Cutting pepper and onions
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:

Explanation:
The gravitational force between two objects is given by

where
G is the gravitational constant
m1, m2 are the two masses
r is the separation between the two masses
In this problem, we have
the mass of each truck
r = 2.0 m is their separation
Substituting,

<span> The meteor accelerates after it is brought into earths force field
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Answer:
The maximum height of the ball is 2 m.
Explanation:
Given that,
Mass of ball = 50 g
Height = 1.0 m
Angle = 30°
The equation is

We need to calculate the velocity
Using conservation of energy

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

Put the value into the formula

Put the value into the formula




We need to calculate the maximum height of the ball
Using again conservation of energy

Here, h = y highest point
Put the value into the formula



Put the value of y in the given equation




Hence, The maximum height of the ball is 2 m.