a car mass200kg movingat 36km/hr is brought to rest after it covered adistance of 10m .find the retarding force acting on the car
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Answer:
Mass of the car (m) = 200 kg
Initial speed (u) = 36 km/h = 10 m/s
Final velocity (v) = 0
Distance covered (S) = 10 m
By applying the equation
V1 - u² = 2aS
=>0 - 100 = 2 x a x 10
=> - 100 = 20 aa = - 100/20 = -5 m/s2
F = ma = 200 x 5 = -1000 N
Retarding force = 1000 N
Mass of the car (m) = 200 kg
Initial speed (u) = 36 km/h = 10 m/s
Final velocity (v) = 0
Distance covered (S) = 10 m
By applying the equation
V1 - u² = 2aS
=>0 - 100 = 2 x a x 10
=> - 100 = 20 aa = - 100/20 = -5 m/s2
F = ma = 200 x 5 = -1000 N
Retarding force = 1000 N
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Answered by
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m = 200kg
u = 36km/hr
= 36 × 5/18
= 10m/s
v = 0
s = 10m
Formula: v² = u² + 2as
0² = 10² + 2 × a × 10
0 = 100 + 20a
20a = -100
a = -100/20
a = -5 m/s²
Force = m × a
= 200 × -5
= -1000 N
negative sign shows that force is acting in opposite direction.
hope it helped.
u = 36km/hr
= 36 × 5/18
= 10m/s
v = 0
s = 10m
Formula: v² = u² + 2as
0² = 10² + 2 × a × 10
0 = 100 + 20a
20a = -100
a = -100/20
a = -5 m/s²
Force = m × a
= 200 × -5
= -1000 N
negative sign shows that force is acting in opposite direction.
hope it helped.
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