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AnnZ [28]
3 years ago
14

A heat engine cycle is executed with steam in the saturation dome. The pressure of steam is 1.1 MPa during heat addition and 0.3

MPa during heat rejection. The highest possible efficiency of this heat engine is:
Physics
1 answer:
klio [65]3 years ago
8 0

Answer:The highest possible efficiency of this heat engine is 11%

Explanation:

Saturated water temperature at  P1, Pressure in Heat addition,

1.1 MPa=185°C +273= 458K

Saturated water temperature at  P2, Pressure in Heat rejection,

0.3MPa=133.5°C+ 273=406.5K

The highest possible efficiency of any heat engine is the Carnot efficiency given as

Carnot efficiency,  ηmax = 1- (T2/ T1)

1- (406.5K/458K)

1-0.88755=0.112

=11%

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A 53 kg crate is at rest on a level floor, and the coefficient of kinetic friction is 0.36. The acceleration of gravity is 9.8 m
tino4ka555 [31]

Answer:

42.6 m

Explanation:

mass of crate m = 53 kg

coefficient of kinetic friction, μ = 0.36

acceleration due to gravity, g = 9.8 m/s^2

Force, F = 372.098 N

Net force, f = F - friction force

f = 372.098 - μ m x g = 372.098 - 0.36 x 53 x 9.8

f = 185.114 N

acceleration, a = f / m = 185.114 / 53 = 3.49 m/s^2

initial velocity, u = 0

time, t = 4.94 s

s = ut + 1/2 at^2

s = 0 + 1/2 x 3.49 x 4.94 x 4.94

s = 42.6 m

6 0
3 years ago
When you rub a plastic rod with fur, the plastic rod becomes negatively charged and the fur becomes positively charged. As a con
Yuri [45]

Answer: C. the rod gains mass and the fur loses mass.

Explanation:Atomic particles have mass. The electron has a mass that is approximately 1/1836 that of the proton and with exchange exchange of charge this is also factored in. The movement of effect described above is known as the triboelectic charging process—charging by friction—which results in a transfer of electrons between the two objects when they are rubbed together. Plastic having a much greater affinity for electrons than animal fur pulls electrons from the atoms of fur, leaving both objects with an imbalance of charge. The plastic rod would have an excess of electrons and the fur has a shortage of electrons. Having an excess of electrons, the plastic is charged negatively and has more mass. In the same vein, the shortage of electrons on the fur leaves it with a positive charge and consequently with lesser mass.

4 0
3 years ago
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
3 years ago
What would the velocity of an object be if it was on a 1.75m
JulijaS [17]

Answer:

T = 4 sec / 2 = 2 sec      period of revolution

S = 2 pi R = 2 * pi * 1.75 m = 11 m

V = S / T = 11 m / 2 sec = 5.5 m/s       speed of object

4 0
2 years ago
Please help me! Some people have proposed a new way to build houses in areas that are likely to experience tsunamis. In this des
Alla [95]

Answer:

house wouldn't have solid walls on all four sides. Instead, some of the wall areas would be replaced by substances that

water can travel through quickly, as shown in the diagram. How would this design help a house survive a tsunami? What

drawbacks might there be to this design?

Explanation:

3 0
3 years ago
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