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Crazy boy [7]
3 years ago
12

An object of mass 8kg moved around the circle of radius 4m. with a constant speed of 15m/s.

Physics
1 answer:
Marianna [84]3 years ago
4 0

Answer:

a. Angular velocity = 0.267rad/s.

b. Centripetal acceleration = 56.25m/s.

Explanation:

<u>Given the following data;</u>

Mass, m = 8kg

Radius, r = 4m

Constant speed, V = 15m/s

a. To find the angular velocity

Angular velocity = radius/speed

Substituting into the equation, we have;

Angular velocity = 4/15

Angular velocity = 0.267rad/s

b. To find the acceleration;

Centripetal acceleration = V²/r

Substituting into the equation, we have;

Centripetal acceleration = 15²/4

Centripetal acceleration = 225/4

Centripetal acceleration = 56.25m/s.

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The bullet travels a horizontal distance of 276.5 m

The bullet is shot forward with a horizontal velocity u_x. It takes a time <em>t</em> to fall a vertical distance <em>y</em> and at the same time travels a horizontal distance <em>x. </em>

The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.

The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.

Calculate the time taken for the bullet to fall through a vertical distance <em>y </em>using the equation,

y=u_yt+\frac{1}{2} gt^2

Substitute 0 m/s for u_y, 9.81 m/s²for <em>g</em> and 1.5 m for <em>y</em>.

y=u_yt+\frac{1}{2} gt^2\\ 1.5 m=(0m/s)t+\frac{1}{2} (9.81m/s^2)t^2\\ t=\sqrt{\frac{2(1.5m)}{9.81m/s^2} } =0.5530s

The horizontal distance traveled by the bullet is given by,

x=u_xt

Substitute 500 m/s for u_x and 0.5530s for t.

x=u_xt\\ =(500m/s)(0.5530s)\\ =276.5m

The bullet travels a distance of 276.5 m.


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