Answer:
600m
Explanation:
30×20 at a constant speed is 600m.
Answer:
d = 69 .57 meter
Explanation:
First case
Speed of car ( v ) = 20.5 mi/h = 9.164 M/S
distance ( d ) = 11.6 meter ( m = mass of the car )
Work done = 0.5 m v² = 0.5 * 9.164² * m J = 41.99 m J
Force = ( workdone /distance ) = ( 41.99 m / 11.6 ) = 3.619 m N
Second case
v = 50.2 mi/h = 22.44135 m/s
d = ?
Work done = 0.5 * 22.44² * m J = 251.7768 * m J
Since the braking force remains the same .
3.619 m = ( 251.7768 m / d )
d = 69 .57 meter
The freezing point ..... :)
Given parameters:
First velocity = 2.50m/s
Time of travel = 3s
Second velocity = 1.50m/s
Unknown:
The displacement during the first interval = ?
Velocity is the displacement of a body with time. Displacement is a distance move in a specific direction by a body.
Velocity = 
So;
Displacement = Velocity x Time taken
Now input the parameter for the first velocity and time of travel;
Displacement = 2.5 x 3 = 7.5m
The displacement id 7.5m
Aswer:
False, the values of the distance traveled and the displacement only coincide when the trayectorie is a straight line. Otherwise, the distance will always be greater than the offset.
Although these terms are used synonymously in other cases, they are totally different. Since the distance that a mobile travels is the equivalent of the length of its trajectory. Whereas, the displacement will be a vector magnitude.
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