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Nady [450]
3 years ago
7

A heat engine is designed to do work. This is possible only if certain relationships between the heats and temperatures at the i

nput and output hold true. Which of the following sets of statements must apply for the heat engine to do work?A) Qh < Qc and Th < TcB) Qh > Qc and Th < TcC) Qh < Qc and Th > TcD) Qh > Qc and Th > Tc
Physics
1 answer:
ryzh [129]3 years ago
5 0

Answer: option D

Explanation: let us first define an heat engine.

An heat engine is that device that converts partly heat energy into mechanical energy.

Inside the heat engine is a substance that undergoes compression and expansion, intake and outtake of heat, this is known as a working substance.

For heat engines to work perfectly, the working substance has to take a substantial large amount of heat energy from a source, this source is the hot reservoir, the amount of heat energy given out by this reservoir is Qh and the temperature at this region is Th. The heat from the hot reservoir is accepted by the working substances and sent to a region that will discard it out, this region is known as the cold reservoir and the heat energy at this point is Qc and temperature is Tc.

Judging by the direction of heat flow, heat energy moves from hot reservoir to the cold reservoir for the heat engine to work perfectly fine, hence Qh must be greater than Qc and Th must be greater than Tc.

Hence Qh>Qc and Th>Tc

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Distance Between Two Cars A car leaves an intersection traveling west. Its position 4 sec later is 19 ft from the intersection.
faltersainse [42]

Answer:

The answer to the question is;

The rate at which the distance between the two cars is changing is equal to 14.4 ft/sec.

Explanation:

We note that the distance  traveled by each car after 4 seconds is

Car A = 19 ft in the west direction.

Car B = 26 ft in the north direction

The distance between the two cars is given by the length of the hypotenuse side of a right angled triangle with the north being the y coordinate and the  west being the x coordinate.

Therefore, let the distance between the two cars be s

we have

s² = x² + y²

= (19 ft)² + (26 ft)² = 1037 ft²

s = \sqrt{1037 ft^2} = 32.202 ft.

The rate of change of the distance from their location 4 seconds after they commenced their journeys is given by;

Since s² = x² + y² we have

\frac{ds^{2} }{dt} = \frac{dx^{2} }{dt}  + \frac{dy^{2} }{dt}

→ 2s\frac{ds }{dt} = 2x\frac{dx}{dt}  + 2y\frac{dy }{dt} which gives

s\frac{ds }{dt} = x\frac{dx}{dt}  + y\frac{dy }{dt}

We note that the speeds of the cars were given as

Car B moving north = 12 ft/sec, which is the y direction and

Car A moving west = 8 ft/sec which is the x direction.

Therefore

\frac{dy }{dt} =  12 ft/sec and

\frac{dx}{dt} = 8 ft/sec

s\frac{ds }{dt} = x\frac{dx}{dt}  + y\frac{dy }{dt} becomes

32.202 ft.\frac{ds }{dt} = 19 ft \times 8 \frac{ft}{sec}  + 26ft\times 12\frac{ft}{sec}  = 464 ft²/sec

\frac{ds }{dt} = \frac{464\frac{ft^{2} }{sec} }{32.202 ft.} = 14.409 ft/sec ≈ 14.4 ft/sec to one place of decimal.

3 0
3 years ago
square root A 1400 kg car is coasting on a horizontal road with a speed of 18 m/s . After passing over an unpaved, sandy stretch
Lina20 [59]

Answer:

The net force on the car is 2560 N.

Explanation:

According to work energy theorem, the amount of work done is equal to the change of kinetic energy by an object. If 'W' be the work done on an object to change its kinetic energy from an initial value 'K_{i}' to the final value 'K_{f}', then mathematically,

W = K_{f} - K_{i} = \dfrac{1}{2}~m~(v_{f}^{2} - v_{i}^{2})........................................(I)

where 'm' is the mass of the object and 'v_{i}' and 'v_{f}' be the initial and final velocity of the object respectively. If 'F_{net}' be the net force applied on the car, as per given problem, and 's' is the displacement occurs then we can write,

W = F_{net}~.~s.......................................................(II)

Given, m = 1400~Kg,~v_{i} = 18~m~s^{-1}~v_{f} = 14~m~s^{-1}~and~s = 35~m.

Equating equations (I) and (II),

&& - F_{net} \times 35~m = \dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})~m^{2}~s^{-2}\\&or,& F_{net} = \dfrac{\dfrac{1}{2} \times 1400~Kg~\times(14^{2} - 18^{2})}{35}~N\\&or,& F_{net} = 2560~N

6 0
3 years ago
What happens overtime to rocks that are stressed
malfutka [58]

Answer:

Stress is the force applied to an object. In geology, stress is the force per unit area that is placed on a rock. Four types of stresses act on materials.

A deeply buried rock is pushed down by the weight of all the material above it. Since the rock cannot move, it cannot deform. This is called confining stress.

Compression squeezes rocks together, causing rocks to fold or fracture (break) (Figure below). Compression is the most common stress at convergent plate boundaries.

Explanation:

4 0
3 years ago
A boat having a length 3m and breadth 2m is floating on a
brilliants [131]

Answer:

60 kg

Explanation:

The man's weight is equal to the weight of the water he displaces.

mg = ρVg

m = ρV

m = (1000 kg/m³) (3 m × 2 m × 0.01 m)

m = 60 kg

The man's mass is 60 kg.

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3 years ago
Magnetic levitation has been used to innovate which aspect of life? A. communication B. mapping C. transportation D. space explo
bekas [8.4K]
I think the correct answer among the choices listed above is option C. Magnetic levitation has been used to innovate transportation. This innovation is commonly known as maglev. It is a new transportation technology where noncontacting vehicles travel above a guideway by magnetic fields.
4 0
4 years ago
Read 2 more answers
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