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3 years ago

Rewrite this formula to solve for GGi: DX=fL/GGi

Physics

1 answer:

sashaice [31]3 years ago

3 0

Comment if you need clarification!

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What is the instantaneous speed of the monkey at time t=5s?

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I’m assuming we’re suppose to get some kind of graph but, Instantaneous speed is the speed that is happening right now. Like driving a car at 15k/h. The instantaneous speed of the car 15k/h. On the graph, at 5s. Wherever the line is, will tell you what the speed is.

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3 years ago

StarsA. begin as protostars, which fire up when they collapse and become denser and hotter.B. create elements by splitting the n

Colt1911 [192]

Answer:

Explanation:

Begin as protostars, which fire up when they collapse and become denser and hotter.

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3 years ago

A ball of mass m on a string of length L is attached to a pivot. The ball is released from rest while the string is parallel to

mash [69]

Answer:

L/2

Explanation:

Neglect any air or other resistant, for the ball can wrap its string around the bar, it must rotate a full circle around the bar. This means the ball should be able to swing to the top position where it's directly above the bar. By the law of energy conservation, this happens when the ball is at the same level as where it's previously released vertically. It means the swinging radius around the bar must be at least half of the string length.

So the distance d between the bar and the pivot should be at least L/2

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3 years ago

Two charges separated by 1 m exert 1 N forces on each other. If the charges are pushed to 1/4m separation, the force on each cha

Sloan [31]

Answer:

The force on each charge = 16 N

Explanation:

From coulombs law,

F = 1/4πε₀(q₁q₂)/d²........................ Equation 1

q₁q₂ = F4πε₀d²...................... Equation 2

Where F = force on the two charges, q₁ = charge on the first body, q₂ = charge on the second body, d = distance of separation, 1/4πε₀ = constant of proportionality.

When d = 1 m, F = 1 N,

Constant: 1/4πε₀ = 9×10⁹ Nm²/C²

Substituting these values into equation 2,

q₁q₂ = 1×1²/9×10⁹

q₁q₂ = 1/9×10⁹ C²

When d = 1/4 m, q₁q₂ = 1/9×10⁹ C² and 1/4πε₀ = 9×10⁹ Nm²/C²

Substituting these values into equation 1

F = 9×10⁹×1/9×10⁹ /(1/4)²

F = 1/(1/16)

F = 16 N

Therefore the force on each charge = 16 N

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3 years ago

Two identical positive charges exert a repulsive force of 6.1 × 10−9 n when separated by a distance 3.6 × 10−10 m. calculate the

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