Explanation:
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Answer:
Explanation:
If you ignore air resistant, then nothing affects the package horizontal motion. In Newton's first law it would keep the package at a constant speed, the speed of the airplane.
So to the eyes of the pilot who is also moving at the same horizontal speed, the lateral position of the package does not change. He can only perceive that the package is getting further away from him as it's dropping vertically.
To a person on the ground then the package is travelling in a parabolic path, where its horizontal speed is constant but vertical speed is increasing toward the ground at the rate of g.
The best and most correct answer among the choices provided by the question is the first choice, larger.
Rankine is Fahrenheit + 460 , while Kelvin is Celsius + 273. We all know that Fahrenheit has larger number compared to kelvin , thus rankine is much larger.
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.
The efficiency of the turbine is 50%.
The given parameters:
- <em>Kinetic energy of the wind, E = 3000 J</em>
- <em>Output electrical energy, E(out) = 750 J</em>
- <em>Energy lost to heat, E(lost) = 750 J</em>
- <em>Kinetic energy of the turbine, E(in) = ?</em>
The kinetic energy of the turbine which is the input energy is calculated as follows;
E(in) = 3000 - (750 + 750)
E(in) = 1500 J
The efficiency of the turbine is calculated as follows;
Thus, the efficiency of the turbine is 50%.
Learn more about efficiency of turbine here: brainly.com/question/2009210
Answer:
a) h'= 5/7 h
, b) h ’= h
Explanation:
Let's use energy conservation for this exercise
Starting point. Upper left side
Em₀ = mg h
Final point. Lower left side
Emf = K = ½ m v² + ½ I w²
Em₀ = emf
mgh = ½ m v² + ½ I w²
Angular and linear velocity are related
v = r w
w = v / r
The moment of inertia of the marble that we take as a solid sphere is
I = 2/5 m r²
We substitute
m g h = ½ m v² + ½ 2/5 m r² (v / r)²
g h = ½ v2 (1 + 2/5)
v = √(g h 10/7)
This is the speed at the bottom of the bowl
Now let's apply energy conservation to the right side
a) right side if rubbing
Em₀ = K
Emf = U = mg h’
½ m v² = mg h’
h’= ½ (g h 10/7) / g
h'= 5/7 h
b) right side with rubbing
Em₀ = K
Emf = K + U = -½ I w² + m g h
Emf = -½ 2/5 m r² v² / r² + m gh
Em₀ = emf
½ v² = -1/5 v² + g h’
h’= (1/2 +1/5) (gh 10/7) / g
h ’= h
c) When there is friction, an energy of rotation is accumulated that must be dissipated, by local it goes higher
