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:
a) 24
b) 3.3 sec
c) 29.8 m/s
d) 48.85 m
Explanation:
a)
α = angular acceleration = - 28.4 rad/s²
r = radius of the tire = 0.32 m
w₀ = initial angular velocity = 93 rad/s
w = final angular velocity = 0 rad/s
θ = angular displacement
Using the equation
w² = w₀² + 2αθ
0² = 93² + 2 (- 28.4) θ
θ = 152.3 rad
n = number of revolutions
Number of revolutions are given as
b)
t = time taken to stop
using the equation
w = w₀ + αt
0 = 93 + (- 28.4) t
t = 3.3 sec
c)
v₀ = initial velocity of the car
initial velocity of the car is given as
v₀ = r w₀ = (0.32) (93) = 29.8 m/s
d)
v = final velocity = 0 m/s
a = linear acceleration = rα = (0.32) (- 28.4) = - 9.09 m/s²
d = distance traveled by car before stopping
Using the equation
v² = v₀² + 2 a d
0² = 29.8² + 2 (- 9.09) d
d = 48.85 m
Wind is the result of convection currents. :)
Answer:
ITS THE LAST ONE(4TH), I THINK
Explanation:
