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
<span>Most objects tend to contain the same numbers of positive and negative charge because this is the most stable situation. In fact, if an object has an excess of positive charge, it tends to attract an equal number of negative charges to balance this effect and restore neutrality: the attracted negative charges combine with the excess of positive charges, leaving the object electrically neutral.</span>
Answer:
23.1 N/C
Explanation:
OP = 3 m , OQ = 4 m
q = - 8 nC, Q = 75 nC
Electric field at P due to the charge Q is
Electric field at P due to the charge q is
According to the diagram, tanθ = 3/4
Resolve the components of E1 along x axis and along y axis.
So, Electric field along X axis, Ex = - E1 Cos θ
Ex = - 27 x 4 / 5 = - 21.6 N/C
Electric field along y axis, Ey = E1 Sinθ - E2
Ey = 27 x 3 /5 - 8 = 8.2 N/C
The resultant electric field at P is given by
Answer: Mass of the planet, M= 8.53 x 10^8kg
Explanation:
Given Radius = 2.0 x 106m
Period T = 7h 11m
Using the third law of kepler's equation which states that the square of the orbital period of any planet is proportional to the cube of the semi-major axis of its orbit.
This is represented by the equation
T^2 = ( 4π^2/GM) R^3
Where T is the period in seconds
T = (7h x 60m + 11m)(60 sec)
= 25860 sec
G represents the gravitational constant
= 6.6 x 10^-11 N.m^2/kg^2 and M is the mass of the planet
Making M the subject of the formula,
M = (4π^2/G)*R^3/T^2
M = (4π^2/ 6.6 x10^-11)*(2×106m)^3(25860s)^2
Therefore Mass of the planet, M= 8.53 x 10^8kg
Answer:
4. The direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period
Explanation:
The creosote bush depends on sunlight to produce the food they require through photosynthesis. The shade from the solar panels would reduce the amount of sunlight that the bush receives. This would increase the mortality of the bush.
In order to test the hypothesis the student must record the direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period. If the plants receive sunlight less than the above amount the plants should start dying. If not then the hypothesis is false.
Hence, the answer is 4. The direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period.
