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Anton [14]
3 years ago
9

Just how strong is the electric force? suppose you had two small boxes, each containing 1.0 g of protons. (a) if one were placed

on the moon by an astronaut and the other were left on the earth, and if they were connected by a very light (and very long!) string, what would be the tension in the string? express your answer in newtons and in pounds. do you need to take into account the gravitational forces of the earth and moon on the protons? why? (b) what gravitational force would each box of protons exert on the other box?
Physics
1 answer:
vlabodo [156]3 years ago
3 0
Let's use Newton's Law of Second Motion: F=ma. When no other direct force is acting on the system, the acceleration is due to the gravity. The modified equation becomes: F = mg. So, yes, you need to take into account the gravitational accelerations in the moon and on Earth. 

g,moon = 1.622 m/s²
g,Earth = 9.81 m/s²

The net force is the tension of the string:

F,Earth - F,moon = Tension
Tension = (1/1000 kg)(9.81 m/s²) - (1/1000 kg)(1.622 m/s²)
Tension = 8.188×10⁻³ N

To convert, 1 pound force is equal to 4.45 Newtons:

Tension = 8.188×10⁻³ N * 1 lbf/4.45 N
Tension = 1.84×10⁻³ lbf
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Answer:

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g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 k
Salsk061 [2.6K]

Answer:

h = 3.5 m

Explanation:

First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:

2gh = v_f^2 - v_i^2\\

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 3.5 m

vf = final speed = ?

vi = initial speed = 0 m/s

Therefore,

(2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s

Now, we will apply the law of conservation of momentum:

m_1v_1 = m_2v_2

where,

m₁ = mass of colliding ball = 3.6 kg

m₂ = mass of ball on the other end = 3.6 kg

v₁ = vf = final velocity of ball while collision = 8.3 m/s

v₂ = vi = initial velocity of other end ball = ?

Therefore,

(3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s

Now, we again use the third equation of motion for the upward motion of the ball:

2gh = v_f^2 - v_i^2\\

where,

g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)

h = height = ?

vf = final speed = 0 m/s

vi = initial speed = 8.3 m/s

Therefore,

(2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\

<u>h = 3.5 m</u>

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True: Friction depends on the types of surfaces involved and how hard the surfaces push together.

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Under feelings of weightlessness, you are still being pulled by gravity, but your perpendicular velocity and distance from the source can cancel each other out.
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2 years ago
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