. Find the magnitude of the force between two charges, where the first charge is -2µc and the second charge is 3.5µc separated b
1 answer:
You might be interested in
Answer:
21.8°
Explanation:
Let's call θ the angle between BC and the horizontal.
Draw a free body diagram for each block.
There are 4 forces acting on block D:
Weight force P pulling down,
Normal force N₁ pushing perpendicular to AB,
Friction force N₁μ pushing parallel up AB,
and tension force T pushing parallel up AB.
There are 4 forces acting on block E:
Weight force P pulling down,
Normal force N₂ pushing perpendicular to BC,
Friction force N₂μ pushing parallel to BC,
and tension force T pulling parallel to BC.
Sum of forces on D in the perpendicular direction:
∑F = ma
N₁ − P sin θ = 0
N₁ = P sin θ
Sum of forces on D in the parallel direction:
∑F = ma
T + N₁μ − P cos θ = 0
T = P cos θ − N₁μ
T = P cos θ − P sin θ μ
T = P (cos θ − sin θ μ)
Sum of forces on E in the perpendicular direction:
∑F = ma
N₂ − P cos θ = 0
N₂ = P cos θ
Sum of forces on E in the parallel direction:
∑F = ma
N₂μ + P sin θ − T = 0
T = N₂μ + P sin θ
T = P cos θ μ + P sin θ
T = P (cos θ μ + sin θ)
Set equal:
P (cos θ − sin θ μ) = P (cos θ μ + sin θ)
cos θ − sin θ μ = cos θ μ + sin θ
1 − tan θ μ = μ + tan θ
1 − μ = tan θ μ + tan θ
1 − μ = tan θ (μ + 1)
tan θ = (1 − μ) / (1 + μ)
Plug in values:
tan θ = (1 − 0.4) / (1 + 0.4)
θ = 23.2°
∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.
Answer:
Part A:
Distance=864000 m=864 km
Part B:
Energy Used=ΔE=8638000 Joules
Part C:
Explanation:
Given Data:
v=20m/s
Time =t=12 hours
In Secs:
Time=12*60*60=43200 secs
Solution:
Part A:
Distance = Speed**Time
Distance=v*t
Distance= 20*43200
Distance=864000 m=864 km
Part B:
Energy Used=ΔE= Energy Required-Kinetic Energy of swans
Energy Required to move= Power Required*time
Energy Required to move=200*43200=8640000 Joules
Kinetic Energy=
Energy Used=ΔE=8640000 -2000
Energy Used=ΔE=8638000 Joules
Part C:
Fraction of Mass used=Δm/m
For This first calculate fraction of energy used:
Fraction of energy=ΔE/Energy required to move
ΔE is calculated in part B
Fraction of energy=8638000/8640000
Fraction of energy=0.99977
Kinetic Energy=
Now, the relation between energies ratio and masses is: