Answer:
Explanation:
The point at which magnetic field is to be found lies outside wire so while applying Ampere's law we shall take the whole of current . If B be magnetic field which is circular around conductor.
Applying Ampere's law :-
∫ B dl = μ₀ I ; I is current passing through ampere's loop
B x 2π x 2.00 = 4 x π x 10⁻⁷ x 2
B = 2 x 10⁻⁷ T.
Answer:
minimum mass of the neutron star = 1.624 × 10^30 kg
Explanation:
For a material to remain on the surface of a rapidly rotating neuron star, the magnitude oĺf the gravitational acceleration on the material must be equal to the magnitude of the centripetal acceleration of the rotating neuron star.
This can be represented by the explanations in the attached document.
minimum mass of the neutron star = 1.624 × 10^30 kg
Answer:
The energy in its ground state is 10 meV.
Explanation:
It is given that,
The energy of the electron in its first excited state is 40 meV.
Energy of the electron in any state is given by :

For ground state, n = 1
.............(1)
For first excited state, n = 2
.............(2)
Dividing equation (1) and (2), we get :


So, the energy in its ground state is 10 meV. Hence, this is the required solution.
Answer: vl = 2.75 m/s vt = 1.5 m/s
Explanation:
If we assume that no external forces act during the collision, total momentum must be conserved.
If both cars are identical and also the drivers have the same mass, we can write the following:
m (vi1 + vi2) = m (vf1 + vf2) (1)
The sum of the initial speeds must be equal to the sum of the final ones.
If we are told that kinetic energy must be conserved also, simplifying, we can write:
vi1² + vi2² = vf1² + vf2² (2)
The only condition that satisfies (1) and (2) simultaneously is the one in which both masses exchange speeds, so we can write:
vf1 = vi2 and vf2 = vi1
If we call v1 to the speed of the leading car, and v2 to the trailing one, we can finally put the following:
vf1 = 2.75 m/s vf2 = 1.5 m/s
Answer:

Explanation:
Velocity of the ship is given as

the direction of the velocity of the ship is making an angle of 11 degree with the current
so we will have two components of the velocity
1) along the direction of the current
2) perpendicular to the direction of the current
so here we know that the component of the ship velocity along the direction of the current is given as


