Answer:
521.8 V
Explanation:
The RMS value of the voltage of an AC signal is given by
where
is the peak voltage, which corresponds to the amplitude of the AC waveform
In this problem, we know the RMS voltage
Therefore, we can re-arrange the previous equation to find the peak voltage (the amplitude of the waveform):
It would most definitely not be a hill or a cliff. I believe it would be a plain.
Answer:
Air resistance
Explanation:
Despite the law of conservation of energy stating that energy can neither be created nor destroyed but can only be transformed from one state to another, some energy is usually lost in the process of transformation and its majorly attributed to frictional loss. Friction opposes normal movement hence in air, air resistance tends to reduce the original energy compared to the initial. That is why the final energy in this case is slightly less than the original energy.
3 - b) weight decreases
4 -a) 50 kg as mass is same everywhere
5 - b) 200 dynes
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
