Types of energy transfer involve
Conduction, Convection, Evaporation,and Condensation <span />
Answer:
the distance from charge A to C is r₁₃= 1.216 m
Explanation:
following Coulomb's law , the force exerted by 2 point charges between themselves is:
F= k*q₁*q₂/r₁₂² , where q is charge , r is distance and 1 and 2 represents the charge A and charge B respectively , k=constant
since C ( denoted as 3) is at equilibrium
F₁₃=F₂₃
k*q₁*q₃/r₁₃²=k*q₂*q₃/r₂₃²
q₁/r₁₃²=q₂/r₂₃²
r₁₃²/q₁=r₂₃²/q₂
r₂₃=r₁₃*√(q₂/q₁)
since C is at rest and is co linear with A and B ( otherwise it would receive a net force in either vertical or horizontal direction) , we have
r₁₃+r₂₃=d=r₁₂
r₁₃+r₁₃*√(q₂/q₁)=d
r₁₃*(1+√(q₂/q₁))=d
r₁₃=d/(1+√(q₂/q₁))
replacing values
r₁₃=d/(1+√(q₂/q₁)) = 3.00 m/(1+√(3.10 C/1.44 C)) = 1.216 m
thus the distance from charge A to C is r₁₃= 1.216 m
Answer:
94
Explanation:
f = 2.57 x 10^13 Hz
E = 10 eV = 10 x 1.6 x 10^-19 J = 1.6 x 10^-18 J
Energy of each photon = h f
Where, h is Plank's constant
Energy of each photon = 6.63 x 10^-34 x 2.57 x 10^13 = 1.7 x 10^-20 J
Number of photons = Total energy / energy of one photon
N = (1.6 x 10^-18) / (1.7 x 10^-20) = 94.11 = 94
Answer:
The object would weight 63 N on the Earth surface
Explanation:
We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:
Now, if the body is on the surface of the Earth, its weight (w) would be:
Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion:
