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attashe74 [19]
2 years ago
14

A dancer moves around a path like the one

Physics
1 answer:
mash [69]2 years ago
7 0

Given what we know, despite not having the figure attached to the question, we can still confirm that the magnitude for the acceleration of the dancer will be zero.

<h3>Why is the dancer's acceleration equal to zero?</h3>

This has to do with how the question clarifies the speed of the dancer. Though it does not give us an exact value, we are told that the speed is constant. This is an indicator that the acceleration is zero because with any other value for acceleration the speed <u>cannot remain</u> constant.

Therefore, given that any value for acceleration will increase or decrease the speed of the dancer, but we are told that the dancer's speed is constant throughout the trip, we can confirm that the magnitude for the acceleration of the dancer is zero.

To learn more about acceleration visit;

brainly.com/question/12134554?referrer=searchResults

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pochemuha

Answer:

the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

Explanation:

Given;

height of the cliff, h = 210 m

initial horizontal velocity of the cannonball, Ux = 50 m/s

initial vertical velocity of the cannonball, Uy = 0

The time for the cannonball to reach the ground is calculated as;

h = u_yt - \frac{1}{2} gt^2\\\\h = 0 - \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 210}{9.8} }\\\\t  = 6.55 \ s

The horizontal distance covered by the cannonball before it hits the ground is calculated as;

X = U_x \times \ t\\\\X = 50 \times \ 6.55\\\\X = 327.5 \ m

Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m

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gulaghasi [49]
<h3>Answer;</h3>

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