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:
resistor R₂ has the lowest current density
Explanation:
The current density is
j = I / A
now let's analyze each case
a) R₂ has an area 2A₀ and a length L₀ that R₁
b) R₃ has an area Ao and a length 3L₀ what R₁
we can see that all the area is given in relation to the resistance R₁
the current density in R₁ is
j₁ = I / A₀
the current density in R₂
j₂ = I / 2A₀
j₂ 2 = ½ I/A₀
the current density in R₃
j₃ = I / A₀
j₂ < j₁ = j₃
therefore resistor R₂ has the lowest current density
Answer:
C) The restoring force
Explanation:
Hooke's Law states that the restoring force acting on a spring is given by the equation:

where
k is the spring constant
x is the displacement of the spring from its equilibrium position
The negative sign in the equation tells the direction of the restoring force. In fact, this force tends to bring the spring back to its equilibrium position: so, the force is always in opposite direction to the displacement.
This means that when the spring is stretched to the right, the restoring force tends to bring it back to the left, to the equibrium position; if the spring is compressed to the left, the restoring force tends to bring it back to the right, to the equilibrium position.
So the correct option is
C) The restoring force
Answer:
1302 K or 1029 C
Explanation:
Air at atmospheric pressure has pressure of 1 atm
20 C = 20 + 273 = 293 K
Assume ideal gas, according to the ideal gas law:

Where P1, V1 and T1 are the pressure, volume and temperature of the gas before the compression and P2, V2 and T2 are the pressure, volume and temperature of the gas after the compression

Since the gas is compressed to 1/9 of its original volume, V2/V1 = 1/9:
or 1029 C