Answer:
W=76.55 miles.metric tons
Explanation:
Given that
Weight on the earth = 12 tons
So weight on the moon =12/6 = 2 tons
( because at moon g will become g/6)
As we know that
![F=\dfrac{K}{x^2}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7BK%7D%7Bx%5E2%7D)
Here x= 1100 miles
F 2 tons
![2=\dfrac{K}{1100^2}](https://tex.z-dn.net/?f=2%3D%5Cdfrac%7BK%7D%7B1100%5E2%7D)
So
![K=2.4\times 10^6](https://tex.z-dn.net/?f=K%3D2.4%5Ctimes%2010%5E6)
We know that
Work = F. dx
![W=\int_{x_1}^{x_2}F.dx](https://tex.z-dn.net/?f=W%3D%5Cint_%7Bx_1%7D%5E%7Bx_2%7DF.dx)
![W=\int_{1100}^{1140}\dfrac{2.4\times 10^6}{x^2}.dx](https://tex.z-dn.net/?f=W%3D%5Cint_%7B1100%7D%5E%7B1140%7D%5Cdfrac%7B2.4%5Ctimes%2010%5E6%7D%7Bx%5E2%7D.dx)
![W=-2.4\times 10^6\left[\dfrac{1}{x}\right]_{1100}^{1140}](https://tex.z-dn.net/?f=W%3D-2.4%5Ctimes%2010%5E6%5Cleft%5B%5Cdfrac%7B1%7D%7Bx%7D%5Cright%5D_%7B1100%7D%5E%7B1140%7D)
![W=-2.4\times 10^6\left[\dfrac{1}{1140}-\dfrac{1}{1100}\right]](https://tex.z-dn.net/?f=W%3D-2.4%5Ctimes%2010%5E6%5Cleft%5B%5Cdfrac%7B1%7D%7B1140%7D-%5Cdfrac%7B1%7D%7B1100%7D%5Cright%5D)
W=76.55 miles.metric tons
Answer:
The pitching speed of the ball is 19.7 m/s
Explanation:
- Here, we can use the third equation of motion,
![v^{2} = u^{2} - 2as](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%20u%5E%7B2%7D%20-%202as)
- whereas v represents the final velocity, u represents initial velocity, a is the acceleration due to gravity and s is the displacement or distance an object traveled
- Here, the initial velocity of the the ball is given as zero and the acceleration due to gravity is 9.8 , the distance 's' is given as 20 m
- Using the equation,
![v^{2} = 2 * 9.8 * 20 = 392\\v = \sqrt{392} = 19.7m/s](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%202%20%2A%209.8%20%2A%2020%20%3D%20392%5C%5Cv%20%3D%20%5Csqrt%7B392%7D%20%3D%2019.7m%2Fs)
- Hence, the pitching speed of the ball is 19.7 m/s
Explanation:
Rhythm, Harmony, Timbre, Dynamics, Texture, and Form
Answer:
h' = 603.08 m
Explanation:
First, we will calculate the initial velocity of the pellet on the surface of Earth by using third equation of motion:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity on the surface of earth = - 9.8 m/s² (negative sign due to upward motion)
h = height of pellet = 100 m
Vf = final velocity of pellet = 0 m/s (since, pellet will momentarily stop at highest point)
Vi = Initial Velocity of Pellet = ?
Therefore,
(2)(-9.8 m/s²)(100 m) = (0 m/s)² - Vi²
Vi = √(1960 m²/s²)
Vi = 44.27 m/s
Now, we use this equation at the surface of moon with same initial velocity:
2g'h' = Vf² - Vi²
where,
g' = acceleration due to gravity on the surface of moon = 1.625 m/s²
h' = maximum height gained by pellet on moon = ?
Therefore,
2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²
h' = (1960 m²/s²)/(3.25 m/s²)
<u>h' = 603.08 m</u>
Answer:
It cannot be constant because if it does not change and each time it increases its strength and speed.
Explanation: