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gregori [183]
3 years ago
9

You have been hired by a venture capitalist to evaluate a concept engine proposed by an inventor, who claims that the engine con

sumes 100 MW at a temperature of 500 K, rejects 20 MW at a temperature of 300 K, and delivers 50 MW of mechanical work. Does this claim violate the first law of thermodynamics, the second law, both, or neither?
Physics
1 answer:
soldier1979 [14.2K]3 years ago
4 0

Answer:

It violates neither of them

Explanation:

first law of thermodynamics, also known as Law of Conservation of Energy, states that energy can neither be created nor destroyed; energy can only be transferred or changed from one form to another.

As we can see, 100MW work and temp of 500K was initial.

Now, the summation of output and lost work and temp = 70MW and 300k, do it doesn't violates first law.

The Second Law of Thermodynamics says that processes that involve the transfer or conversion of heat energy are irreversible, from the question, it is irreversible.

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A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta
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Answer:

t_t=4.131\ s

Explanation:

Given:

height above the horizontal form where the ball is hit, y=1\ m

angle of projectile above the horizontal, \theta=30^{\circ}

initial speed of the projectile, u=40\ m.s^{-1}

<u>Firstly we find the </u><u>vertical component of the initial velocity</u><u>:</u>

u_y=u.\sin\theta

u_y=40\times \sin30^{\circ}

u_y=20\ m.s^{-1}

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

  • v_y=0\ m.s^{-1}

Using eq. of motion:

v_y^2=u_y^2-2g.h (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

0^2=20^2-2\times 9.8\times h

h=20.4082\ m (from the height where it is thrown)

<u>Now we find the time taken to ascend to this height:</u>

v_y=u_y-g.t

0=20-9.8t

t=2.041\ s

<u>Time taken to descent the total height:</u>

  • we've total height, h'=h+y =20.4082+1

h'=u_y'.t'+\frac{1}{2} g.t'^2

  • during the course of descend its initial vertical velocity is zero because it is at the top height, so u_y'=0\ m.s^{-1}

21.4082=0+4.9t'^2

t'=2.09\ s

<u>Now the total time taken by the ball to hit the ground:</u>

t_t=t'+t

t_t=2.09+2.041

t_t=4.131\ s

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