Answer:
why should we do , do by your own , no sense
Explanation:
Answer:
The frequency of the photon decreases upon scattering
Explanation:
Here we note that when a photon is scattered by a charged particle, it is referred to as Compton scattering.
Compton scattering results in a reduction of the energy of the photon and hence an increase in the wavelength (from λ to λ') of the photon known as Compton effect.
Therefore, since the wavelength increases, we have from
λf = λ'f' = c
f = c/λ
Where:
f and f' = The frequency of the motion of the photon before and after the scattering
c = Speed of light (constant)
We have that the frequency, f, is inversely proportional to the wavelength, λ as follows;
f = c/λ
As λ = increases, and c is constant, f decreases, therefore, the frequency of the photon decreases upon scattering.
A bathroom scales works due to gravity. Under normal
conditions, a reading can be obtained when your body is pushing some force on
the scale. However in this case, since you and the scale are both moving
downwards, so your body is no longer pushing on the scale. Therefore the answer
is:
<span>The reading will drop to 0 instantly</span>
The motion of the buoy consists of two independent motions on the horizontal and vertical axis.
On the horizontal axis, the motion of the buoy is a uniform motion with constant speed
. On the vertical axis, the motion of the buoy is a uniformly accelerated motion with constant acceleration
. The vertical position of the buoy at time t is given by
where h is the initial heigth of the buoy when it is released from the plane. At the time t=21 s, the buoy reaches the ground, so y(21 s)=0. If we substitute these two numbers inside the equation, we can find the value of h, the vertical displacement from the plane to the ocean:
Answer:
C
Explanation:
To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):
d = d₀ + d₀αT
for the sphere, we were given
D₀ = 4.000 cm
α = 1.1 x 10⁻⁵/degrees celsius
we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T EQN 1
Similarly for the Aluminium ring we have
we were given
d₀ = 3.994 cm
α = 2.4 x 10⁻⁵/degrees celsius
we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T EQN 2
Since @ the temperature T at which the sphere fall through the ring, d=D
Eqn 1 = Eqn 2
4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms
0.006=5.18x10⁻⁵T
T=115.7K
