Answer:
3.4 x 10^-4 T
Explanation:
A = 1.5 x 10^-3 m^2
N = 50
R = 180 ohm
q = 9.3 x 106-5 c
Let B be the magnetic field.
Initially the normal of coil is parallel to the magnetic field so the magnetic flux is maximum and then it is rotated by 90 degree, it means the normal of the coil makes an angle 90 degree with the magnetic field so the flux is zero .
Let e be the induced emf and i be the induced current
e = rate of change of magnetic flux
e = dФ / dt
i / R = B x A / t
i x t / ( A x R) = B
B = q / ( A x R)
B = (9.3 x 10^-5) / (1.5 x 10^-3 x 180) = 3.4 x 10^-4 T
Explanation:
Let acceleration due to Gravity for a planet is given by:
Here,
Escape velocity is given by:
Here, 
and g_X = 2g
Therefore,
Answer:
0.0953125 N
Explanation:
Applying,
F = kq'q/r²................. Equation 1
Where F = electrostatic force, k = coulomb's constant, q' and q = first and second charge respectively, r = distance between the charge.
From the equation,
If both charges remain constant,
Therefore,
F = C/r²
C = Constant = product of the two charge(q' and q) and k
Fr² = F'r'²................ Equation 2
From the question,
Given: F = 6.10 N
Assume: r = x m, r' = 8x
Substitute these value into equation 2
6.1(x²) = F'(8x)²
F' = 6.1/64
F' = 0.0953125 N
Hence the new force will become 0.0953125 N
Answer:
a) S = 1.69 10⁹ W/m², b) P = 5.63 Pa
, c) F = 20.6 10⁻¹² N
Explanation:
a) The intensity defined as the energy per unit area
S = U / A
Area of a circle is
W = 6.2 mw = 6.2 10-3 W
R = 1080 nm = 1080 10⁻⁹ m = 1.080 10⁻⁶ m
A = π R2
A = π (1,080 10⁻⁶)²
A = 3.66 10 -12 m²
S = 6.2 10-3 / 3.66 10-12
S = 1.69 10⁹ W / m²
b) The radiation pressure
P = 1 / c (dU / dt) / A
S = (dU / dt) / A
P = S / c
P = 1.69 10 9 / 3. 108
P = 5.63 Pa
c) the definition of pressure is force over area
P = F / A
F = P A
F = 5.63 3.66 10⁻¹²
F = 20.6 10⁻¹² N
d) for this we use Newton's second law
F = ma
a = F / m
<span>The common designations are radio waves, microwaves, infrared (IR), visible light, ultraviolet (UV), X-rays and gamma-rays. Visible light falls in the rangeof the EM spectrum between infrared (IR) and ultraviolet (UV).</span>
