Answer:
The velocity of the ball before it hits the ground is 381.2 m/s
Explanation:
Given;
time taken to reach the ground, t = 38.9 s
The height of fall is given by;
h = ¹/₂gt²
h = ¹/₂(9.8)(38.9)²
h = 7414.73 m
The velocity of the ball before it hits the ground is given as;
v² = u² + 2gh
where;
u is the initial velocity of the on the root = 0
v is the final velocity of the ball before it hits the ground
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 7414.73 )
v = 381.2 m/s
Therefore, the velocity of the ball before it hits the ground is 381.2 m/s
The Beams And Joints That Hold It .
Answer:
(a) 
(b) 
(c) 
(d) 
Solution:
As per the question:
Refractive index of medium 1, 
Angle of refraction for medium 1, 
Angle of refraction for medium 2, 
Now,
(a) The expression for the refractive index of medium 2 is given by using Snell's law:
where
= Refractive Index of medium 2
Now,
(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:
(c) To calculate the velocity of light in medium 1:
We know that:
Thus for medium 1
(d) To calculate the velocity of light in medium 2:
For medium 2:
1. electrons
2. positive to negative
3. insulator
4. TRUE
5. closed circuit
6. TRUE
7. series
8. TRUE
9. v=ir
10. TRUE
Hope this helps! :)
Answer:
437.5Kjoules
Explanation:
K.E=half multiply by mass multiply by square of velocity
=437.5Kjoules
