The speed of light to be slightly less in atmosphere then in vacuum because of absorption and re-emission of light by the atmospheric molecules occurred when light travels through a material
<u>Explanation:</u>
When light passes through atmosphere, it interacts or transmits through the transparent molecules in atmosphere. In this process of transmission through atmosphere, the light will be getting absorbed by them and some will get re-emitted or refracted depending upon wavelength.
But in vacuum the absence of any kind of particles will lead to no interaction and no energy loss, thus the speed of the light will be same in vacuum while due to interactions with molecules of atmosphere, there speed will be slightly less compared to in vacuum.
Answer:
Explanation:
mass m = 3 kg
spring constant be k
k x .8 = 40 N
k = 40 / .8 = 50 N /m
angular frequency ω = √ ( k / m )
= √ ( 50 / 3 )
= 4.08 rad /s
Let amplitude of oscillation be A .
1/2 k A² = 1/2 m v²
50 A² = 3 x 1²
A = .245 m = 24.5 cm
For displacement , the equation of SHM is
x = A sinωt
= 24.5 sin4.08 t
x = 24.5 sin4.08 t
Here, angle 4.08 t is in radians .
Answer:
A and B
Explanation:
The data sets that depict an accelerating object is Data Set A & Data Set B.
The both data sets show that the body is accelerating. Also, they show that the body started from rest (0m/s) at a 0sec.
Data Set A shows a non-constant acceleration which has changing amount of velocity with change in time. While Data Set B shows a constant acceleration which has constant amount of velocity with change in time.
Answer:0.669
Explanation:
Given
mass of clock 93 kg
Initial force required to move it 610 N
After clock sets in motion it requires a force of 514 N to keep moving it with a constant velocity
Initially static friction is acting which is more than kinetic friction
thus 613 force is required to overcome static friction


Answer:

Explanation:
<u>Given Data:</u>
Mass = m = 4 kg
Acceleration due to gravity = g = 9.8 m/s²
Height = h = 1 m
<u>Required:</u>
Potential Energy = P.E. = ?
<u>Formula:</u>
P.E. = mgh
<u>Solution:</u>
P.E. = (4)(9.8)(1)
P.E. = 39.2 Joules
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>