The solution for this problem is:
(10 x 9.8) = 98.1 m/sec^2 acceleration. Time, to travel 9.4cm or (.094m.), at acceleration of 98m/sec^2
= sqrt(2d/a), = sqrt (98.1 m/sec^2/0.094m) = 32.3050619 sec per cycle
Frequency = (w/2pi), = 32.3050619/2pi
= 32.3050619/6.28318531
= 5.14 Hz would be the answer
To finish one orbit it will take 98 x 60 seconds. So; <span>(2 x pi)/(98 x 60) = 1.07 x 10^-3 rad/sec. </span><span>
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•wind
•snow
•high tide/low tide
•thunder/lightning storms
Given that,
Radius of track, r = 50 m
time , t = 9 s
velocity, v = ?
Distance covered by car in one lap around a track is equal to the circumference of the track.
C = 2 π r = 2 * 3.14 * 50
C = 314.159 m
Distance covered by car, s = 314.159 m
Velocity = distance/ time
V = 314.159 / 9
V = 34.9 m/s
The average velocity of car is 34.9 m/s.
Nope, I disagree with the former answer. The answer is definitely Z. <u>W area</u> (boxed with red outline) is represented as the hot reservoir while <u>Z area</u> is the cold reservoir (boxed with blue outline). X area is the heat engine itself and Y area is the work produced from thermal energy from hot reservoir. Typically, all heat engines lose some heat to the environment (based from the second law of thermodynamics) that is symbolically illustrated by the lost energy in the cold reservoir. This lost thermal energy is basically the unusable thermal energy. The higher thermal energy lost, the less efficient your heat engine is.
