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Olegator [25]
4 years ago
7

If you swing a bucket of water fast enough in a vertical circle, at the highest point the water does not spill out. This happens

because an outward force balances the pull of gravity on the water.
True or False?
Physics
1 answer:
amm18124 years ago
3 0

Answer:

True

Explanation:

If we swing a bucket of water fast enough in a vertical circle the water does not spill out even at the top-most position of the bucket. This happens because the centrifugal force acting away from the center in a circular motion neutralizes or overcomes the gravitational force on the water particles.

<u>Centrifugal force is mathematically related as:</u>

F=m.r.\omega^2

where:

m = mass of the revolving body

r = radius of revolution

\omega= angular velocity in radians per second

This force F acts in radially outward direction.

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Correct the sentence below to reflect an “I” statement. 
Anarel [89]
The word “you” shouldn’t be used, and I learned the I statement to have your feelings stated, then when, then because, and finally a want. So, to correct this, an appropriate statement would be “ I feel misunderstood when I am treated like a child because my curfew is set for 9:00 pm on the weekends and I want to be seen as your equal and as though you are not superior to me. You can use the word “you” after stating you want something out of the other person.
5 0
3 years ago
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How does water flowing over a waterfall involve both kinetic energy and potential energy?
Vinvika [58]

Answer:

While the water falls v increases and h decreases, so the kinetic energy increases and the gravitational potential energy decreases, and this happens in a way that the total energy is always the same. (If there is no friction)

Explanation:

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3 years ago
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A spaceship travels at approximately 0. 25c with respect to earth. this means 24 hours at the spaceship will be x hours to us on
Blababa [14]

The time passed on earth is  mathematically given as

t' = 24.79 hrs

<h3>What is the time passed on earth?</h3>

Generally, the equation for is time  mathematically given as

t' = \gamma t

Where

\gamma = Lorentz\ factor \\\\\gamma = 1/ \sqrt {(1 - v^2/c^2)}

t' = t/ \sqrt {(1 - v^2/c^2)}

Therefore

t' = 24/ \sqrt {(1 - (0.25c/c)^2) }

t'= 24/ \sqrt {(1 - 0.25^2)

t' = 24.79 hrs

In conclusion, the time passed on earth

t' = 24.79 hrs

Read more about the time

brainly.com/question/28050940

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4 0
2 years ago
How much would you weigh on an imaginary planet that has no gravitational force?<br><br>​
GarryVolchara [31]

0kg

If the gravitational pull is zero and I multiply by mass I get a zero

5 0
3 years ago
An proton-antiproton pair is produced by a 2.20 × 10 3 MeV photon. What is the kinetic energy of the antiproton if the kinetic e
timama [110]

Answer:

K = 80.75 MeV    

Explanation:

To calculate the kinetic energy of the antiproton we need to use conservation of energy:

E_{ph} = E_{p} + E_{ap} = E_{0p} + K_{p} + E_{0ap} + K_{ap} = m_{0p}c^{2} + K_{p} + m_{0ap}c^{2} + K_{ap}

<em>where E_{ph}: is the photon energy, E_{0p} and E_{0ap}: are the rest energies of the proton and the antiproton, respectively, equals to m₀c², K_{p} and K_{ap}: are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.</em>        

Therefore the kinetic energy of the antiproton is:    

K_{ap} = E_{ph} - m_{0p}c^{2} - K_{p} - m_{0ap}c^{2}

<u>The proton mass is equal to the antiproton mass, so</u>:

K_{ap} = E_{ph} - 2m_{0p}c^{2} - K_{p}  

K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2} - 242.85MeV

K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2}(\frac{1eV}{1.602 \cdot 10^{-19}J})(\frac{1 MeV}{10^{6}eV}) - 242.85MeV

K_{ap} = 80.75 MeV              

Hence, the kinetic energy of the antiproton is 80.75 MeV.

I hope it helps you!

3 0
3 years ago
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