Explanation:
The principle of an electric motor is based on the current carrying conductor which produces magnetic field around it. A current carrying conductor is placed perpendicular to the magnetic field so that it experiences a force.
The largest electric motors are used for ship propulsion, pipeline compression and pumped-storage applications with ratings reaching 100 megawatts. Electric motors are found in industrial fans, blowers and pumps, machine tools, household appliances, power tools and disk drives.
Solution :
Acceleration due to gravity of the earth, g 

Acceleration due to gravity at 1000 km depths is :




= 8.23 m/s
Acceleration due to gravity at 2000 km depths is :




= 6.73 m/s
Acceleration due to gravity at 3000 km depths is :



= 5.18 m/s
Acceleration due to gravity at 4000 km depths is :




= 3.64 m/s
Heat rises therefore the heat from the fire rises up to your hand... i didnt have any answer choices to work with sorry
Answer:
35.35 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 20 m/s
Angle of projection (θ) = 30°
Acceleration due to gravity (g) = 9.8 m/s²
Range (R) =.?
The range (i.e how far away) of the ball can be obtained as follow:
R = u² Sine 2θ /g
R = 20² Sine (2×30) / 9.8
R = 400 Sine 60 / 9.8
R = (400 × 0866) / 9.8
R = 346.4 / 9.8
R = 35.35 m
Therefore, the range (i.e how far away) of the ball is 35.35 m
Answer:
7.45 s.
Explanation:
Given:
h = 68.1 m
vi = 0 m/s
vf = 42.4 m/s
g = 9.81 m/s^2
Using,
h = vi*t +1/2*(a*t^2)
68.1 = 1/2 * (9.81*t^2)
t = sqrt((68.1*2)/9.81)
= 3.726 s.
Total time of flight = 2*t
= 2 * 3.726
= 7.45 s.