

6 0
3 years ago
What are the Properties of Transverse Waves and Longitudinal wave
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Answer:

Key terms

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The answer you are looking for is a leech! 

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8 0
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Modern vehicles are designed ________ in a crash to absorb kinetic energy?
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<span>Modern vehicles are designed to crush or crumple to absorb kinetic energy.</span>
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a steel sphere and brass ring have diameter 25cm and 24.9cm at 15°C.If the sphere and the ring are heated together.what is the t
Oksi-84 [34.3K]

Answer:

Explanation:

Due to heat energy , metal expands . Formula for linear expansion is as follows .

L = l ( 1 + α Δt )

where L is expanded length , l is original length , α is coefficient of linear expansion and Δt is increase in temperature .

To pass the sphere through the ring , the diameter of both ring and sphere should be same after heating . Let after increase of temperature Δt , their diameter becomes same as L  . The linear coefficient of brass and steel are

20 x 10⁻⁶ and 12 x 10⁻⁶ respectively .

For steel sphere ,

L = 25 ( 1 + 12 x 10⁻⁶ Δt )

For brass ring

L = 24.9 ( 1 + 20 x 10⁻⁶ Δt )

25 ( 1 + 12 x 10⁻⁶ Δt ) = 24.9 ( 1 + 20 x 10⁻⁶ Δt )

1.004( 1 + 12 x 10⁻⁶ Δt ) =  ( 1 + 20 x 10⁻⁶ Δt )

1.004 + 12.0482 x 10⁻⁶ Δt  =   1 + 20 x 10⁻⁶ Δt

.004 = 7.9518 x 10⁻⁶ Δt

Δt  = 4000 / 7.9518

= 503⁰C.

final temp = 503 + 15 = 518⁰C  .

6 0
3 years ago
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