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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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Keith_Richards [23]

Answer:

1777.92 m/s

Explanation:

R = Radius of asteroid = 545 km

M = Mass of planet

g = Acceleration due to gravity = 2.9 m/s²

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Acceleration due to gravity is given by

g=\dfrac{GM}{R^2}\\\Rightarrow M=\dfrac{gR^2}{G}

The expression of escape velocity is given by

v=\sqrt{\dfrac{2GM}{R}}\\\Rightarrow v=\sqrt{\dfrac{2G}{R}\dfrac{gR^2}{G}}\\\Rightarrow v=\sqrt{2gR}\\\Rightarrow v=\sqrt{2\times 2.9\times 545000}\\\Rightarrow v=1777.92\ m/s

The escape speed is 1777.92 m/s

3 0
3 years ago
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Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5
schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

3 0
2 years ago
A speaker vibrates at a frequency of 200 hz what is its period
Norma-Jean [14]

Period = (1) / (frequency)

Period = (1) / (200 per second)  =  0.005 second  =  5 milliseconds

5 0
3 years ago
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A 7.00 kg bowling ball is held 2.00 m above the ground. Using g- 9.8 m/s^2, how much energy does the bowling ball have due to it
Alexeev081 [22]
It should be B.137 just took the test.
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7. Nancy has a mass of 60 kg and sits on the very end of a 3.00 m long plank pivoted in the middle. How much torque must her co-
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Answer:

Torque = 882Nm

Explanation:

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Torque= 60×9.8×1.5

Torque= 882Nm

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3 years ago
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