Answer:
1470 W
Explanation:
Power: This can be defined as the rate at which work is done or energy is used up. The S.I unit of power is Watt (W).
The expression for power is given as,
P = Energy/time
P = mgh/t ...................... Equation 1
Where P = power, m = mass, h = height, t = time, g = acceleration due to gravity.
Given: m = 75 kg, g =9.8 m/s², h = 1 m, t = 1 s.
Substitute into equation 1
P = (75×1×9.8)/1
P = 735 W.
From the above,
1 hp = 735 W
2 hp = (2×735) W
2 hp = 1470 W.
Hence 2 hp = 1470 W
Since momentum is a vector quantity, take any direction as positive and other as negative. Answer won't change.
Answer:
False
Explanation:
In miles per hour, light speed is about 670,616,629 mph
"The equation can be used to calculate the power absorbed by any surface" statement concerning the Stefan-Boltzmann equation is correct.
Answer: Option A
<u>Explanation:</u>
According to Stefan Boltzmann equation, the power radiated by black body radiation source is directly proportionate to the fourth power of temperature of the source. So the radiation transferred is absorbed by another surface and that absorbed power will also be equal to the fourth power of the temperature. So the equation describes the relation of net radiation loss with the change in temperature from hotter temperature to cooler temperature surface.
So this law is application for calculating power absorbed by any surface.
Answer:
1.24 C
Explanation:
We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.
The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m
So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.
Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.
So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²
So, 4iρD/d² = 0.750πD²/4Δt.
iΔt = 0.750πD²/4 ÷ 4iρD/d²
iΔt = 0.750πD²d²/16ρ.
So the charge Q = iΔt
= 0.750π(Dd)²/16ρ
= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)
= 123.76 × 10⁻² C
= 1.2376 C
≅ 1.24 C
