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melisa1 [442]
3 years ago
7

A 15 g ball at the end of 2.4 m string is swung in the horizontal circle. It revolves once every 2.09 s. What is the magnitude o

f the
string's tension ?​
Physics
1 answer:
Mrrafil [7]3 years ago
3 0

Answer:

0.325 N

Explanation:

From the question,

T = 4π²rm/t²............................ Equation 1

Where T = Tension, r = radius or length of the string, m = mass of the string, t = time.

Given: r = 2.4 m, m = 15 g = 0.015 kg, t = 2.09 s.

Constant: π = 3.14

Substitute these values into equation 1

T = 4(3.14²)(2.4)(0.015)/2.09²

T = 1.4198/4.3681

T = 0.325 N

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Answer:

37.125 m

Explanation:

Using the equation of motion

s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration

<u>Distance during acceleration</u>

Acceleration, a=\frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.

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s=0*4.5+ 0.5*1*4.5^{2}=0+10.125 =10.125 m

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At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time

Distance=4.5 m/s*6 s=27 m

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