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harkovskaia [24]
3 years ago
7

Which formula can be used to calculate average speed (v) for steady motions?

Physics
1 answer:
Liono4ka [1.6K]3 years ago
3 0
I’m pretty sure it’s speed=distance/time
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mrs_skeptik [129]
I would use miles Because miles will give you less of an answer
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In order to open the clam it catches, a seagull will drop the clam repeatedly onto a hard surface from high in the air until the
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Answer:

2.2 s

Explanation:

Hi!

Let's consider the origin of the coordinate system at the ground, and consider that the clam starts with zero velocity, the equation of motion of the clam is given by

x(t) = 23.1 m - \frac{1}{2}(9.8 m/s^2) t^2

We are looking for a time t for which x(t) = 0

0 = 23.1 m - (4.9 m/s^2) t^2

Solving for t:

t = \sqrt{\frac{23.1}{4.9}} s = 2.17124 s

Rounding at the first decimal:

t = 2.2 s

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3 years ago
Consider two celestial objects with masses m1 and m2 with a separation distance between their centers r. If the first mass m1 we
Scilla [17]

The new magnitude of the force of attraction will be 6 times the original force of attraction

<h3>How to determine the initial force </h3>
  • Mass 1 = m₁
  • Mass 2 = m₂
  • Gravitational constant = G
  • Distance apart = r
  • Initial force (F₁) = ?

F = Gm₁m₂ / r²

F₁ = Gm₁m₂ / r²

<h3>How to determine the new force </h3>
  • Mass 1 = 2m₁
  • Mass 2 = 3m₂
  • Gravitational constant = G
  • Distance apart (r) = r
  • New force (F₂) =?

F = Gm₁m₂ / r²

F₂ = G × 2m₁ × 3m₂ / r²

F₂ = 6Gm₁m₂ / r²

But

F₁ = Gm₁m₂ / r²

Therefore

F₂ = 6Gm₁m₂ / r²

F₂ = 6F₁

Thus, the new magnitude of the force of attraction will be 6 times the original force of attraction

Learn more about gravitational force:

brainly.com/question/21500344

#SPJ1

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You drop a ball from the top of a building, how long does it take to reach the ground
irakobra [83]
Need more details to the question
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A 5.3 kg cat and a 2.5 kg bowl of tuna fish are at opposite ends of the 4.0-m-long seesaw. how far to the left of the pivot must
nika2105 [10]
Moment about the pivot must be equal for the seesaw to balance. Initially, the first cat and the bowl are at 2 m from the pivot.

The moment due to cat = 5.3*2 = 10.6 kg.m
The moment due to bowl = 2.5*2 = 5 kg.m
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Therefore, the 3.7 kg cat should stand at a distance x from the pivot in left to balance the 5.6 kg.m.
That is,
3.7*x = 5.6 => x = 5.6/3.7 = 1.5134 m to the left (on the side of the bowl)
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