Answer:
The surface gravity g of the planet is 1/4 of the surface gravity on earth.
Explanation:
Surface gravity is given by the following formula:
So the gravity of both the earth and the planet is written in terms of their own radius, so we get:
The problem tells us the radius of the planet is twice that of the radius on earth, so:
If we substituted that into the gravity of the planet equation we would end up with the following formula:
Which yields:
So we can now compare the two gravities:
When simplifying the ratio we end up with:
So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.
acceleration = Velocity changes ÷ time of the velocity changes
4 m/s^2 =
4 × 10^(-3) × 3600 km / h =
4 × 3.6 =
14.4 km / h
Thus :
14.4 = V(2) - V(1) / t(2) - t(1)
14.4 = V(2) - 20 / 10
Multiply both sides by 10
10 × 14.4 = 10 × ( V(2) - 20 ) / 10
144 = V(2) - 20
Add both sides 20
144 + 20 = V(2) - 20 + 20
V(2) = 164 Km/h
Thus the final velocity after 10 seconds is 164 Km/h .
Answer:
they're more inclined to be violent so A
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