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Anna71 [15]
2 years ago
15

swings a 5.5 kg cup of water in a vertical circle of radius 1.9 m. (a) What minimum speed must the cup have in this demo if the

front row is not to get wet or spill any water
Physics
1 answer:
Tanzania [10]2 years ago
6 0

Answer:

4.32

Explanation:

The centripetal acceleration of any object is given as

A(cr) = v²/r, where

A(c) = the centripetal acceleration

v = the linear acceleration

r = the given radius, 1.9 m

Since we are not given directly the centripetal acceleration, we'd be using the value of acceleration due to gravity, 9.8. This means that

9.8 = v²/1.9

v² = 1.9 * 9.8

v² = 18.62

v = √18.62

v = 4.32 m/s

This means that, the minimum speed the cup must have so as not to get wet or any spill is 4.32 m/s

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Answer:

a=2.9\ m/sec^2

Explanation:

<u>Net Forces and Acceleration</u>

The second Newton's Law relates the net force F_r acting on an object of mass m with the acceleration a it gets. Both the net force and the acceleration are vector and have the same direction because they are proportional to each other.

\vec F_r=m\vec a

According to the information given in the question, two forces are acting on the swimming student: One of 256 N pointing to the south and other to the west of 104 N. Since those forces are not aligned, we must add them like vectors as shown in the figure below.

The magnitude of the resulting force F_r is computed as the hypotenuse of a right triangle

|F_r|=\sqrt{256^2+104^2}

|F_r|=276.32\ Nw

The acceleration can be obtained from the formula

F_r=ma

Note we are using only magnitudes here

\displaystyle a=\frac{F_r}{m}

\displaystyle a=\frac{276.32Nw}{95.3Kg}

\boxed{a=2.9\ m/sec^2}

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D=vt

52 = V 21

V= 52/21=2.47m/s

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