Two events are observed in a frame of reference S to occur at the same space point, with the second event occurring after a time
of 1.70 s . In a second frame S' moving relative to S, the second event is observed to occur after a time of 2.25 s . What is the difference between the positions of the two events as measured in S'?
1 answer:
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Answer:
40 MJ (D)
Explanation:
Quantity of heat (Qh) = 100 MJ
temperature of steam (Th) = 450°c = 450 + 273 = 723 K
emperature of water (TI) = 20 °c = 20 + 273 = 293 k
efficiency = (Qh-Qi)/Qh = (Th-Ti)/Th
- Qi= 0.5947 x
- (0.5947 x
) = Qi
Qi = 40.5 MJ equivalent to 40 MJ (D)
The climber move 0.19 m/s faster than surfer on the nearby beach.
Since both the person are on the earth, and moves with the constant angular velocity of earth, however there linear velocity is different.
Number of seconds in a day, t=24*60*60=86400 sec
The linear speed on the beach is calculated as
V1=
Here, t is the time
Plugging the values in the above equation
V1==465.421 m/s
Velocity on the mountain is calculated as
V2=
Plugging the values in the above equation
V2==465.61 m/s
Therefore person on the mountain moves faster than the person on the beach by 465.61-465.421=0.19 m/s
B. velocity at position x, velocity at position x=0, position x, and the original position
In the equation
=
+2 a x (x - x₀)
= velocity at position "x"
= velocity at position "x = 0 "
x = final position
= initial position of the object at the start of the motion