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wolverine [178]
3 years ago
10

In 3 meters a person running 0.5 m/s accelerates 1.2 m/s 2. How fast were they going afterward? Choose the right equation for th

is word problem from below.
Physics
1 answer:
Feliz [49]3 years ago
7 0

Answer:

2.72 m/s

Explanation:

In 3 meters a person running 0.5 m/s accelerates 1.2 m/s².

It means,

Distance, s = 3 m

Initial velocity, u = 0.5 m/s

Acceleration, a = 1.2 m/s²

We need to find the final velocity of the person. Using equation of motion to find it as follows :

v^2-u^2=2as\\\\v^2=2as+u^2\\\\v^2=2\times 1.2\times 3+(0.5)^2\\\\v=2.72\ m/s

So, the final velocity of the person is 2.72 m/s.

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Answer:

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Sand dunes during the Dust Bowl are an example of windward.

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3 years ago
A wall, acted upon by a force of 20 N, does not move. The work done on the wall in this process is
prisoha [69]

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Explanation:

A wall that is acted upon by a force of 20N does not move suggests that no work has been done on the wall.

Work done is the force applied to move in the direction of the applied force.

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5 0
3 years ago
A cannonball is fired perfectly horizontally from the top of a 210 m tall cliff. It is fired with an initial velocity of 50 m/s.
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

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Answer:

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