The energy delivered to the resistive coil is dissipated as heat at a rate equal to the power input of the circuit. However, not
all of the energy in the circuit is dissipated by the coil. Because the emf source has internal resistance, energy is also dissipated by the battery as heat. Calculate the rate of dissipation of energy PbatPbatP_bat in the battery.
1 answer:
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Answer:
The value is
Explanation:
From the question we are told that
The weight of the block is
The dimension of the block is
Generally two atmosphere is equivalent to
Generally 1 atm =
The area of the block would be evaluated using width and height because we need for the smaller surface to be in contact with the ground in order to maximize the pressure and minimize number of blocks
So
=>
Generally the force due to this blocks is mathematically represented as
Here N is the number of blocks
So
=>
Answer:
v = 29.4 m / s
Explanation:
For this exercise we can use the conservation of mechanical energy
Lowest starting point.
Em₀ = K = ½ m v²
final point. Higher
= U = m g h
Let's use trigonometry to lock her up
cos 60 = y / L
y = L cos 60
Height is the initial length minus the length at the maximum angle
h = L - L cos 60
h = L (1- cos 60)
energy is conserved
Em₀ = Em_{f}
½ m v² = mgL (1 - cos 60)
v = 2g L (1- cos 60)
let's calculate
v² = 2 9.8 3.0 (1- cos 60)
v = 29.4 m / s