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:
Height of cliff = S = 20 m (Approx)
Explanation:
Given:
Initial velocity = 8 m/s
Distance s = 16 m
Starting acceleration (a) = 0
Computation:
s = ut + 1/2a(t)²
16 = 8t
t = 2 sec
Height of cliff = S
Gravitational acceleration = 10 m/s
S = 1/2a(t)²
S = 1/2(10)(2)²
Height of cliff = S = 20 m (Approx)
Answer:
4 hoop, disk, sphere
Explanation:
Because
We are given data that
Hoop, disk, sphere have Same mass and radius
So let
And Initial angular velocity, = 0
The Force on each be F
And Time = t
Also let
Radius of each = r
So let's find the inertia shall we!!
I1 = m r² /2
= 0.5 mr² the his is for dis
I2 = m r² for hoop
And
Moment of inertia of sphere wiil be
I3 = (2/5) mr²
= 0.4 mr²
So
ωf = ωi + α t
= 0 + ( τ / I ) t
= ( F r / I ) t
So we can see that
ωf is inversely proportional to moment of inertia.
And so we take the
Order of I ( least to greatest ) :
I3 (sphere) , I1 (disk) , I2 (hoop) , ,
Order of ωf: ( least to greatest)
That of omega xf is the reverse of inertial so
hoop, disk, sphere
Option - 4
Actually Welcome to the concept of Efficiency.
Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%
The efficiency is => 22% => 22/100.
so we get as,
E = W(output) /W(input)
hence, W(output) = E x W(input)
so we get as,
W(output) = (22/100) x 2.2 x 10^7
=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7
hence, W(output) = 4.84 x 10^6 J
The useful work done on the mass is 4.84 x 10^6 J
