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:
P = 4000 [Pa]
Explanation:
Pressure is defined as the relationship between Force and the area where the body rests.
The support area is equal to:
![A=50*20=1000[cm^{2} ]](https://tex.z-dn.net/?f=A%3D50%2A20%3D1000%5Bcm%5E%7B2%7D%20%5D)
But we must convert from square centimeters to square meters.
![1000[cm^{2}]*\frac{1^{2}m^{2} }{100^{2}m^{2} }=0.1[m^{2} ]](https://tex.z-dn.net/?f=1000%5Bcm%5E%7B2%7D%5D%2A%5Cfrac%7B1%5E%7B2%7Dm%5E%7B2%7D%20%20%7D%7B100%5E%7B2%7Dm%5E%7B2%7D%20%20%7D%3D0.1%5Bm%5E%7B2%7D%20%5D)
And the pressure is:
![P=\frac{F}{A} \\P=400/0.1\\P=4000[N/m^{2} ]or 4000[Pa]](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BF%7D%7BA%7D%20%5C%5CP%3D400%2F0.1%5C%5CP%3D4000%5BN%2Fm%5E%7B2%7D%20%5Dor%204000%5BPa%5D)
Answer:
a)
b)
c)
d)
m
e)λ=∞
Explanation:
De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

with h the Plank's constant (
) and P the momentum of the object that is mass (m) times velocity (v).
a)

b)

c)

d)
m
e) 
λ=∞
Answer:
-0.912 m/s
Explanation:
When the package is thrown out, momentum is conserved. The total momentum after is the same as the total momentum before, which is 0, since the boat was initially at rest.

where
are the mass of the child, the boat and the package, respectively.
are the velocity of the package and the boat after throwing.


