Answer:
Explanation:
horizontal is zero slope, and the vertical is undefined
The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.
<h3>What do you mean by electric potential? </h3>
The amount of work needed to move a unit charge from a reference point to a specific point against an electric field. It's SI unit is volt.
V = kq/r
Where V represents electric potential, K is coulomb constant, q is Charge and r is distance between any two around charge to the point charge.
Electric potential at O due to four charges is given by,
V = 4KQ/ r
where, r = √2a/2 = a/√2
V = 4k × Q√2/a
V = √2Q/πÈa
The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.
To learn more about electric potential refer to:
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A.
According to the equation: v=λ*f
λ stands for wavelength. wavelength increases, the frequency will decrease. lower frequency deserves low energy.
Answer:
7.00 m
Explanation:
Given:
v₀ = 2.00 m/s
v = 5.00 m/s
a = 1.50 m/s²
Find: Δx
v² = v₀² + 2aΔx
(5.00 m/s)² = (2.00 m/s)² + 2(1.50 m/s²)Δx
Δx = 7.00 m
Answer:
Explanation:
(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)
ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .
ΔK = 0
ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .
ΔUg = positive
ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .
ΔUs = 0
ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence
ΔEth = negative .
b )
work done by athlete
= 400 x 2 = 800 J
energy output = 800 J
c )
It is 25% of metabolic energy output of his body
so metalic energy output of body
= 4x 800 J .
3200 J
power = energy output / time
= 3200 / 1.6
= 2000 W .
d )
1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .
2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .
