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:
1 mole of H2O is 18 grams (2 g H + 16 g Oxygen)
36 / 18 = 2
So 2 moles = 2 * 6.02E23 = 12.04E23 = 1.204E24
Answer:
The path of an object in uniform motion is a straight line.
<span>A van is traveling on a road at a speed of 55 km/h relative to a
stationary observer on the side of the road. A girl sitting near the
driver of the van throws a paper airplane to a boy at the back of the
van with a speed of 2 km/h relative to the girl, the boy, and the van.
The speed of the paper airplane, relative to the same stationary observer
on the side of the road, is (55 - 2) = 53 km/h. No rounding is necessary.</span>
For an object`s motion, the Kinematic equation is,
Here, v is the final velocity and h is stands for the height of the object and a is the acceleration of the object.
As according to question,
and 
Thus, putting these values in above equation, we get
or
Therefore, initial velocity is 2.8 m/s
