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Kryger [21]
3 years ago
15

One horsepower is a unit of power equal to 746 W. How much energy can a 150 horsepower engine transform in 10.0 seconds

Physics
1 answer:
11111nata11111 [884]3 years ago
7 0
1,119,000 basically 746 times 150 then multiply that answer by 10
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In a certain process, the energy change of the system is 250 \rm kJ. The process involves 480 \rm kJ of work done by the system.
alisha [4.7K]

Answer:

\Delta Q=-230kJ

Explanation:

Using the first law of thermodynamics:

\Delta U=\Delta Q-W

Where \Delta U is the change in the internal energy of the system, in this case  \Delta U=250kJ, \Delta Q is the heat tranferred, and W is the work,  W=-480kJ with a negative sign since the work is done by the system.

From the previous equation we solve for heat, because it is the unknown variable in this problem

\Delta Q=\Delta U +W

And replacing the known values:

\Delta Q=250kJ +(-480kJ)

\Delta Q=250kJ -480kJ

\Delta Q=-230kJ

The negative sign shows us that the heat is tranferred from the system into the surroundings.

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4 years ago
Stars combine Hydrogen to make Helium during nuclear fusion. Living things are made of heavier elements like Carbon, Oxygen, Iro
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8 0
3 years ago
A pulling force is called what?<br> A. normal<br> B. tension<br> C. balanced<br> D. compression
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3 0
3 years ago
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A ball of moist clay falls 17.3 m to the ground. It is in contact with the ground for 24.0 ms before stopping. (a) What is the a
gizmo_the_mogwai [7]

Answer:

Acceleration,  767.08\ m/s^2

Explanation:

Given that,

Height from a ball falls the ground, h = 17.3 m

It is in contact with the ground for 24.0 ms before stopping.

We need to find the average acceleration the ball during the time it is in contact with the ground.

Firstly, find the velocity when it reached the ground. So,

v^2=u^2+2ah

u = initial velocity=0 m/s

a = acceleration=g

v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 17.3} \\\\v=18.41\ m/s

It is in negative direction, u = -18.41 m/s

Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18.41}{24\times 10^{-3}}\\\\a=767.08\ m/s^2

So, the average acceleration of the ball during the time it is in contact is 767.08\ m/s^2.

4 0
3 years ago
Based on the law of conservation of energy, how can we reasonably improve a machine’s ability to do work? Move the machine to a
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Reduce the friction. Since the total energy is conserved, the only way to improve its work capacity is by reducing energy that doesnt go into work.
7 0
3 years ago
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