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:
First of all, let's remind that:
- The kinetic energy of an object is given by 
, where m is the mass and v is the speed
- The momentum of an object is given by 
- The inertia of an object is proportional to its mass, so we can write 
, where k just indicates a constant of proportionality
In this problem, we have:
- 
(the two objects have same kinetic energy)
- 
(A has three times the momentum of B)
Re-writing both equation we have:
If we divide first equation by second one we get
And if we substitute it into the first equation we get
So, B has 9 times more mass than A, and so B has 9 times more inertia than A, and their ratio is:
<span>A series circuit has one path for electrons, but a parallel circuit has more than one path.</span>
Explanation:
Given that,
The optical power of the equivalent single lens is 45.4 diopters.
(a) The relationship between the focal length and the focal length is given by:
f = 0.022 m
or
f = 2.2 cm
(b) We need to find how far in front of the retina is this "equivalent lens" located. It is given by using lens formula as :
Here, u = infinity
v = 2.2 cm
So, at 2.2 cm in front of the retina is this "equivalent lens" located.
Hence, this is the required solution.
