Answer:
The magnitude of the magnetic torque on the coil is 1.98 A.m²
Explanation:
Magnitude of magnetic torque in a flat circular coil is given as;
τ = NIASinθ
where;
N is the number of turns of the coil
I is the current in the coil
A is the area of the coil
θ is the angle of inclination of the coil and magnetic field
Given'
Number of turns, N = 200
Current, I = 7.0 A
Angle of inclination, θ = 30°
Diameter, d = 6 cm = 0.06 m
A = πd²/4 = π(0.06)²/4 = 0.002828 m²
τ = NIASinθ
τ = 200 x 7 x 0.002828 x Sin30
τ = 1.98 A.m²
Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²
Answer:
Option C. Objects 1 and 3 will not move, and objects 2 and 4 will accelerate
upward.
Explanation:
The following data were obtained from the question:
OBJECT >>>>>>>>> WEIGHT (N)
1 >>>>>>>>>>>>>>>> 35
2 >>>>>>>>>>>>>>>> 23
3 >>>>>>>>>>>>>>>> 26
4 >>>>>>>>>>>>>>>> 18
Force (F) applied = 25 N
From the above, the force applied to each object is 25N. Thus the following can be concluded based on the data given above:
For object 1:
Weight = 35 N
Force applied = 25 N
Thus, the object will not move since the weight of the object is greater than the force applied
For object 2:
Weight = 23 N
Force applied = 25 N
Thus, the object will move since the force applied is greater than the weight of the object.
For object 3:
Weight = 26 N
Force applied = 25 N
Thus, the object will not move since the weight of the object is greater than the force applied.
For object 4:
Weight = 18 N
Force applied = 25 N
Thus, the object will move since the force applied is greater than the weight of the object.
From the above illustrations, Object 1 and 3 will not move, and objects 2 and 4 will accelerate i.e move
Answer:
a) w = 7.27 * 10^-5 rad/s
b) v1 = 463.1 m/s
c) v1 = 440.433 m/s
Explanation:
Given:-
- The radius of the earth, R = 6.37 * 10 ^6 m
- The time period for 1 revolution T = 24 hrs
Find:
What is the earth's angular speed?
What is the speed of a point on the equator?
What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?
Solution:
- The angular speed w of the earth can be related with the Time period T of the earth revolution by:
w = 2π / T
w = 2π / 24*3600
w = 7.27 * 10^-5 rad/s
- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.
v1 = R*w
v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)
v1 = 463.1 m/s
- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.
π/2 ........... s
x ............ 1/5 s
x = π/2*5 = 18°
- The radius of the earth R' at point where θ = 18° from the equator is:
R' = R*cos(18)
R' = (6.37 * 10 ^6)*cos(18)
R' = 6058230.0088 m
- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.
v2 = R'*w
v2 = (6058230.0088)*(7.27 * 10^-5)
v2 = 440.433 m/s
Answer:
False
Explanation:
This proposition is false because by example the sun exerts a force over the earth and them are not in contact
