Answer:
The magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T
Explanation:
Given;
radius of the wire, r = 0.45 m
current on the loop, I = 2.4 A
angle of inclination, θ = 36⁰
torque on the coil, τ = 1.5 N.m
The torque on the coil is given by;
τ = NIBAsinθ
where;
B is the magnetic field
Area of the loop is given by;
A = πr² = π(0.45)² = 0.636 m
τ = NIBAsinθ
1.5 = (1 x 2.4 x 0.636 x sin36)B
1.5 = 0.8972B
B = 1.5 / 0.8972
B = 1.67 T
Therefore, the magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T
Answer:
U2 = 47.38m/s = initial velocity of B before impact
Explanation:
An example of the diagram is shown in the attached file because of missing angle of direction in the question
Mass A, B are mass of cars
A = 1965
B =1245
U1 = initial velocity of A = 52km/hr
U2 = initial velocity of B
V = common final velocity of two cars
BU2 = (A + B)*V sin ¤ ...eq1 y plane
AU1 = (A + B) *V cos ¤ ....equ 2plane
From equ 2
V = AU1/(A + B)*cos ¤
Substitute V into equation 1
We have
U2 = (AU1/B)tan ¤ where ¤ = angle of direction which is taken to be 30°
Substitute all parameters to get
U2 = (1965/1245)*52 * tan 30°
U2 = 47.38m/s
Answer:
C.
Explanation:
If there are the same number of protons (+) as electrons (-) they will cancel out and make the atom neutral.
Answer:
1. is the age group 35 and 44
2. is 2006 i think its 2006 i cant really tell in the picture but its the one before the last one!