Answer:
remains the same, but the apparent brightness is decreased by a factor of four.
Explanation:
A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.
It is typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).
The luminosity of a star refers to the total amount of light radiated by the star per second and it is measured in watts (w).
The apparent brightness of a star is a measure of the rate at which radiated energy from a star reaches an observer on Earth per square meter per second.
The apparent brightness of a star is measured in watts per square meter.
If the distance between us (humans) and a star is doubled, with everything else remaining the same, the luminosity remains the same, but the apparent brightness is decreased by a factor of four (4).
Some of the examples of stars are;
- Canopus.
- Sun (closest to the Earth)
- Betelgeuse.
- Antares.
- Vega.
Answer:
1410 Hz
Explanation:
Capacitance is reduced by 2, so the angular frequency will increase by a factor of
.
1) 
When both the electric field and the magnetic field are acting on the electron normal to the beam and normal to each other, the electric force and the magnetic force on the electron have opposite directions: in order to produce no deflection on the electron beam, the two forces must be equal in magnitude

where
q is the electron charge
E is the magnitude of the electric field
v is the electron speed
B is the magnitude of the magnetic field
Solving the formula for v, we find

2) 4.1 mm
When the electric field is removed, only the magnetic force acts on the electron, providing the centripetal force that keeps the electron in a circular path:

where m is the mass of the electron and r is the radius of the trajectory. Solving the formula for r, we find

3) 
The speed of the electron in the circular trajectory is equal to the ratio between the circumference of the orbit,
, and the period, T:

Solving the equation for T and using the results found in 1) and 2), we find the period of the orbit:

Answer:
v₂ = v/1.5= 0.667 v
Explanation:
For this exercise we will use the conservation of the moment, for this we will define a system formed by the two students and the cars, for this isolated system the forces during the contact are internal, therefore the moment conserves.
Initial moment before pushing
p₀ = 0
Final moment after they have been pushed
= m₁ v₁ + m₂ v₂
p₀ = 
0 = m₁ v₁ + m₂ v₂
m₁ v₁ = - m₂ v₂
Let's replace
M (-v) = -1.5M v₂
v₂ = v / 1.5
v₂ = 0.667 v
9.8 ms^-2 is acceleration