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Nat2105 [25]
2 years ago
12

: The maximum theoretical efficiency of a Carnot engine operating between reservoirs at the steam point and at room temperature

is about A :
Physics
1 answer:
Hunter-Best [27]2 years ago
5 0

Answer:

The value is   \eta  =  0.2145  or  21.45%

Explanation:

From the question we are told that

    The first reservoir is at steam point  T_s =  100^o C =  100 + 273 = 373 \ K  

    The  second reservoir is at room temperature T_r  =  20^o C = 293 \ K

Generally the  maximum theoretical efficiency of a Carnot engine  is mathematically evaluated as

     \eta  =  1- \frac{T_r}{T_s}

=>    1 - \frac{ 293}{373}

=>    \eta  =  0.2145

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Explanation: A sneaker is a want because you don't actually need it to survive

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2 years ago
At NASA's Zero Gravity Research Facility in Cleveland, Ohio, experimental payloads fall freely from rest in an evacuated vertica
Diano4ka-milaya [45]

Answer:

(a). Energy is 64,680 J

(b) velocity is 51.43m/s

(c) velocity in mph is 115.0mph

Explanation:

(a).

The potential energy P of the payload of mass m is at a vertical distance h is  

P =mgh.

Therefore, for the payload of mass m = 50kg at a vertical distance of h = 132 m, the potential energy is

P = (50kg)(9.8m/s^2)(132m)

\boxed{P = 64,680J}

(b).

When the payload reaches the bottom of the shaft, all of its potential energy is converted into its kinetic energy; therefore,

mgh= \dfrac{1}{2}mv^2

v= \sqrt{2gh}

v = \sqrt{2*9.8*135}

\boxed{v = 51.43m/s}

(c).

The velocity in mph is

\dfrac{51.43m}{s} * \dfrac{3600s}{hr} * \dfrac{1mile}{1609.34m}

\boxed{v= 115.0mph}

5 0
3 years ago
Hot air balloon ( look at pic)
kramer

Answer:

I think that the answer is convection.

Explanation:

Hope this helps.

6 0
3 years ago
What type of energy requires the motion of an object?
hichkok12 [17]

Answer:

Mechanical Energy.

Explanation:

This can occur as either kinetic or potential energy.

8 0
2 years ago
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A cons
Virty [35]

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

H_m=1.65m

H_E=1.16307m

Explanation:

From the question we are told that

Mass of ball M=2kg

Length of string L= 2m

Wind force F=13.2N

Generally the equation for \angle \theta is mathematically given as

tan\theta=\frac{F}{mg}

\theta=tan^-^1\frac{F}{mg}

\theta=tan^-^1\frac{13.2}{2*2}

\theta=73.14\textdegree

Max angle =2*\theta= 2*73.14=>146.28\textdegree

Generally the equation for max Height H_m is mathematically given as

H_m=L(1-cos146.28)

H_m=0.9(1+0.8318)

H_m=1.65m

Generally the equation for Equilibrium Height H_E is mathematically given as

H_E=L(1-cos73.14)

H_E=0.9(1+0.2923)

H_E=1.16307m

8 0
2 years ago
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