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Allushta [10]
4 years ago
5

An object is moving with an initial velocity of 9m/s. It accelerates at a rate 1.5m/sec2 over a distance of 20m. What is it’s ne

w velocity?
Physics
1 answer:
puteri [66]4 years ago
4 0

Answer:

11.87

Explanation:

final velocity^2= initial velocity ^2+ 2* Acceleration* distance

Final Velocity^2= 9*9+2*1.5.20

final velocity^2 = 141

final velocity = 11.87

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Answer:

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Explanation:

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*brainlest* An object accelerates from rest with a constant acceleration of 2m/s^2. How far will it have moved after 9s? Equatio
boyakko [2]

Answer: You throw an object upwards from the ground with a velocity of 20m/s and it is subject to a downward

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Given Formula Set up Solution

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6 0
3 years ago
No matter what values of m and k you used for the spring, what is the ratio of the period to k m.
Leto [7]

The relationship between the period of an oscillating spring and the attached mass determines the ratio of the period to \sqrt{\dfrac{m}{k} }.

Response:

  • The ratio of the period to  \sqrt{\dfrac{m}{k} } is always approximately<u> 2·π : 1</u>

<u />

<h3>How is the value of the ratio of the period to \sqrt{\dfrac{m}{k} } calculated?</h3>

Given:

The relationship between the period, <em>T</em>, the spring constant <em>k</em>, and the

mass attached to the spring <em>m</em> is presented as follows;

T =  \mathbf{2 \cdot \pi \cdot \sqrt{\dfrac{m}{k} }}

Therefore, the fraction of of the period to \sqrt{\dfrac{m}{k} }, is given as follows;

\mathbf{\dfrac{T}{ \sqrt{\dfrac{m}{k} }}} = 2 \cdot \pi

2·π ≈ 6.23

Therefore;

T :{ \sqrt{\dfrac{m}{k} }} = 2 \cdot \pi : 1

Which gives;

  • The ratio of the period to  \sqrt{\dfrac{m}{k} } is always approximately<u> 2·π : 1</u>

Learn more about the oscillations in spring here:

brainly.com/question/14510622

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3 years ago
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The cloud chamber, the Geiger counter, and a piece of
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8 0
4 years ago
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