The same braking force does work on these objects to slow them down. The work done is equal to their change in kinetic energy:
FΔx = 0.5mv²
F = force, Δx = distance traveled, m = mass, v = speed
Isolate Δx:
Δx = 0.5mv²/F
Calculate Δx for each object.
Object 1: m = 4.0kg, v = 2.0m/s
Δx = 0.5(4.0)(2.0)²/F = 8/F
Object 2: m = 1.0kg, v = 4.0m/s
Δx = 0.5(1.0)(4.0)²/F = 8/F
The two objects travel the same distance before stopping.
Answer:
same
Explanation:
Acc. to Einstien's postulate of special theory of
Relativity ,
Velocity of the light beam is same in all frames of references
(a) If the freight car is at rest
The frame we can assumed as Non - inertial frame of reference
s
In the inertial frame of reference , velocity of the light beam has its own value as : 3 x 10^8 m/s
(b) If the freight car is moving , the frame we can assumed as Non -inertial frame of reference
In thus case also , The velocity of the light beam will also have the same value as ; 3 x 108 m/s
Answer:
It traveled 24 centimeters
Explanation:
The displacement of the object is equal to the area under the velocity vs time graph.
We can split this graph into two shapes, a triangle and a rectangle. So the total area is:
A = ½bh + wh
A = ½ (4 s − 0 s) (4 cm/s) + (8 s − 4 s) (4 cm/s)
A = 8 cm + 16 cm
A = 24 cm
The answer is C.) 10 m East
