<span>As it is descended from a vertical height h,
The lost Potential Energy = Mgh
The gained Kenetic Energy = (1/2)Mv^2; The rotational KE = (1/2)Jw^2
The angular speed w = speed/ Radius = v/R
So Rotational KE = (1/2)Jw^2 = (1/2)J(v/R)^2; J is moment of inertia
Now Mgh = (1/2)Mv^2 + (1/2)J(v/R)^2 => 2gh/v^2 = 1 + (J/MR^2)
As v = (5gh/4)^1/2, (J/MR^2) = 2gh/v^2 - 1 => (J/MR^2) = (8gh/5gh) - 1
so (J/MR^2) = 3/5 and therefore J = (3/5)MR^2.</span>
<h2>
Answer: 117.626m/s</h2>
Explanation:
The escape velocity 
is given by the following equation:
(1)
Where:
is the Gravitational Constant and its value is 
is the mass of the asteroid
is the radius of the asteroid
On the other hand, we know the density of the asteroid is 
and its volume is 
.
The density of a body is given by:
(2)
Finding :
(3)
(4) This is the mass of the spherical asteroid
In addition, we know the volume of a sphere is given by the following formula:
(5)
Finding :
![R=\sqrt[3]{\frac{3V}{4\pi}}](https://tex.z-dn.net/?f=R%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3V%7D%7B4%5Cpi%7D%7D)
(6)
![R=\sqrt[3]{\frac{3(2.17(10)^{12}m^{3})}{4\pi}}](https://tex.z-dn.net/?f=R%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B3%282.17%2810%29%5E%7B12%7Dm%5E%7B3%7D%29%7D%7B4%5Cpi%7D%7D)
(7)
(8) This is the radius of the asteroid
Now we have all the necessary elements to calculate the escape velocity from (1):
(9)
Finally:
This is the minimum initial speed the rocks need to be thrown in order for them never return back to the asteroid.
The magnitude of the force<span> a 1.5 x 10-3 C charge exerts on a 3.2 x 10-4 C charge located 1.5 m away is 1920 Newtons. The formula used to solve this problem is:
F = kq1q2/r^2
where:
F = Electric force, Newtons
k = Coulomb's constant, 9x10^9 Nm^2/C^2
q1 = point charge 1, C
q2 = point charge 2, C
r = distance between charges, meters
Using direct substitution, the force F is determined to be 1920 Newtons.</span>
Answer:
= 1,386 m / s
Explanation:
Rocket propulsion is a moment process that described by the expression
- v₀ = 
ln (M₀ / Mf)
Where v are the velocities, final, initial and relative and M the masses
The data they give are the relative velocity (see = 2000 m / s) and the initial mass the mass of the loaded rocket (M₀ = 5Mf)
We consider that the rocket starts from rest (v₀ = 0)
At the time of burning half of the fuel the mass ratio is that the current mass is
M = 2.5 Mf
- 0 = 2000 ln (5Mf / 2.5 Mf) = 2000 ln 2
= 1,386 m / s
When the grasshoppers vertical velocity is exactly zero.
v = -g•t + v0.
v: vertical part of velocity. Is zero at maximum height.
g: 9.81
t: time you are looking for
v0: initial vertical velocity
Find the vertical part of the initial velocity, by using the angle at which the grasshopper jumps.
