Answer: hello options related to your question is missing attached below is the missing part of your question
answer: No charge of the length of the bonds expected because the rod did not touch the charge source ( option A )
Explanation:
When the Charge is first, Furthest away and second and closest to the source charge. <em>The spring like bonds can be said to have No charge of the length of the bonds expected because the rod did not touch the charge source </em><em>when Furthest away the bond with charge will be less effective </em>
Answer:
Transverse
Explanation:
There are two types of waves, according to the direction of their oscillation:
- Transverse waves: in a transverse wave, the direction of the oscillation is perpendicular to the direction of motion of the wave. Examples of transverse waves are electromagnetic waves
- Longitudinal waves: in a longitudinal wave, the direction of the oscillation is parallel to the direction of motion of the wave. Examples of longitudinal waves are sound waves.
Light waves corresponds to the visible part of the electromagnetic spectrum, which includes all the different types of electromagnetic waves (which consist of oscillations of electric and magnetic fields that are perpendicular to the direction of propagation of the wave): therefore, they are transverse waves.
Answer:
true
Explanation:
this the nucleus is located at the centre and contains protons and neutrons
Answer:
a) v = 2,9992 10⁸ m / s
, b) Eo = 375 V / m
, B = 1.25 10⁻⁶ T,
c) λ = 3,157 10⁻⁷ m, f = 9.50 10¹⁴ Hz
, T = 1.05 10⁻¹⁵ s
, UV
Explanation:
In this problem they give us the equation of the traveling wave
E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]
a) what the wave velocity
all waves must meet
v = λ f
In this case, because of an electromagnetic wave, the speed must be the speed of light.
k = 2π / λ
λ = 2π / k
λ = 2π / 1.99 10⁷
λ = 3,157 10⁻⁷ m
w = 2π f
f = w / 2 π
f = 5.97 10¹⁵ / 2π
f = 9.50 10¹⁴ Hz
the wave speed is
v = 3,157 10⁻⁷ 9.50 10¹⁴
v = 2,9992 10⁸ m / s
b) The electric field is
Eo = 375 V / m
to find the magnetic field we use
E / B = c
B = E / c
B = 375 / 2,9992 10⁸
B = 1.25 10⁻⁶ T
c) The period is
T = 1 / f
T = 1 / 9.50 10¹⁴
T = 1.05 10⁻¹⁵ s
the wavelength value is
λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm
this wavelength corresponds to the ultraviolet
Answer:
FB = 0.187 N
Explanation:
To find the magnetic force FB in the wire you use the following formula:
the angle between B and L is given by:
Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:
![|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N](https://tex.z-dn.net/?f=%7C%5Cvec%7BF_B%7D%7C%3DIsin%5Ctheta%5Cint_0%5E%7B0.25%7DB%28y%29dy%3DIsin%5Ctheta%5Cint_0%5E%7B0.25%7D%280.5y%29dy%5C%5C%5C%5C%7C%5Cvec%7BF_B%7D%7C%3D%282.0%2A10%5E%7B-3%7DA%29%28sin36.86%5C%C2%B0%29%280.5T%29%5B%5Cfrac%7B0.25%5E2%7D%7B2%7Dm%5D%3D0.187%5C%20N)
hence, the magnitude of the magnetic force is 0.187N
