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1 answer:
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A) The net force on the motorbike is 205.9 N
B) The acceleration of the motorbike is
C) The final speed is 5.2 m/s
D) The elapsed time is 3.80 s
Explanation:
A)
There are two forces acting on the motorbike:
- The applied force, F = 250 N, forward
- The frictional force, , backward
The frictional force can be written as
where
is the coefficient of kinetic friction
is the mass of the motorbike
is the acceleration of gravity
Therefore the net force is given by
And substituting, we find
2)
The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:
where
m is the mass
a is the acceleration
In this problem, we have
is the net force
m = 150 kg is the mass
Solving for a, we find the acceleration:
C)
Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance covered
For this motorbike, we have:
u = 0 (it starts from rest)
s = 350 m
Solving for v,
4)
For this part of the problem, we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the elapsed time
Here we have:
v = 5.2 m/s
u = 0
Solving for t, we find
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Answer:
Work done by the frictional force is
Explanation:
It is given that,
Mass of the car, m = 1000 kg
Initial velocity of car, u = 26.1 m/s
Finally, it comes to rest, v = 0
We have to find the work done by the frictional forces. Work done is equal to the change in kinetic energy as per work - energy theorem i.e.
W = −340605 J
or
Hence, the correct option is (a).