Answer:
9.8m/s^2 down (option C)
Explanation:
The only acceleration acting on this motion case in the acceleration due to gravity: 9.8 m/s^2 in the downwards direction.
There exists the same question that has the following choices.
<span>1 the car's friction is reduced to zero
2 the car's centripetal force drops to zero
3 the car continues in a straight path from the point at which it encountered the ice
4 the car continues in a circular path from the point at which it encountered the ice
</span>
The correct answer is choice 3 <span>the car continues in a straight path from the point at which it encountered the ice</span>
Work is calculated by multiplying the amount of force exerted and the distance that the object had moved. From the given above, the work that the forklift does is calculated by,
W = F x d
W = (2.00 x 10^3 N)(5 m)
V = 10 x 10^3 N.m
Thus, the work done is approximately 10,000 J.
Answer:
2.65m/s
Explanation:
Using the equation of motion:
v² = u²+2a∆S where
v is the final velocity
u is the initial velocity
∆S is the change in distance
a is the acceleration
Given
u = 0m/s
a = 9.8m/s²
∆S = 1.3-0.943
∆S = 0.357m
Substituting the given parameters into the formula
v² = 0²+2(9.8)(0.357)
v² = 0+6.9972
v² = 6.9972
v=√6.9972
v = 2.65m/s
Hence the velocity at which it hit the ground is 2.65m/s