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stepan [7]
3 years ago
11

A steam turbine operates at a boiler temperature of 450 k and an exhaust temperature of 300 k. what is the maximum theoretical e

fficiency of this system?
Physics
1 answer:
Helga [31]3 years ago
5 0
The maximum theoretical efficiency of the system is the one corresponding to the efficiency of a Carnot cycle operating between the same temperatures of the system:
\eta=1- \frac{T_c}{T_h}
where T_c and T_h are the cold and hot temperatures, respectively.
In our problem, T_c=300 K and T_h=450 K, therefore the maximum theoretical efficiency is
\eta=1- \frac{300 K}{450 K}=0.33
So, 33%.
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A 37 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar
Ne4ueva [31]

Answer:

The frictional force acting on the block is 14.8 N.

Explanation:

Given that,

Weight of block = 37 N

Coefficients of static = 0.8

Kinetic friction = 0.4

Tension = 24 N

We need to calculate the maximum friction force

Using formula of friction force

f=\mu mg

Put the value into the formula

f=0.8\times37

f=29.6\ N

So, the tension must exceeds 29.6 N for the block to move

We need to calculate the frictional force acting on the block

Using formula of frictional force

f = \mu N

Put the value in to the formula

f=0.4\times37

f=14.8\ N

Hence, The frictional force acting on the block is 14.8 N.

6 0
3 years ago
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
aivan3 [116]

Answer:

Explanation:

Given that,

Initial angular velocity is 0

ωo=0rad/s

It has angular velocity of 11rev/sec

ωi=11rev/sec

1rev=2πrad

Then, wi=11rev/sec ×2πrad

wi=22πrad/sec

And after 30 revolution

θ=30revolution

θ=30×2πrad

θ=60πrad

Final angular velocity is

ωf=18rev/sec

ωf=18×2πrad/sec

ωf=36πrad/sec

a. Angular acceleration(α)

Then, angular acceleration is given as

wf²=wi²+2αθ

(36π)²=(22π)²+2α×60π

(36π)²-(22π)²=120πα

Then, 120πα = 8014.119

α=8014.119/120π

α=21.26 rad/s²

Let. convert to revolution /sec²

α=21.26/2π

α=3.38rev/sec

b. Time Taken to complete 30revolution

θ=60πrad

∆θ= ½(wf+wi)•t

60π=½(36π+22π)t

60π×2=58πt

Then, t=120π/58π

t=2.07seconds

c. Time to reach 11rev/sec

wf=wo+αt

22π=0+21.26t

22π=21.26t

Then, t=22π/21.26

t=3.251seconds

d. Number of revolution to get to 11rev/s

∆θ= ½(wf+wo)•t

∆θ= ½(0+11)•3.251

∆θ= ½(11)•3.251

∆θ= 17.88rev.

5 0
3 years ago
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Sloan [31]


For the answer to the question above
Forecasting how a business might do in the future. 
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--Going past 5-- 

Inventory tracking 
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3 years ago
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How is most of the electricity we use at home generated?
Sveta_85 [38]
Nuclear power plants, wind farms, water farms, and geothermal heating
3 0
3 years ago
At what phase would you expect to find extremely high and low tides?
creativ13 [48]

Answer:

full moon and new moon

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3 years ago
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