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Vitek1552 [10]
3 years ago
9

A 500 gram mass is connected to a spring and undergoing uniform circular motion. The radius is at 14.5 cm and the applied centri

petal force is 5.88N. The velocity of the spinning mass is measured to be 1.31 m/s. Keeping the force and mass constant the radius of the bob was increased by 1 cm. What should be the expected velocity of the spinning mass at the new radius?
Physics
1 answer:
swat323 years ago
6 0

To solve the problem it is necessary to apply the concepts related to the Centripetal Force.

By definition the centripetal force is given by

F_c = \frac{mv^2}{R}

Our values are defined by

m=0.5Kg\\F_c = 5.88N \\R = 0.145m+1cm = 0.155m\\v = 1.31m/s

Therefore replacing in the equation we have to,

F_c = \frac{mv^2}{R}

5.88=\frac{0.5*v^2}{0.155}

Re-arrange to find V,

V=\sqrt{\frac{5.88*0.155}{0.5}}

V= 1.35m/s

Therefore the expected velocity of the spinning mass at the new radius is 1.35m/s

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A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it
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Answer:

t = 1.82

Explanation:

Given

u = 7.70m/s -- initial velocity

s = 30.2m --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

s = ut + \frac{1}{2}gt^2

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Subtract 30.2 from both sides

30.2 -30.2  = 7.70t + 4.9 * t^2 - 30.2

0  = 7.70t + 4.9 * t^2 - 30.2

0  = 7.70t + 4.9t^2 - 30.2

7.70t + 4.9t^2 - 30.2  = 0

4.9t^2 + 7.70t - 30.2  = 0

Solve using quadratic formula:

t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}

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a = 4.9;\ b = 7.70;\ c = -30.2

t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}

t = \frac{-7.70\±\sqrt{651.21}}{9.8}

t = \frac{-7.70\±25.52}{9.8}

Split the expression

t = \frac{-7.70+25.52}{9.8} or t = \frac{-7.70-25.52}{9.8}

t = \frac{17.82}{9.8} or t = -\frac{33.22}{9.8}

Time can't be negative;  So, we have:

t = \frac{17.82}{9.8}

t = 1.82

Hence, the time to hit the ground is 1.82 seconds

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