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adelina 88 [10]
3 years ago
9

What is the centripetal force that would be required to keep a 4.0 kg mass moving in a horizontal circle with a radius of 0.80 m

eters at a speed of 6.0 meters/second?
A. 3.9 × 101 newtons tangent to the circle B. -3.0 × 101 newtons tangent to the circle C. 1.4 × 102 newtons radially outward D. 1.8 × 102 newtons radially inward E. 1.8 × 102 newtons radially outward
Physics
2 answers:
trasher [3.6K]3 years ago
6 0

Answer:

other guy is right

Explanation:

KIM [24]3 years ago
4 0

Answer:

D. 1.8 × 102 newtons radially inward

Explanation:

The magnitude of the centripetal force is given by:

F=m\frac{v^2}{r}

where

m is the mass of the object

v is the tangential speed

r is the radius of the circular trajector

In this problem, we have m = 4.0 kg, v = 6.0 m/s and r = 0.80 m, therefore substituting into the equation we get

F=(4.0 kg)\frac{(6.0 m/s)^2}{0.80 m}=180 N

The centripetal force is the force that keeps the object in a circular trajectory, so it is a force that is always directed inward (towards the centre of the circular path) and radially. Therefore, the correct answer is

D. 1.8 × 102 newtons radially inward

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Answer:

<em> The space in seconds that will be kept = 2.27 seconds</em>

Explanation:

S = d/t..................... Equation 1

making t the subject of formula in the equation above,

t = d/S.................... Equation 2

Where S = speed, d = distance, t = time.

<em>Conversion: (i)if 1 mph = 0.44704 m/s,</em>

<em>                 then, 30 mph = 30×0.44704    </em>

<em>                = 13.41 m/s</em>

<em>               (ii) If 1 foot = 0.3048 m</em>

<em>            then, 100 foot = 30.48 m.</em>

<em>Given: S = 30 mph = 13.41 m/s, d = 100 foot = 30.48 m</em>

<em>Substituting these values into equation 2</em>

<em>t = 30.48/13.41</em>

<em>t = 2.27 seconds.</em>

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Explanation:

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