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Leya [2.2K]
3 years ago
15

When a constant force acts on an object, what does the object's change in momentum depend upon?

Physics
1 answer:
Eva8 [605]3 years ago
8 0
You don’t need time (1) or the force (4) either so just 3,5
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A ball is let go and falls to the ground. It bounces a few times before
Ksenya-84 [330]

Answer:

I just want the points but you should pay attention in class jk

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3 years ago
Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec
AysviL [449]
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
E= \frac{\sigma}{2\epsilon_0}
where
\sigma is the charge density
\epsilon_0 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:
E=30 N/C
7 0
3 years ago
An automobile engine has an efficiency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine
katrin [286]

Answer:

If efficiency is .22 then  W = .22 * Q   where Q is the heat input

Heat Input    Q = 2510 / .22 = 11,400 J

Heat rejected = 11.400 - 2510 = 8900  J of heat wasted

Also, 8900 J / (4.19 J / cal) = 2120 cal

5 0
3 years ago
Two crates, one with mass 5.4 kg and the other with mass 8.2 kg, connected by a light rope. The coefficient of kinetic friction
Gnom [1K]

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

3 0
3 years ago
How does a wind turbine make electricity?
Marysya12 [62]

Answer:

1

Explanation:

7 0
3 years ago
Read 2 more answers
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