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IRISSAK [1]
4 years ago
6

Anna Litical is practicing a centripetal force demonstration at home. She fills a bucket with water, ties it to a strong rope, a

nd spins it in a circle. Anna spins the bucket when it is half-full of water and when it is quarter-full of water. In which case is more force required to spin the bucket in a circle? Explain using an equation as a "guide to thinking.
Physics
1 answer:
dlinn [17]4 years ago
6 0

Answer:

Explanation:

Suppose when bucket is half full it has a mass of 2 m rotating in a circle of  radius r

When Bucket is quarter full then it has a mass of m rotating in a circle of radius r.

When an object moves in a circular path then it experiences an inward force which is given by

F_1=\frac{2mv^2}{r}

where v=velocity of bucket

Force in case 2 is given by

F_2=\frac{mv^2}{r}

Thus F_1=2F_2

therefore force required in half bucket is more than force required in quarter bucket full.

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5 0
3 years ago
A student is standing in an elevator that is
daser333 [38]
The force that the student exerts on the floor of the elevator must be "(1) less than the weight of the student when at <span>rest" since the elevator is "beating" him downward.</span>
4 0
4 years ago
A swimmer pushing off from the wall of a pool exerts a force of 1 newton on the wall. What is the reaction force of the wall on
8090 [49]

Answer: 1 Newton

Explanation:

"Every action has an equal and opposite reaction."

Please mark as Brainliest if it is correct.

3 0
3 years ago
Select the correct answer.
love history [14]

Answer:

A

Explanation:

Givens

d = 100 meters plus the depth of the well

d = 100 + x                          where x is the depth of the well.

vi = 0 m/s                           The object is dropped. It was not thrown.

t = 5 seconds

a = 9.81 m/s^2

formula

d = vi*t  +  1/2 a t^2

solution

100 + x = 0 + 1/2 * 9.81 * 5^2

100 + x = 122.625             Subtract 100 from both sides

x = 122.625 - 100

x = 22.6  m

The well is 22.6 meters deep.

5 0
3 years ago
A 65-kg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the
Colt1911 [192]

Answer:

(a). The average force exerted by the wall on him is 406.25 N.

(b). The work done by the wall on him is zero.

(c). The change in the kinetic energy of his center of mass is 203.1 J.

Explanation:

Given that,

Mass of ice = 65 kg

Distance = 0.50 m

Speed = 2.5 m/s

(a). We need to calculate the average force exerted by the wall on him

Using work energy theorem

W=\Delta K

Fd=\dfrac{1}{2}mv^2

F=\dfrac{\dfrac{1}{2}mv^2}{d}

F=\dfrac{mv^2}{2d}

Put the value into the formula

F=\dfrac{65\times(2.5)^2}{2\times0.50}

F=406.25\ N

(b). The wall does not move so the displacement is zero.

We need to calculate the work done by the wall on him

Using formula of work done

W=Fd

Put the value into the formula

W=406.25\times0

W=0

(c). We need to calculate the change in the kinetic energy of his center of mass

Using formula of change in kinetic energy

\Delta K=\dfrac{1}{2}mv^2

Put the value into the formula

\Delta K=\dfrac{1}{2}\times65\times(2.5)^2

\Delta K=203.1\ J

Hence, (a). The average force exerted by the wall on him is 406.25 N.

(b). The work done by the wall on him is zero.

(c). The change in the kinetic energy of his center of mass is 203.1 J.

8 0
4 years ago
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