N= energy efficiency pout means output and pin means input the reason this would show efficiency is because your output should be greater then your input and because depending on how small your number is after your division will tell you how efficient it is you want a big number.
Answer:
E = 440816.32 N/C
Explanation:
Given data:
Three point charge of charge equal to +3.0 micro coulomb
fourth point charge = - 3.0 micro coulomb
side of square = 0.50 m
N.m^2/c^2
Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value
So we have
[
[
]
plugging all value
E = 440816.32 N/C
The <span>asthenosphere is under the lithosphere.</span>
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
