The first law of thermodynamics says that the variation of internal energy of a system is given by:
where Q is the heat delivered by the system, while W is the work done on the system.
We must be careful with the signs here. The sign convention generally used is:
Q positive = Q absorbed by the system
Q negative = Q delivered by the system
W positive = W done on the system
W negative = W done by the system
So, in our problem, the heat is negative because it is releaed by the system:
Q=-1275 J
while the work is positive because it is performed by the surrounding on the system:
W=+855 J
So, the variation of internal energy of the system is
Answer:
d = 3.54 x 10⁴ Km
Explanation:
Given,
The distance between the two objects, r = 2.5 x 10⁴ Km
The gravitational force between them, F = 580 N
The gravitational force between the two objects is given by the formula
F = GMm/r² newton
When the gravitational force becomes half, then the distance between them becomes
Let us multiply the above equation by 1/2 on both sides
( 1/2) F = (1/2) GMm/r²
= GMm/2r²
= GMm/(√2r)²
Therefore, the distance becomes √2d, when the gravitational force between them becomes half
d = √2r = √2 x 2.5 x 10⁴ Km
= 3.54 x 10⁴ Km
Hence, the two objects should be kept at a distance, d = 3.54 x 10⁴ Km so that the gravitational force becomes half.
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:
Explanation:
The electric field produced by a single point charge is given by:
where
k is the Coulomb's constant
q is the charge
r is the distance from the charge
In this problem, we have
E = 1.0 N/C (magnitude of the electric field)
r = 1.0 m (distance from the charge)
Solving the equation for q, we find the charge:
Answer:
a) W = 643.5 J, b) W = -427.4 J
Explanation:
a) Work is defined by
W = F. x = F x cos θ
in this case they ask us for the work done by the external force F = 165 N parallel to the ramp, therefore the angle between this force and the displacement is zero
W = F x
let's calculate
W = 165 3.9
W = 643.5 J
b) the work of the gravitational force, which is the weight of the body, in ramp problems the coordinate system is one axis parallel to the plane and the other perpendicular, let's use trigonometry to decompose the weight in these two axes
sin θ = Wₓ / W
cos θ = Wy / W
Wₓ = W sinθ = mg sin θ
Wy = W cos θ
the work carried out by each of these components is even Wₓ, it has to be antiparallel to the displacement, so the angle is zero
W = Wₓ x cos 180
W = - mg sin 34 x
let's calculate
W = -20 9.8 sin 34 3.9
W = -427.4 J
The work done by the component perpendicular to the plane is ero because the angle between the displacement and the weight component is 90º, so the cosine is zero.
