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nordsb [41]
3 years ago
8

A ball is thrown vertically upward with an initial velocity of 29.4 m/s². What is the maximum height reached by the ball? How lo

ng is the ball in the air? What is the final velocity of the ball before it goes back to its original position?​
Physics
1 answer:
Elena L [17]3 years ago
4 0

Answer:

Most likely it will reach 29.4 up in the air before coming back down.

Explanation:

The force of the ball will be the same on how high it will go.

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Answer:

The human activity that uses the most water worldwide is for agriculture. In other words, watering crops.

Explanation:

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2 years ago
2 H2O2-> 2H20+02
Luden [163]

Answer:

This reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen. Add 2 molecules of hydrogen peroxide and 2 molecules of water. Since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

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2 years ago
What is the relationship between area, force, and pressure? what is the relationship between area, force, and pressure? pressure
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7 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

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3 years ago
The diagram pictured shows a loop of wire being rotated at a constant rate about an axis in a uniform magnetic field.
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