To develop this problem use the concept of the sum given pressure in the tank. At the bottom of the tank the pressure of this will be given by atmospheric pressure, the pressure given by the oil and the pressure by the water, that is to say that mathematically the pressure would be
<em>Note: Here the pressures are expressed in terms of density (), gravity (g) and thickness (t) or height (h). If we rearrange this equation to find the oil thickness we will have to,</em>
<em></em>
Our values are given as,
Replacing we have that the thickness of the oil is:
Therefore the thick of the oil is 0.5947m
I can confirm it is not b.
Answer:
Makes it hard to go to sleep at night
Explanation:
Answer:
The coil radius of other generator is 5.15 cm
Explanation:
Consider the equation for induced emf in a generator coil:
EMF = NBAω Sin(ωt)
where,
N = No. of turns in coil
B = magnetic field
A = Cross-sectional area of coil = π r²
ω = angular velocity
t = time
It is given that for both the coils magnetic field, no. of turn and frequency is same. Since, the frequency is same, therefore, the angular velocity, will also be same. As, ω = 2πft.
Therefore, EMF for both coils or generators will be:
EMF₁ = NBπr₁²ω Sin(ωt)
EMF₂ = NBπr₂²ω Sin(ωt)
dividing both the equations:
EMF₁/EMF₂ = (r₁/r₂)²
r₂ = r₁ √(EMF₂/EMF₁)
where,
EMF₁ = 1.8 V
EMF₂ = 3.9 V
r₁ = 3.5 cm
r₂ = ?
Therefore,
r₂ = (3.5 cm)√(3.9 V/1.8 V)
<u>r₂ = 5.15 cm</u>
Answer:
Explanation:
The PE equation for a mass/spring system is
Δx² and filling in:
and
so
k = 26000 N/m
If the displacement from equilibrium changes more, the PE needed to compress it will also change.
gives us that
PE = 520J