I think its c or d but im not quite sure.....
Answer:
51.85m/s
Explanation:
Given parameters:
Mass of ball = 0.0459kg
Force = 2380N
Time taken = 0.001s
Unknown:
Speed of the ball afterwards = ?
Solution:
To solve this problem, we use Newton's second law of motion:
F = m x
F is the force
m is the mass
v is the final velocity
u is the initial velocity
t is the time taken
2380 = 0.0459 x
0.0459v = 2.38
v = 51.85m/s
Answer:
The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.
Explanation:
Given that,
The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.
A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.
We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.
Using Snell's law for red light as :

Again using Snell's law for blue light as :

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.
Answer:
hmax = 1/2 · v²/g
Explanation:
Hi there!
Due to the conservation of energy and since there is no dissipative force (like friction) all the kinetic energy (KE) of the ball has to be converted into gravitational potential energy (PE) when the ball comes to stop.
KE = PE
Where KE is the initial kinetic energy and PE is the final potential energy.
The kinetic energy of the ball is calculated as follows:
KE = 1/2 · m · v²
Where:
m = mass of the ball
v = velocity.
The potential energy is calculated as follows:
PE = m · g · h
Where:
m = mass of the ball.
g = acceleration due to gravity (known value: 9.81 m/s²).
h = height.
At the maximum height, the potential energy is equal to the initial kinetic energy because the energy is conserved, i.e, all the kinetic energy was converted into potential energy (there was no energy dissipation as heat because there was no friction). Then:
PE = KE
m · g · hmax = 1/2 · m · v²
Solving for hmax:
hmax = 1/2 · v² / g
Answer:
Different and better?
Explanation:
i dont think that helps lol