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Elza [17]
3 years ago
10

A Carnot engine has a power output of 150 kW. The engine operates between two reservoirs at 20.0°C and 500°

Physics
2 answers:
Nookie1986 [14]3 years ago
8 0
Efficiency η of a Carnot engine is defined to be: 
<span>η = 1 - Tc / Th = (Th - Tc) / Th </span>
<span>where </span>
<span>Tc is the absolute temperature of the cold reservoir, and </span>
<span>Th is the absolute temperature of the hot reservoir. </span>

<span>In this case, given is η=22% and Th - Tc = 75K </span>
<span>Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). </span>

<span>Th = (Th - Tc) / η </span>
<span>Th = 75 / 0.22 = 341 K (rounded to closest number) </span>
<span>Tc = Th - 75 = 266 K </span>

<span>Lower temperature is Tc = 266 K </span>
<span>Higher temperature is Th = 341 K</span>
rjkz [21]3 years ago
5 0

Answer:

8.8 × 10³³J

Explanation:

efficiency of an engine η  = net work output ÷ heat input = 1 - \frac{T_{c} }{T_{H} }

T_{H} = 500 + 273 =773K

T_{C} = 20 + 273 =293K

n_{c} = 1 - \frac{293}{773}

  =0.62

power output= work output = 150kW

power output per hour = 150 × 1000×3600

                                     =5.4×10⁸J

recall, net work output ÷ heat input = 1 - \frac{T_{c} }{T_{H} }

∴ 0.62= 5.4×10⁸ ÷ heat input

heat input= 5.4×10⁸÷0.62

=8.81×10⁸J

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A man wishes to lift a stone weighing 1440 N, using a first-class lever that measures 5 meters. What force should it perform if
Crazy boy [7]

Answer:

3360 N

Explanation:

In a first-class lever, the effort force and load force are on opposite sides of the fulcrum.

The lever is 5 m long.  The load force is 1.50 m from the fulcrum, so the effort force must be 3.50 m from the fulcrum.

The torques are equal:

Fr = Fr

(1440 N) (3.5 m) = F (1.5 m)

F = 3360 N

4 0
3 years ago
Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t
Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

M=-\frac{s'}{s}

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

M=-\frac{15 cm}{12 cm}=-1.25

D. Real and inverted

The magnification equation can be also rewritten as

M=\frac{y'}{y}

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

y'=My

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

M=\frac{5}{9}=-\frac{s'}{s}

which can be rewritten as

s'=-M s = -\frac{5}{9}s

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

s'=-Ms

and then we can substitute it into the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}

and we can solve for s:

\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

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3 years ago
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irina1246 [14]

Answer:

Explanation:

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2. What are the forces that act on the block when it is placed on the ramp and is held in place by the spring scale? What is the net force acting on the block? Explain. (Assume that the ramps are frictionless surfaces.)

There are three forces acting on the block when it is placed on the ramp and is held in place by the spring scale: as in 1, there are tension and gravity but there is a third force - reaction force from the ramp surface on the block that is perpendicular to the surface. Again the block is not moving so the net force is zero.

3. What is the magnitude of normal force acting on the block when it is resting on the flat surface? How does the normal force change as the angle of the ramp increases? Explain. (Assume that the ramps are frictionless surfaces.)

On flat surface, the normal force is equal to the gravity force of the block i.e. its weight. On a vertical surface, the normal force is equal to zero. For the angle of ramp, θ, the normal force = weight * cos θ.

6 0
3 years ago
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