Answer:
v = 23.66 m/s
Explanation:
recall that one of the equations of motion may be expressed:
v² = u² + 2as,
Where
v = final velocity (we are asked to find this)
u = initial velocity = 0 m/s since we are told that it starts from rest
a = acceleration = 0.56m/s²
s = distance traveled = given as 500m
Simply substitute the known values into the equation:
v² = u² + 2as
v² = 0 + 2(0.56)(500)
v² = 560
v = √560
v = 23.66 m/s
Answer:
Li has less mass and therefore less inertia, so he can change his motion more easily than Raj.
Explanation:
Inertia describes the resistance of an object to any change in its state of motion, and it depends on the mass of the object only. In particular:
- if an object has a large inertia (large mass), then it is more difficult to change its state of motion (i.e. to put it in motion, or to slow it down, or to change its direction of motion)
- if an object has small inertia (small mass), then it is more easy to change its state of motion
In this problem, Li has less mass than Raj, so he has less inertia, therefore he can change his motion more easily than Raj.
An object in motion will stay in motion, therefore the person will still be going the same speed as the car was going before the collision
As per the question the wavelength of spectral line of the neutral hydrogen atom is 21 cm.
Spectral line of hydrogen atom will be observed when an electron will jump from higher excited to the ground state. If E is the energy of the ground state and E' is the energy of the higher excited state,the energy emitted when electron will jump from higher excited to ground state is E'-E.
As per Planck's quantum theory -

Where f is the frequency of emitted radiation and h is the Planck's constant.Actually this energy emitted is also electromagnetic in nature. We know that all the electromagnetic radiation will move with the same velocity which is equal to the velocity of light.
The velocity of light c= 3×10^8 m/s.Hence the velocity of wave corresponding to this spectral line is 3×10^8 m/s.
We know that velocity of a wave is the product of frequency and wavelength of that corresponding wave. Mathematically we can write it as-

[
is the wavelength of the wave ]
It is given that wavelength =21 cm=0.21 m
The frequency is calculated as -




Hence the antenna will be tuned to a frequency of 14.2857×10^8 Hz
Here Hertz[ Hz] is the unit of frequency.