Solution :
Given :
Mass attached to the spring = 4 kg
Mass dropped = 6 kg
Force constant = 100 N/m
Initial amplitude = 2 m
Therefore,
a). 
= 10 m/s
Final velocity, v at equilibrium position, v = 5 m/s
Now, 
A' = amplitude = 1.4142 m
b). 
m' = 2m
Hence, 
c). 
Therefore, factor
Thus, the energy will change half times as the result of the collision.
40 seconds I’m pretty sure sorry if I’m wrong
The best use of an atomic model to explain the charge of the particles in Thomson's beams is:
<u>An atom's smaller negative particles are at a distance from the central positive particles, so the negative particles are easier to remove.</u>
<u>Explanation:</u>
In Thomson's model, an atom comprises of electrons that are surrounded by a group of positive particles to equal the electron's negative particles, like negatively charged “plums” that are surrounded by positively charged “pudding”.
Atoms are composed of a nucleus that consists of protons and neutrons . Electron was discovered by Sir J.J.Thomson. Atoms are neutral overall, therefore in Thomson’s ‘plum pudding model’:
-
atoms are spheres of positive charge
- electrons are dotted around inside
Thomson's conclusions made him to propose the Rutherford model of the atom where the atom had a concentrated nucleus of positive charge and also large mass.
Answer:
3.1 m/s
Explanation:
First, find the time it takes for the cat to land. Take down to be positive.
Given:
Δy = 0.61 m
v₀ = 0 m/s
a = 9.81 m/s²
Find: t
Δy = v₀ t + ½ at²
(0.61 m) = (0 m/s) t + ½ (9.81 m/s²) t²
t = 0.353 s
Now find the horizontal velocity needed to travel 1.1 m in that time.
Given:
Δx = 1.1 m
a = 0 m/s²
t = 0.353 s
Find: v₀
Δx = v₀ t + ½ at²
(1.1 m) = v₀ (0.353 s) + ½ (0 m/s²) (0.353 s)²
v₀ = 3.1 m/s
Answer:
Same magnitude of the 10 nc charge cause the electric field is external.
Explanation:
To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.
As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.
F = qE
If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.
