Answer:
True.
Explanation:
Hypotheses are supposed to be tested, to be proven correct. They usually start from a small thought without 100% correct, but as you do further testing, they become stronger.
Answer:
Elevation =31.85[m]
Explanation:
We can solve this problem by using the principle of energy conservation. This consists of transforming kinetic energy into potential energy or vice versa. For this specific case is the transformation of kinetic energy to potential energy.
We need to first identify all the input data, and establish a condition or a point where the potential energy is zero.
The point where the ball is thrown shall be taken as a reference point of potential energy.
![E_{p} = E_{k} \\where:\\E_{p}= potential energy [J]\\ E_{k}= kinetic energy [J]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3D%20E_%7Bk%7D%20%5C%5Cwhere%3A%5C%5CE_%7Bp%7D%3D%20potential%20energy%20%5BJ%5D%5C%5C%20E_%7Bk%7D%3D%20kinetic%20energy%20%5BJ%5D)
m = mass of the ball = 300 [gr] = 0.3 [kg]
v = initial velocity = 25 [m/s]
![E_{k}=\frac{1}{2} * m* v^{2} \\E_{k}= \frac{1}{2} * 0.3* (25)^{2} \\E_{k}= 93.75 [J]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20m%2A%20v%5E%7B2%7D%20%5C%5CE_%7Bk%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A%200.3%2A%20%2825%29%5E%7B2%7D%20%5C%5CE_%7Bk%7D%3D%2093.75%20%5BJ%5D)
![93.75=m*g*h\\where:\\g = gravity = 9.81 [m/s^2]\\h = elevation [m]\\replacing\\h=\frac{E_{k}}{m*g} \\h=\frac{93.75}{.3*9.81} \\h=31.85[m]](https://tex.z-dn.net/?f=93.75%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cg%20%3D%20gravity%20%3D%209.81%20%5Bm%2Fs%5E2%5D%5C%5Ch%20%3D%20elevation%20%5Bm%5D%5C%5Creplacing%5C%5Ch%3D%5Cfrac%7BE_%7Bk%7D%7D%7Bm%2Ag%7D%20%5C%5Ch%3D%5Cfrac%7B93.75%7D%7B.3%2A9.81%7D%20%5C%5Ch%3D31.85%5Bm%5D)
Answer:
15√2 N
Explanation:
The acceleration is given by ...
a = F/m = 5t/5 = t . . . . meters/second^2
The velocity is the integral of acceleration:
v = ∫a·dt = (1/2)t^2
This will be 9 m/s when ...
9 = (1/2)t^2
t = √18 . . . . seconds
And the force at that time is ...
F = 5(√18) = 15√2 . . . . newtons
Answer:

Explanation:
Given data:
mass of block is 
radius of block = 0.061 m
moment of inertia is 
D is distance covered by block = 0.65 m
speed of block is 1.705 m/s
From conservation of momentum we have

![0.84 \times 9.81 \times 0.65 = \frac{1}{2}\times 0.84 \times 1.705^2 +\frac{1}{2} \times 6.2 \times 10^{-3} [\frac{1.705}{0.061}]^2 + E_l](https://tex.z-dn.net/?f=0.84%20%5Ctimes%209.81%20%5Ctimes%200.65%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%200.84%20%5Ctimes%201.705%5E2%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%206.2%20%5Ctimes%2010%5E%7B-3%7D%20%5B%5Cfrac%7B1.705%7D%7B0.061%7D%5D%5E2%20%2B%20E_l)
solving for energy loss

Answer:
D. s=3x+4y
Explanation:
The line is at the 3rd column in the 4th row.