Answer:
Explanation:
Work done in lifting the weight once = mgh
= 20 x 9.8 x (1.9+1.7)
= 705.6 J
= 705.6 / 4.2 calorie
= 168 cals
Total energy to be spent = 600 x 10³ cals
No of times weight is required to be lifted
= 600 x 10³ / 168
= 3.57 x 10³ times
Total time to be taken = 2 x 3.57 x 10³
= 7.14 x 10³ s
=119 minutes .
Answer:
B. decreases while his angular speed remains unchanged.
Explanation:
His angular speed will always be the same as the wheel's angular speed, which remains constant as it's in uniform motion. As for linear speed, which is defined as the product of angular speed and distance r to the center of rotation, and his distance to center is decreasing, his linear speed must be decreasing as well.
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Answer:
Mechanical would have been conserved if only the force of gravity (the weight of the object does work on the system). The tension force does work also on the system but negative work instead. The net force acting of the system is zero since the upward tension in the string suspending the object is equal to the weight of the object but acting in the opposite direction. As a result they cancel out. In the equation above the effect of the tension force on the object has been neglected or not taken into consideration. For the mechanical energy E to be conserved, the work done by this tension force must be included into the equation. Otherwise it would seem as though energy has been generated in some manner that is equal in magnitude to the work done by the tension force.
The conserved form of the equation is given by
E = K + Ug + Wother.
In this case Wother = work done by the tension force.
In that form the total mechanical energy is conserved.
Answer:
48 m
Explanation:
Two trains traveling towards one another on a straight track are 300m apart when the engineers on both trains become aware of the impending Collision and hit their brakes. The eastbound train, initially moving at 97.0 km/h Slows down at 3.50ms^2. The westbound train, initially moving at 127 km/h slows down at 4.20 m/s^2.
The eastbound train
First convert km/h to m/s
(97 × 1000)/3600
97000/3600
26.944444 m/s
As the train is decelerating, final velocity V = 0 and acceleration a will be negative. Using third equation of motion
V^2 = U^2 - 2as
O = 26.944^2 - 2 × 3.5 S
726 = 7S
S = 726/7
S1 = 103.7 m
The westbound train
Convert km/h to m/s
(127×1000)/3600
127000/3600
35.2778 m/s
Using third equation of motion
V^2 = U^2 - 2as
0 = 35.2778^2 - 2 × 4.2 × S
1244.52 = 8.4S
S = 1244.52/8.4
S2 = 148.2 m
S1 + S2 = 103.7 + 148.2 = 251.86
The distance between them once they stop will be
300 - 251.86 = 48.14 m
Therefore, the distance between them once they stop is 48 metres approximately.
