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Paul [167]
3 years ago
6

Please help ASAP please ASAP

Physics
1 answer:
alina1380 [7]3 years ago
5 0
The answer to your question is 185
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An engine absorbs 2150 J as heat from a hot reservoir and gives off 750 J as heat to a cold reservoir during each cycle. What is
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65.2 %

Explanation:

Let Q1 = Heat absorbed by the system

Q2 = Heat released by the system

e= (1 - (Q2/Q1)) x 100

e= (1 - (750/2150)) x 100

e= (1 - 0.348) x 100

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A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on
MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

Ec=\frac{1}{2} *m*v^{2}

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}

W=\frac{1}{2} *m*(v2^{2}-v1^{2})

In this case:

  • W=?
  • m= 2,145 kg
  • v2= 12 \frac{m}{s}
  • v1= 25 \frac{m}{s}

Replacing:

W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

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3 years ago
PLEASE help can you give me the answers to the Light Unit Review Practice quiz I'm totally lost
solniwko [45]
There’s nothing to see so there for I can’t help you
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3 years ago
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