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Katen [24]
3 years ago
6

A ball is falling at terminal velocity. Terminal velocity occurs when the ball is in equilibrium and the forces are balanced. Wh

ich free body diagram shows the ball falling at terminal velocity? A free body diagram with one force pointing downward labeled F Subscript g Baseline 20 N. A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N. A free body diagram with 2 forces. The first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 5 N. A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 30 N.
Physics
2 answers:
Tema [17]3 years ago
8 0

Answer:

b

Explanation:

Greeley [361]3 years ago
4 0

Answer:

A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N.

Explanation:

This is because at terminal velocity, the ball stops accelerating and the net force on the ball is zero. For the net force to be zero, equal and opposite forces must act on the ball, so that their resultant force is zero. That is F₁ + F₂ = 0 ⇒ F₁ = -F₂

Since F₁ = 20 N, then F₂ = -F₁ = -20 N

So, if F₁ points upwards since it is positive, then F₂ points downwards since it is negative.

So, a free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N best describes the ball falling at terminal velocity.

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A hot iron ball of mass 200 g is cooled to a temperature of 22°C. 6.9 kJ of heat is lost to the surroundings during the process.
qaws [65]
Heat lost or gained, H = mc(θ₂ - θ₁) 
Where m = mass, c = Specific heat capacity, θ₂= final temperature, θ₁ = initial temperature

m = 200g, c = 0.444 J/g°C, θ₁ = 22 °C  (Since it was cooled).

H = 6.9 kj = 6.9 *1000J = 6900 J

6900 = 200*0.444* (θ₂ - 22)

6900/(200*0.444)  =  θ₂ - 22

77.70 = θ₂ - 22

θ₂ - 22 = 77.7

θ₂      =  77.7 + 22 = 99.7

So initial temperature before cooling ≈ 100°C .  Option C.


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Svetach [21]
The best and most correct answer among the choices provided by your question is the third choice or letter C.

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