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butalik [34]
3 years ago
6

"Heat engines and heat pumps are similar in that they both operate on the principle that heat flows naturally from a hot substan

ce to a colder one and, in the process, can be made to do work. However, there is an importance difference between them."
Physics
1 answer:
iris [78.8K]3 years ago
3 0

Answer:

The answer is below

Explanation:

The difference between a heat engine and a heat pump is that a heat engine is a device that converts Thermal energy into Mechanical Energy while a heat pump is a system that extracts heat from a cold body and delivers it to a hot body.

A Heat Engine generates mechanical energy (work) from the temperature difference between two thermal reservoirs while a  Heat pump extract heat from a low temperature reservoir and delivers the heat energy to a high temperature reservoir.

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3 years ago
Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 48.0° counterclockwise from the x ‑axis,
Ivanshal [37]

Answer:

Fn: magnitude of the net force.

Fn=30.11N , oriented 75.3 ° clockwise from the -x axis

Explanation:

Components on the x-y axes of the 17 N force(F₁)

F₁x=17*cos48°= 11.38N

F₁y=17*sin48° = 12.63 N

Components on the x-y axes of the  the second force(F₂)

F₂x= −19.0 N

F₂y=   16.5 N

Components on the x-y axes of the net force (Fn)

Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N

Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N

Magnitude of the net force.

F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }

F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }

F_{n} = 30.11N

Direction of the net force (β)

\beta =tan^{-1} (\frac{29.13}{7.62} )

β=75.3°

Magnitude and direction of the net force

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In the attached graph we can observe the magnitude and direction of the net force

6 0
3 years ago
An airplane touches down on the runway with a speed of 70 m/s2. Determine the airplane after each second of its deceleration.
ivann1987 [24]
<span>vf^2 = vi^2 + 2*a*d
---
vf = velocity final
vi = velocity initial
a = acceleration
d = distance
---
since the airplane is decelerating to zero, vf = 0
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0 = 55*55 + 2*(-2.5)*d
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5 0
3 years ago
An electron initially has a speed 16 km/s along the x-direction and enters an electric field of strength 27 mV/m that points in
weqwewe [10]

Answer:

a)t=1.4\times 10^{-5}\ s

b)S= 46.4 cm

Explanation:

Given that

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E= 27 mV/m

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lets time taken by electron is t

d = V x t

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b)

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F = m a = E q                    ------------1

Mass of electron ,m

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Charge on electron

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So now by putting the values in equation 1

a=\dfrac{E q}{m}

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S= ut+\dfrac{1}{2}at^2

Here initial velocity u= 0 m/s

S= \dfrac{1}{2}\times 4.74\times 10^{9}\times (1.4\times 10^{-5})^2\ m

S=0.464 m

S= 46.4 cm

S is the deflection of electron.

4 0
4 years ago
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