Answer:
Explanation:
Given that,
Radius of a spherical shell, r = 0.7 m
Torque acting on the shell, 
Angular acceleration of the shell, 
We need to find the rotational inertia of the shell about the axis of rotation. The relation between the torque and the angular acceleration is given by :
I is the rotational inertia of the shell
So, the rotational inertia of the shell is 
.
Density<span> is the </span>mass<span> of an object </span>divided<span> by its </span>volume<span>. So the answer would be Yes. Hope it helps! (:</span>
Answer:
k = 17043.5 N/m = 17.04 KN/m
Explanation:
First we need to find the force applied by safe pn the spring:
F = Weight of Safe
F = mg
where,
F = Force Applied by the safe on the spring = ?
m = mass of safe = 800 kg
g = 9.8 m/s²
Therefore,
F = (800 kg)(9.8 m/s²)
F = 7840 N
Now, using Hooke's Law:
F = kΔx
where,
K = Spring Constant = ?
Δx = compression = 46 cm = 0.46 m
Therefore,
7840 N = k (0.46 m)
k = 7840 N/0.46 m
<u>k = 17043.5 N/m = 17.04 KN/m</u>
Answer:
See below
Explanation:
Vertical position is given by
df = do + vo t - 1/2 a t^2 df = final position = 0 (on the ground)
do =original position = 2 m
vo = original <u>VERTICAL</u> velocity = 0
a = acceleration of gravity = 9.81 m/s^2
THIS BECOMES
0 = 2 + 0 * t - 1/2 ( 9.81)t^2
to show t =<u> .639 seconds to hit the ground </u>
During this .639 seconds it flies horizontally at 10 m/s for a distance of
10 m/s * .639 s =<u> 6.39 m </u>
According to the statement " Collision <span>between two bodies in which the total kinetic energy of the two bodies after the collision is equal to their total kinetic energy before the collision."
The best answer is :
Option A " </span><span>BODY A COMES TO REST BODY B STARTS MOVING WITH INITIAL VELOCITY OF BODY A "</span>
