Answer
given,
Length of the string, L = 2 m
speed of the wave , v = 50 m/s
string is stretched between two string
For the waves the nodes must be between the strings
the wavelength is given by
where n is the number of antinodes; n = 1,2,3,...
the frequency expression is given by
now, wavelength calculation
n = 1
λ₁ = 4 m
n = 2
λ₂ = 2 m
n =3
λ₃ = 1.333 m
now, frequency calculation
n = 1
f₁ = 12.5 Hz
n = 2
f₂= 25 Hz
n = 3
f₃ = 37.5 Hz
The answer is D. Either absorbed or reflected. The reason is because if no light is being shown on the other side, the substance is not letting any light pass through. Since we do not know anything else about the substance, we do not know which one of the two it is doing. The scientist would not see any light on the detector if 100% of the light is reflected and the same thing would happen if 100% of the light was absorbed.
Answer:
on the first shell (ring) there will be 2 electrons
and on the 2nd shell there will be only one electron
while in the nucleus (the middle of the diagram) there will be 4 neutrons and 3 protons
Explanation:
you can see the picture attached
Answer:
speed of the charge electric is v = - (Eo q/m) cos t
Explanation:
The electric charge has a very small mass so it follows the oscillations of the electric field. We force ourselves on the load,
F = q Eo sint
a) To find the velocity of the particle, let's use Newton's second law to find the acceleration and of this by integration the velocity
F = ma
q Eo sint = ma
a = Eo q / m sint
a = dv / dt
dv = adt
∫ dv = ∫ a dt
v-vo = I (Eoq / m) sin t dt
v- vo = Eo q / m (-cos t)
We evaluate the integral from the initial point, as the particle starts from rest Vo = 0, for t = 0
v = - (Eo q / m) cos t
b) Kinetic energy
K = ½ m v2
K = ½ m (Eoq / m)²2 (sint)²
K = ¹/₂ Eo² q² / m sin² t
c) The average kinetic energy over a period
K = ½ m v2
<v2> = (Eoq / m) 2 <cos2 t>
The average of cos2 t = ½, substitute and calculate
K = ½ m (Eoq / m)² ½
K = ¼ Eo² q² / m
