Two point charges exert a 5.50 N force on each other. What will the force become if the distance between them is increased by a
1 answer:
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Answer:
v=12.5 i + 12.5 j m/s
Explanation:
Given that
m₁=m₂ = m
m₃ = 2 m
Given that speed of the two pieces
u₁=- 25 j m/s
u₂ =- 25 i m/s
Lets take the speed of the third mass = v m/s
From linear momentum conservation
Pi= Pf
0 = m₁u₁+m₂u₂ + m₃ v
0 = -25 j m - 25 i m + 2 m v
2 v=25 j + 25 i m/s
v=12.5 i + 12.5 j m/s
Therefore the speed of the third mass will be v=12.5 i + 12.5 j m/s
Answer:
Explanation:
The change in potential energy can be expressed as:
where K is a constant with a value of , q1 and q2 are the charges of the proton and the electron and r is the distance between them.
The charge for the proton is and the charge for the electron is
.
Converting r=1.0nm to m:
Replacing values:
Answer:
The volume of water displaced by the raft is 0.233 m³
Explanation:
The question relates to Archimedes' principle which states that the buoyant force experienced by an object immersed in a fluid is equal to the weight of (the force of gravity on) the displaced fluid
The given parameters are;
The combined mass of the person and the raft, m = 233 kg
The liquid on which the raft is located = Water
The density of water, = 1000 kg/m³
Weight = Mass, m × g
Where;
m = The mass of the object
g = The acceleration due to gravity = 9.8 m/s²
Given that the raft is on the surface of the water (floating), the buoyant force is equal to the combined weight of the person and the raft = 233 kg
The combined weight of the person and the raft, = 233 kg × 9.8 m/s² = 2,283.4 N
Therefore;
The buoyant force = 2,283.4 N = The weight of the water displaced
The mass of the water displaced, , = 2,283.4 N/(9.8 m/s²) = 233 kg
Density = Mass/Volume
The volume of water displaced by the raft = The mass of the water displaced/(The density of the water) = 233 kg/(1,000 kg/m³) = 0.233 m³.