Explanation:
thet amplify DC, because of the voltage ( small current input signal)
F = qE + qV × B
where force F, electric field E, velocity V, and magnetic field B are vectors and the × operator is the vector cross product. If the electron remains undeflected, then F = 0 and E = -V × B
which means that |V| = |E| / |B| and the vectors must have the proper geometrical relationship. I therefore get
|V| = 8.8e3 / 3.7e-3
= 2.4e6 m/sec
Acceleration a = V²/r, where r is the radius of curvature.
a = F/m, where m is the mass of an electron,
so qVB/m = V²/r.
Solving for r yields
r = mV/qB
= 9.11e-31 kg * 2.37e6 m/sec / (1.60e-19 coul * 3.7e-3 T)
= 3.65e-3 m
If it's a mechanical wave, then its speed depends on the physical characteristics of the medium.
If it's an electromagnetic wave, then its speed depends on the
electrical characteristics of the medium.
Either way, the properties of the medium determine the wave speed.
You want to change the speed ? You have to change the properties
of the medium.
The Moon is 3.8 108 m from Earth and has a mass of 7.34 1022 kg. 5.97 1024 kg is the mass of the Earth.
<h3>What kind of gravitational pull does the moon have on the planet?</h3>
On the surface of the Moon, the acceleration caused by gravity around 1.625 m/s2 which is 16.6% greater than on the surface of the Earth 0.166.
<h3>What does the Earth's center's gravitational pull feel like?</h3>
Gravity is zero if you are in the centre of the earth since everything around you is pulling "up" (up is the only direction).
<h3>Where is the Earth's and the moon's gravitational centre?</h3>
It is around 1700 kilometres below Earth's surface.
To know more about gravitational force visit:-
brainly.com/question/12528243
#SPJ4
Answer:
angular range is ( 0.681 rad , 0.35 rad )
Explanation:
given data
wavelength λ = 380 nm = 380 × 
m
wavelength λ = 700 nm = 700 × m
to find out
angular range of the first-order
solution
we will apply here slit experiment equation that is
d sinθ = m λ ...........1
here m is 1 for single slit and d is = 
so put here value in equation 1 for 380 nm
we get
d sinθ = m λ
sinθ = 1 × 380 ×
θ = 0.35 rad
and for 700 nm
we get
d sinθ = m λ
sinθ = 1 × 700 ×
θ = 0.681 rad
so angular range is ( 0.681 rad , 0.35 rad )
