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MrMuchimi
3 years ago
14

A ball is thrown vertically upward. What are its velocity and acceleration when it reaches its maximum altitude? What is its acc

eleration just before it hits the ground?
Physics
1 answer:
butalik [34]3 years ago
8 0
<h2>Velocity at maximum height is zero.</h2><h2>Acceleration at maximum height is 9.81 m/s² downward.</h2><h2>Acceleration when it just hits the ground is 9.81 m/s² downward.</h2>

Explanation:

For a ball thrown vertically upward, the velocity at maximum height is zero.

Acceleration is a constant for a ball thrown vertically upward, its is acceleration due to gravity.

Acceleration at maximum height is 9.81 m/s² downward.

Acceleration when it just hits the ground is 9.81 m/s² downward.

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Answer:

Kinda? Depends what the question is fully asking

Explanation:

Acceleration is a change in velocity. So I guess if the velocity of something is -2 m/s and its positively accelerating at a value of +1 m/s, then that means every second its velocity changes by +1m/s.

So that -2 m/s thing after one second will be going -1 m/s.

After another second it'll be going 0 m/s.

After another itll be going +1 m/s and so on.

So at one point for a brief moment, it can have an acceleration but be at 0 m/s velocity.

5 0
3 years ago
What happens to a radioactive isotope as it decays?
kupik [55]
It becomes a different element
8 0
3 years ago
You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand at 30 m/s. What is the kinetic energy (KE) of the volley
FinnZ [79.3K]

awnser :

Explanation:

8 0
2 years ago
Read 2 more answers
Our milky way galaxy is 100000 lyly in diameter. a spaceship crossing the galaxy measures the galaxy's diameter to be a mere 1.
Sidana [21]

The speed of the spaceship relative to the galaxy is 0.99999995c.

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The Milky way galaxy is 100,000 light years in diameter.

The galaxy's diameter is a mere 1. 0 ly.

We know that ;

L = L_0 \sqrt{1-\frac{v^2}{c^2} }

L = 1 light year

L₀ = 100,000 light year

1 = 100,000 \sqrt{1-\frac{v^2}{c^2} }

1 = 100,000 \sqrt{1-\frac{v^2}{(3*10^8)^2} }

\frac{1}{100,000}  = \sqrt{1-\frac{v^2}{c^2} }

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Therefore, the speed of the spaceship relative to the galaxy is 0.99999995c.

Learn more about a light year here:

brainly.com/question/17423632

#SPJ4

5 0
1 year ago
Solve 1 for x if a=-9.8, v=2.7, and t= 35​
aliina [53]

Explanation:

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x=6097

6 0
3 years ago
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