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Alexxandr [17]
3 years ago
15

a furnace supplies 28kW of thermal power at 300C to an engine and exhausts waste energy at 20C. At the very best, how much work

could we expect to get out of the system per second
Physics
1 answer:
DochEvi [55]3 years ago
6 0

Answer:

The amount of work we could expect to get out of the system per second = 28,000J/s

Explanation:

Given the power supplied to the system as  28kW;

Energy = power / time

At very best, the amount of work we could expect to get out of the system per second = 28,000 W / 1 second =  28,000J/s

Therefore, for a a furnace which supplies 28kW of thermal power at 300C to an engine and exhausts waste energy at 20C.

At the very best, the amount of work we could expect to get out of the system per second = 28,000J/s

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8 0
3 years ago
The tallest sequoia sempervirens tree in California’s redwood national parks is 111 m tall. Suppose an object is thrown downward
ch4aika [34]

The object's <u>initial velocity</u> is equal to -10.59\frac{m}{s}

Why?

From the statement we know the height of the tree and the time it takes to reach the ground, so, if we need to calculate its initial velocity, we can use the following formula:

y=y_o-v_{o}*t-\frac{1}{2}g*t^{2}

Where,

y, is the final height (0 meters in this case)

yo, is the initial height (111 meters in this case)

t, is the time elapsed (3.8 seconds in this case)

vo, is the initial speed.

g, is the acceleration due to gravity (-9.81 m/s2)

Now, let's set the origin at the top of the tree, so, rewriting the formula, we have:

y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}

So, isolating the initial velocity, we have:

y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}

y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}\\\\v_{o}*t=y-y_o-\frac{1}{2}g*t^{2}\\\\v_{o}=\frac{y-y_o-\frac{1}{2}g*t^{2}}{t}

Finally, substituting and calculating, we have:

v_{o}=\frac{-111m-0-\frac{1}{2}(-9.8\frac{m}{s^{2}}) *(3.8s)^{2})}{3.8s}\\\\v_{o}=\frac{-111m-\frac{1}{2}(-9.8\frac{m}{s^{2}})*(14.44s^{2})}{3.8s}\\\\v_{o}=\frac{-111m-\frac{1}{2}(-141.51m)}{3.8s}=\frac{-111m+70.75m}{3.8s}\\\\v_{o}=\frac{-40.25m}{3.8s}=-10.59\frac{m}{s}

Hence, we have that the <u>initial velocity</u> of the object is -10.59\frac{m}{s}

Have a nice day!

7 0
4 years ago
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Lunna [17]
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joja [24]

Answer:

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Explanation:

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