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fenix001 [56]
3 years ago
8

A child whirls a ball in a vertical circle. Assuming the speed of the ball is constant (an approximation), when would the tensio

n in the cord connected to the ball be greatest?
a. A little after the bottom of the circle when the ball is climbing.
b. A little before the bottom of the circle when the ball is descending quickly.
c. At the bottom of the circle.
d. Nowhere; the cord is stretched the same amount at all points.
e. At the top of the circle.

Physics
1 answer:
BARSIC [14]3 years ago
8 0

Answer:

C. At the bottom of the circle.

Explanation:

Lets take

Radius of the circle = r

Mass = m

Tension = T

Angular speed = ω

The radial acceleration towards = a

a= ω² r

Weight due to gravity = mg

<h3>At the bottom condition</h3>

T - m g = m a

T =  m ω² r  + m g

<h3>At the top condition</h3>

T + m g = m a

T=  m ω² r -m g

From above equation we can say that tension is grater when ball at bottom of the vertical circle.

Therefore the answer is C.

C. At the bottom of the circle.

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7 0
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I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block
borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

T = 2\pi \sqrt{\frac{m}{k} }

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\

Now, determine the amplitude of oscillation, A;

E = \frac{1}{2} kA^2

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A =  \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm

Therefore, the amplitude of the oscillation is 2.82 cm

8 0
3 years ago
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