(A) We can solve the problem by using Ohm's law, which states:
where
V is the potential difference across the electrical device
I is the current through the device
R is its resistance
For the heater coil in the problem, we know
and
, therefore we can rearrange Ohm's law to find the current through the device:
(B) The resistance of a conductive wire depends on three factors. In fact, it is given by:
where
is the resistivity of the material of the wire
L is the length of the wire
A is the cross-sectional area of the wire
Basically, we see that the longer the wire, the larger its resistance; and the larger the section of the wire, the smaller its resistance.
Answer:
N= 3
Explanation:
For this exercise we must use Faraday's law
E = - dФ / dt
Ф = B . A = B Acos θ
tje bold indicate vectors. As it indicates that the variation of the field is linear, we can approximate the derivatives
E = - A cos θ (B - B₀) / t
The angle enters the magnetic field and the normal to the area is zero
cos 0 = 1
A = π r²
In the length of the wire there are N turns each with a length L₀ = 2π r
L = N (2π r)
r = L / 2π N
we substitute
A = L² / (4π N²)
The magnetic field produced by a solenoid is
B = μ₀ N/L I
for which
B₀ = μ₀ N/L I
The final field is zero, because the current is zero
B = 0
We substitute
E = - (L² / 4π N²) (0 - μ₀ N/L I) / t
E = μ₀ L I / (4π N t)
N = μ₀ L I / (4π t E)
The electromotive force is E = 0.80 mV = 0.8 10⁻³ V
let's calculate
N = 4π 10⁻⁷ 200 1.60 / (4π 0.120 0.8 10⁻³)]
N = 320 10⁻⁷ / 9.6 10⁻⁶
N = 33.3 10⁻¹
N= 3
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
Explanation:
If a positive test charge is placed in an electric field, it will exert the force in the test charge in the direction of electric field vector. We know that the direction of electric field is given by electric field lines. The field lines for a positive charge is outwards. The electric force acting on the charge is given by :
F = q E
Hence, this is the required solution.
37° Celsius is equal to 98.6° Fahrenheit
