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3 years ago

This version of Einstein’s equation is often used directly to find what value?

Physics

2 answers:

stepladder [879]3 years ago

8 0

A. THE ENERGY THAT IS RELEASED IN A NUCLEAR REACTION.

M =MASS

C^2= SPEED OF LIGHT SQUARED

Ainat [17]3 years ago

8 0

C. the mass that is lost in a fusion reaction

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L'intensité de la pesanteur g à la surface d'une planète peut se calculer par la formule : g=G. M R^ 2 M;\\ R: masse de la planè

jarptica [38.1K]

Answer:

Step-by-step explanation:

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2 years ago

How can we avoid water pollution

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Explanation:

1.Pick up litter and throw it away in a garbage can.

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3 years ago

Read 2 more answers

If the force causing a mass to accelerate is multiplied by a number while keeping the mass the same, what happens to the acceler

klasskru [66]

Answer: If mass remains constant and you double the force you will double the acceleration.

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3 years ago

Andy and Charlie are riding on a merry go-round. Andy rides on a horse at the outer rim of the circular platform, four times as

mestny [16]

Answer:

4) the same as Charlie's.

Explanation:

They both make one full circuit (2π radians) of the merry go round in the same amount of time, therefore their angular speed is the same

3 0

3 years ago

A coin dropped in the lift it takes time 0.5 s to reach the floor when lift is staionary it takes time t when lift is moving up

BARSIC [14]

Answer:

t₁ > t₂

Explanation:

A coin is dropped in a lift. It takes time t₁ to reach the floor when lift is stationary. It takes time t₂ when lift is moving up with constant acceleration. Then t₁ > t₂, t₁ = t₂, t₁ >> t₂ , t₂ > t₁

Solution:

Newton's law of motion is given by:

s = ut + (1/2)gt²;

where s is the the distance covered, u is initial velocity, g is the acceleration due to gravity and t is the time taken.

u = 0 m/s, t₁ is the time to reach ground when the light is stationary and t₂ is the time to reach ground when the lift is moving with a constant acceleration a.

hence:

When stationary:

$s=\frac{1}{2}gt_1^2\\\\t_1^2=\frac{2s}{g} \\\\When\ moving\ with\ acceleration(a):\\\\s=\frac{1}{2}(g+a)t_2^2\\\\t_2^2=\frac{2s}{g+a}$

Hence t₂ < t₁, this means that t₁ > t₂.

4 0

3 years ago