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
Answer:
Zero work done,since the body isn't acting against or by gravity.
Explanation:
Gravitational force is usually considered as work done against gravity (-ve) and work by gravity ( +ve ) and also When work isn't done by or against gravity work done in this case is zero.
Gravitational force can be define as that force that attracts a body to any other phyical body or system that have mass.
The planet been considered as our system in this case is assumed to have mass, and ought to demonstrate such properties associated with gravitational force in such system. Such properties include the return of every object been thrown up as a result of gravity acting downwards. The orbiting nature of object along an elliptical part when gravitational force isn't acting on the body and it is assumed to be zero.
Answer:
4.5kgm/s
Explanation:
Change in momentum is expressed as
Change in momentum = m(v-u)
M is the mass
V is the final velocity
u is the initial velocity
Given
m=0.45kg
v = 30m/s
u = 20m/s
Substitute
Change in momentum = 0.45(30-20)
Change in momentum = 0.45×10
Change in momentum = 4.5kgm/s
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