Question:
A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?
Answer:
Time for the race will be t = 9.26 s
Explanation:
Given data:
As the sprinter starts the race so initial velocity = v₁ = 0
Distance = s₁ = 20 m
Acceleration = a = 4.20 ms⁻²
Distance = s₂ = 100 m
We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.
Using 3rd equation of motion
(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)
v₂ = 12.96 ms⁻¹
Time for 20 m distance = t₁ = (v₂ - v ₁)/a
t₁ = 12.96/4.2 = 3.09 s
He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be
Time for 100 m distance = t₂ = s₂/v₂
t₂ = 80/12.96 = 6.17 s
Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s
T = 9.26 s
Answer:
B
Explanation:
kinetic energy (KE) is the energy possessed by moving bodies. It can be expressed as:
KE = 
m
Where: m is the mass of the object, and v its speed.
For example, a stone of mass 2kg was thrown and moves with a speed of 3 m/s. Determine the kinetic energy of the stone.
Thus,
KE = x 2 x 
= 9
KE = 9.0 Joules
Assume that the speed of the stone was 4 m/s, then its KE would be:
KE = x 2 x 
= 16
KE = 16.0 Joules
Therefore, it can be observed that as speed increases, the kinetic energy increases. Thus option B is appropriate.
Answer:
222.22 N
Explanation:
r = 5 cm = 0.05 m, R = 1.5 m, f = ? , F = 200,000 N
Use pascal's law
F / A = f / a
F / 3.14 x R^2 = f / 3.14 x r^2
F / R^2 = f / r^2
f = F x r^2 / R^2
f = 200000 x 0.05 x 0.05 / (1.5 x 1.5)
f = 222.22 N
