Does an object have energy when it is at rest?

melomori [17]

No the answer is energy

4 0

2 years ago

A heat engine performs (245 + A) J of work in each cycle while also delivering (142 + B) J of heat to the cold reservoir. Find t

Ganezh [65]

Answer:

The value is $\eta = 54.4 \%$

Explanation:

From the question we are told that

The work input is $W = ( 245 + A ) \ J$

The heat delivered is $Q = (142 + B) \ J$

The value of A is A = 14

The value of B is B = 72

Generally the efficiency of the heat engine is mathematically represented as

$\eta = \frac{W}{Q_t}$

Here $Q_t$ is the total out energy produce by the heat engine and this is mathematically represented as

$Q_t= Q + W$

=> $Q_t= 245 + A + 142 + B$

=> $Q_t= 390 + A+B$

So

$\eta = \frac{245 + A }{390 + A+ B}$

=> $\eta = 0.544$

=> $\eta = 0.544 *100$

=> $\eta = 54.4 \%$

5 0

2 years ago

12. Which of the following is the property of a system?

Sati [7]

Answer:

A

Explanation:

Pressure, temperature are measurable properties and they are also known as physical properties.

5 0

2 years ago

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$