The acceleration that the same force will provide if both masses are tied together is; 6.0 m/s².
<h3>How to find the Acceleration?</h3>
We are given;
Force; F = 5 N
Acceleration of the first mass, a₁ = 8.0 m/s²
Acceleration of the second mass, a₂ = 24 m/s²
Formula for force is;
F = ma
Let us find both masses; m₁ and m₂.
m₁ = F/a₁
m₂ = F/a₂
Thus;
m₁ = 5/8 kg
m₂ = 5/24 kg
Total mass is; m = m₁ + m₂
m = 5/8 + 5/24
m = 15 + 5/24
m = 20/24 kg
Thus, acceleration if they are both tied together is;
a = F/m
a = 5/(20/24)
a = 6.0 m/s².
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10 kilograms of mass weighs 98.1 newtons on Earth,
16.2 newtons on the Moon, 37.1 newtons on Mars,
and other weights in other places.
Answer:
F = GMmx/[√(a² + x²)]³
Explanation:
The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is
dF = GmdM/L²
Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.
So, the horizontal components add from two symmetrically opposite mass elements dM,
Thus, the horizontal component of the force is
dF' = dFcosФ where Ф is the angle between L and the x axis
dF' = GmdMcosФ/L²
L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.
L = √(a² + x²)
cosФ = x/L
dF' = GmdMcosФ/L²
dF' = GmdMx/L³
dF' = GmdMx/[√(a² + x²)]³
Integrating both sides we have
∫dF' = ∫GmdMx/[√(a² + x²)]³
∫dF' = Gm∫dMx/[√(a² + x²)]³ ∫dM = M
F = GmMx/[√(a² + x²)]³
F = GMmx/[√(a² + x²)]³
So, the force due to the sphere of mass m is
F = GMmx/[√(a² + x²)]³
Answer:
See explaination
Explanation:
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Answer:
b. Relates the electric field at points on a closed surface to the net charge enclosed by that surface
Explanation:
Gauss's law states that the flux of certain fields through a closed surface is proportional to the magnitude of the sources of that field within the same surface. The electric flux expresses the measure of the electric field that crosses a certain surface. Therefore, the electric field on a closed surface is proportional to the net charge enclosed by that surface.
