Answer:
Yes
Explanation:
Eclipses: Eclipses are also known as game of shadows where one object comes between the star(light source) and another object in a straight line such that the shadow of one object falls on other object. This can occur when the apparent size of the star and the object is almost same.
Talking about the Earth, the geometry is such that the Moon and the Sun are of same apparent size as seen from the Earth. Thus Lunar and Solar eclipse can be seen from the Earth. If we were to go on any other planet the same phenomenon can be seen provided the apparent size of moon and the Sun from that planet is same.
We have seen and recorded many such eclipses on Jupiter. These are from the perspective of Earth. When the moons of Jupiter comes exactly between the Sun and Jupiter the shadow of moon will fall on Jupiter. The places where the shadow falls, one will see a solar eclipse.
The displacement would be the final position (37) minus the initial position (50) if using the displacement formula.
Elastic potential energy = 1/2 k * change of x^2
k- coefficient
x - change in length.
to increase energy 1.5 times you have to change x (compress) into
times (it's abot 1.22 or 22%)
I think it is D
Hope my answer help you?
Answer:
<h2>
4.25m/s</h2><h2>
E. None of the option is correct</h2>
Explanation:
Using the law of conservation of momentum to solve the problem. According to the law, the sum of momentum of the bodies before collision is equal to the sum of the bodies after collision. The bodies move with the same velocity after collision.
Mathematically.
mu + MU = (m+M)v
m and M are the masses of the bullet and the block respectively
u and U are their respective velocities
v is their common velocity
from the question, the following parameters are given;
m = 20g = 0.02kg
u = 960m/s
M = 4.5kg
U =0m/s (block is at rest)
Substituting this values into the formula above to get v;
0.02(960)+4.5(0) = (0.02+4.5)v
19.2+0 = 4.52v
4.52v = 19.2
Dividing both sides by 4.52
4.52v/4.52 = 19.2/4.52
v = 4.25m/s
Since they have the same velocity after collision, then the speed of the block immediately after the collision is also 4.25m/s
