For this trajectory, what would the vertical component of acceleration for the module be at time tm=t0−σ=325s? Recall that accel
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Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,
Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?
a = -1.47 m/ (a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/ (since same friction force is applied)
s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
let the distance of pillar is "r" from one end of the slab
So here net torque must be balance with respect to pillar to be in balanced state
So here we will have
here we know that
mg = 19600 N
Mg = 400,000 N
L = 20 m
from above equation we have
so pillar is at distance 10.098 m from one end of the slab