Answer:
Explanation:
We have a metal ring of diameter
d = 4.2cm = 0.042m
r = d/2 = 0.021m
And it is place between the north pole and south pole of a large magnet with the plane of it's area perpendicular to the magnetic field.
Given that the magnetic field is
B = 1.12 T
The rate of decrease of magnetic field is 0.2T/s, since it is decrease then,
dB/dt = -0.2 T/s
The induce electric field is given as,
From faradays law
ε = ∫E•dl = -dΦ/dt
Magnetic flux is given as
Φ = BA
Φ = πr²×B = πr²B
Also, ∫E•dl = E×2πr = 2πrE
So,
∫E•dl = -dΦ/dt
2πrE = -d(πr²B) / dt
r is a constant, then
2πrE = -πr² dB/dt
Divide both side by πr
2E = -r dB/dt
E = -r dB/dt / 2
E = -0.021 × -0.2 / 2
E = 0.0021 V/m
The magnetic field point from north to south pole and it is decreasing and this means that the magnetic flux is also decreasing, so the induce magnetic field must point in the same direction of the original magnetic field, so the induce current circulate counter-clockwise as viewed from the south pole
Answer:
v= 4 m/s
Explanation:
Momenutm is, by definition, the product of mass and velocity.

Let's replace what we know and solve for whatever's left

Answer:
13 m/s
Explanation:
I assume we are ignoring friction.
The boy's PE will all be converted to KE at the bottom of the hill.
to find PE = mgh we need to know h
h = 50 sin 10 = 8.68 meters
then: PE = 20 * 9.81 * 8.68 =<u> 1703.49</u> j
KE = 1/2 m v^2 = <u>1703 .49</u>
v = 13 m/s
Answer:
λ = 102.78 nm
This radiation is in the UV range,
Explanation:
Bohr's atomic model for the hydrogen atom states that the energy is
E = - 13.606 / n²
where 13.606 eV is the ground state energy and n is an integer
an atom transition is the jump of an electron from an initial state to a final state of lesser emergy
ΔE = 13.606 (1 /
- 1 / n_{i}^{2})
the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon
DE = 13.606 (1/1 - 1/3²)
DE = 12.094 eV
let's reduce the energy to the SI system
DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J
let's find the wavelength is this energy, let's use Planck's equation to find the frequency
E = h f
f = E / h
f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴
f = 2.9186 10¹⁵ Hz
now we can look up the wavelength
c = λ f
λ = c / f
λ = 3 10⁸ / 2.9186 10¹⁵
λ = 1.0278 10⁻⁷ m
let's reduce to nm
λ = 102.78 nm
This radiation is in the UV range, which occurs for wavelengths less than 400 nm.