Answer:
U = 1 / r²
Explanation:
In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related
F = - dU / dr
this derivative is a gradient, that is, a directional derivative, so we must have
dU = - F. dr
the esxresion for strength is
F = B / r³
let's replace
∫ dU = - ∫ B / r³ dr
in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product
let's evaluate the integrals
U - Uo = -B (- / 2r² + 1 / 2r₀²)
To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)
U = B / 2r²
we substitute the value of B = 2
U = 1 / r²
Answer : The process of changing a property of a wave to transmit information is called Modulation.
Explanation :
Modulation is the process of changing the property of wave to transmit information. This is done with the help of modulator.
In modulation, the message signal is superimposed on a high frequency signal. A sine wave ( usually high frequency ) is used as a high frequency carrier wave.
Modulation can be done in many ways like :
(1) Frequency modulation
(2) Amplitude modulation
(3) Pulse modulation
Answer: -2.5
Explanation:
1/2(-5)= -2.5
-2.5(1)= -2.5
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This question is incomplete, the complete question is;
A weightlifter holds a 1,300 N barbell 1 meter above the ground. One end of a 2-meter-long chain hangs from the center of the barbell. The chain has a total weight of 400 N. How much work (in J) is required to lift the barbell to a height of 2 m?
What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the process?
Answer: Average force exerted by the weightlifter in the process = 1600N
Explanation:
To find Work done to lift a barbell and half of the hanging chain we say;
W₁ = ( 1300N + (1/2 × 400N)) × 1m
W₁ = (1300 + 200) Nm
W₁ = 1500J
now work done to lift the upper half of the chain we say:
W₂ = (1/2 × 400N) × (1/2 × 1m)
W₂ = 200N × 0.5m
W₂ = 100J
So total work done will be
W = W₁ + W₂
W = 1500J + 100J
W = 1600J
To find the average force exerted by the weight lifter, we say;
F = W/D
F = (1600 / 1m) N
F = 1600N
∴Average force = 1600N
