Answer:
3.18 Nm
Explanation:
Given that:
Radius (r) = 15cm = 15/100 = 0.15m
θ = 45°
Applied force (F) = 30 N
The Torque can be obtained using the relation :
T = rF * sinθ
T = 0.15 * 30 * sin(45)
T = 0.15 * 30 * 0.7071067
T = 3.18198015
T = 3.18 Nm
Answer:
6.20×10⁴ V/m
Explanation:
The magnitude of electric field is:
E = √(Eₓ² + Eᵧ²)
where Eₓ = ∂φ/∂x and Eᵧ = ∂φ/∂y.
φ = 1.11 (x² + y²)^-½ − 429x
Eₓ = -0.555 (x² + y²)^-(³/₂) (2x) − 429
Eᵧ = -0.555 (x² + y²)^-(³/₂) (2y)
Evaluating at (0.003, 0.003):
Eₓ = -44034 V/m
Eᵧ = -43605 V/m
The magnitude is:
E = 61971 V/m
Rounded to three significant figures, the strength of the electric field is 6.20×10⁴ V/m.
Answer: reactants to this system,...
Explanation:
Answer:
a)n= 3.125 x electrons.
b)J= 1.515 x A/m²
c) =1.114 x m/s
d) see explanation
Explanation:
Current 'I' = 5A =>5C/s
diameter 'd'= 2.05 x m
radius 'r' = d/2 => 1.025 x m
no. of electrons 'n'= 8.5 x
a) the amount of electrons pass through the light bulb each second can be determined by:
I= Q/t
Q= I x t => 5 x 1
Q= 5C
As we know that: Q= ne
where e is the charge of electron i.e 1.6 x C
n= Q/e => 5/ 1.6 x
n= 3.125 x electrons.
b) the current density 'J' in the wire is given by
J= I/A => I/πr²
J= 5 / (3.14 x (1.025x )²)
J= 1.515 x A/m²
c) The typical speed'' of an electron is given by:
=
=1.515 x / 8.5 x x |-1.6 x |
=1.114 x m/s
d) According to these equations,
J= I/A
= =
If you were to use wire of twice the diameter, the current density and drift speed will change
Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.
Also drift velocity will decrease as it is inversely proportional to the area
Explanation:
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