When a constant force acts on an object, what does the object's change in momentum depend upon?
1 answer:
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The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
where
is the charge density
is the vacuum permittivity
We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
where
We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
Answer:
R= 2.5 :ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks
Explanation:
We apply Newton's second law:
∑F=m*a
velocity is constant ,then , a=0
Nomenclature
W: weight
m: mass
N : normal force
Ff: Friction force
μk: coefficient of kinetic friction
T: tension force in the rope
F: applied horizontal force
g: acceleration due to gravity.
Force Calculation
W₁=m₁*g=5.4 kg *9.8m/s²=52.92 N
W₂=m₂*g=8.2 kg *9.8m/s²= 80.36N
∑Fy=0
N₁-W₁=0 , N₁=W₁ = 52.92 N
N₂-W₂=0, N₂=W₂=80.36N
Ff₁= μk* N₁=0.4*52.92 N = 21.16N
Ff₂= μk* N₂=0.4*80.36N = 32.14N
Look at the attached graphic
Free-Body diagram m₁=5.4 kg
∑Fx=0
T- Ff₁=0 , T= Ff₁ , T= 21.16N
Free-Body diagram m₂=8.2 kg
∑Fx=0
F-T- Ff₂=0 , F=T+Ff₂= 21.16N+32.14N=53.3N
Ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks (R)
R= F/T= 53.3N/21.16N = 2.5
