The height of the ball above the ground is 38.45 m
First we will calculate the velocity of the ball when it touch the ground by using first equation of motion
v=u+gt
v=0+9.81×2.8
v=27.468 m/s
now the height of the ground can be calculated by the formula
v=√2gh
27.468=√2×9.81×h
h=38.45 m
Explanation:
There's a massive amount, just think of anything everyday. Like a table on the floor, or when your walking around and putting pressure on the floor. When you squeeze something which is solid. Anything like that will do.
I believe the correct answer would be kimberlite. Diamonds are usually found in pipes 50 to 200 m across made of kimberlite. It is an igneous rock that is known to contain traces of diamonds. It is named base on the town where it was discovered which is Kimberley, South Africa.
<em><u>A</u></em><em><u>. </u></em><em><u>R</u></em><em><u>E</u></em><em><u>D</u></em><em><u> </u></em><em><u>W</u></em><em><u>A</u></em><em><u>V</u></em><em><u>E</u></em><em><u>S</u></em><em><u> </u></em><em><u>I</u></em><em><u>S</u></em><em><u> </u></em><em><u>N</u></em><em><u>O</u></em><em><u>T</u></em><em><u> </u></em><em><u>A</u></em><em><u> </u></em><em><u>L</u></em><em><u>I</u></em><em><u>G</u></em><em><u>H</u></em><em><u>T</u></em><em><u> </u></em><em><u>W</u></em><em><u>A</u></em><em><u>V</u></em><em><u>E</u></em><em><u> </u></em><em><u>B</u></em><em><u>E</u></em><em><u>C</u></em><em><u>A</u></em><em><u>U</u></em><em><u>S</u></em><em><u>E</u></em><em><u> </u></em><em><u>THE</u></em>RE<em><u> </u></em><em><u>I</u></em><em><u>S</u></em><em><u> </u></em><em><u>N</u></em><em><u>O</u></em><em><u>T</u></em><em><u> </u></em><em><u>RED</u></em><em><u> </u></em><em><u>W</u></em><em><u>A</u></em><em><u>V</u></em><em><u>E</u></em><em><u>.</u></em>
<em><u>A</u></em><em><u>L</u></em><em><u>S</u></em><em><u>O</u></em><em><u> </u></em><em><u>I</u></em><em><u>F</u></em><em><u> </u></em><em><u>Y</u></em><em><u>O</u></em><em><u>U</u></em><em><u> </u></em><em><u>D</u></em><em><u>O</u></em><em><u>N</u></em><em><u>T</u></em><em><u> </u></em><em><u>B</u></em><em><u>E</u></em><em><u>L</u></em><em><u>I</u></em><em><u>E</u></em><em><u>V</u></em><em><u>E</u></em><em><u> </u></em><em><u>S</u></em><em><u>E</u></em><em><u>A</u></em><em><u>R</u></em><em><u>C</u></em><em><u>H</u></em><em><u> </u></em><em><u>I</u></em><em><u>T</u></em><em><u> </u></em><em><u>F</u></em><em><u>R</u></em><em><u>O</u></em><em><u>M</u></em><em><u> </u></em><em><u>G</u></em><em><u>O</u></em><em><u>O</u></em><em><u>G</u></em><em><u>L</u></em><em><u>E</u></em>
Answer:

Explanation:
Given data
length=100mm
Diameter=5mm
Thermal conductivity=5 W/m.K
Power=50 W
Temperature=25°C
The temperature of heater surface follows from the rate equation written as:

Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium

Substitute the given values
![S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m](https://tex.z-dn.net/?f=S%3D%5Cfrac%7B2%5Cpi%20%280.1m%29%7D%7Bln%5B%5Cfrac%7B4%2A0.1m%7D%7B0.005m%7D%20%5D%7D%5C%5C%20S%3D0.143m)
The temperature of heater is then:

The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.
