Answer:
Diameter of wire = 0.00021 meters
Explanation:
Given the following data;
Number of windings = 30
Length of windings = 6.3 mm to meters = 0.0063 meters
To find the diameter of the wire, we would use this mathematical expression;
Length of windings = number of windings * diameter of wire
Substituting the values into the expression, we have;
0.0063 = 30 * diameter of wire
Diameter of wire = 0.0063/30
Diameter of wire = 0.00021 meters
Answer:
1.67×10^-11 N
Explanation:
m1 = 0.13kg
m2 = 0.34kg
d = 0.42m
G = 6.674 × 10 ^−11
F = Gm1m2/r²
F= (6.674 × 10^ −11×0.13×0.34)/(0.42)²
F = 1.67×10^-11 N
Answer:
The maximum emf that can be generated around the perimeter of a cell in this field is 
Explanation:
To solve this problem it is necessary to apply the concepts on maximum electromotive force.
For definition we know that

Where,
N= Number of turns of the coil
B = Magnetic field
Angular velocity
A = Cross-sectional Area
Angular velocity according kinematics equations is:



Replacing at the equation our values given we have that




Therefore the maximum emf that can be generated around the perimeter of a cell in this field is 
Answer:
W= 1812.6 J
Explanation:
Work (W) is defined as the scalar product of force F by the distance (d) the body travels due to this force.
W= F*d* cosα Formula ( 1)
Where:
F is the force in Newtons (N)
d is the displacemente in meters (m)
α : Angle formed between force and displacement
Data
F = 10 N
d = 200 m
α = 25°
Work done by the pulling force while the passenger walks 200 m
We replace data in the formula (1)
W= F*d* cosα
W= (10 N)*(200 m)* cos25°
W= 1812.6 (N*m)
W= 1812.6 J