Answer: D
Reduced impact time will increase the impact force
Explanation: Collision occurs when two or more bodies collide and exert forces on each other within a short time.
If a body of mass M moving with a velocity V collide with another body, the kinetic energy of the body is equal to the work done by the body.
That is, K.E = 1/2mv^2 = F × s
Where workdone = Force × distance
Make F the subject of formula
Mv^2/2s = F
But V = distance s/time t
Substitute for V
Ms^2/2t^2s = F
Ms/2t^2 = F
From the equation above, we can deduce that F is inversely proportional to the square of time.
Therefore, the reduced impact time will increase the impact force
Answer:
The weight of the wheelbarrow and the road is 784 N and the force required to lift the wheelbarrow is 784 N.
Explanation:
Given that,
The total mass of the wheelbarrow and the road is 80 kg.
The weight of an object is given by :
W = mg
where
g is acceleration due to gravity
So,
W = 80 × 9.8
= 784 N
So, the force required to lift the wheelbarrow is equal to its weight i.e. 784 N.
To solve the problem it is necessary to identify the equation in the manner given above.
This equation corresponds to the displacement of a body under the principle of simple harmonic movement.
Where,
PART A) Our equation corresponds to
Therefore the value of omega is equivalent to that of
From the definition we know that the period as a function of angular velocity is equivalent to
This same point is the equivalent of the maximum point of the speed that the body can reach, since the internal expression of the 
Is equivalent to . So the maximum speed that the body can reach is,
Therefore the maximum felocity will be 5ft / s
PART B) The period of graph is the time taken to reach from one maximum point to next point maximum point, then
Answer:
81.6 m
Explanation:
Answer: 81.6 m.
The time it takes gravity to slow 40 m/s to zero when it teaches maximum height is
-v(initial) / -g = t
-40 m/s / -9.8 m/s^2 = 4.08 s
The height reached is the average velocity times this time 4.08 s, with v(avg) = [v(initial) + v(final)] / 2 with v(final) = 0. v(avg) = v(initial) / 2 = 40 m/s / 2 = 20 m/s.
So the distance d of maximum height is
d = v(avg)•t
d = 20 m/s • 4.08 s = 81.6 m.
m =dm ______ 10.000
Meters
The metre is a unit of length in the metric system, and is the base unit of length in the International System of Units (SI).
As the base unit of length in the SI and other m.k.s. systems (based around metres, kilograms and seconds) the metres is used to help derive other units of measurement such as the newton, for force.
