Answer: 1. the object is moving away from the origin
4. the object started at 2 meters
5. the object is traveling at a constant velocity
Explanation:
The similarities and the differences between gravitational and electric force are listed below
Explanation:
- The magnitude of the gravitational force between two objects is given by Newton's law of gravitation:
where
is the gravitational constant
are the masses of the two objects
r is the separation between them
- Coloumb's law gives instead the strength of the electrostatic force between two charged objects, which is
where:
is the Coulomb's constant
are the two charges
r is the separation between the two charges
By comparing the two equations, we find the following similarities:
- Both the forces are inversely proportional to the square of the distance between the two objects,
- Both the forces are proportional to the product between the "main quantity" of each force, which is the mass for the gravitational force (
) and the charge for the electric force (
Instead, we have the following differences:
- The gravitational force is always attractive, since the sign of
is always positive, while the electric force can be either attractive or repulsive, since the sign of 
can be either positive or negative - The value of the gravitational costant G is much smaller than the value of the Coulomb's constant, so the gravitational force is much weaker than the electric force
Learn more about gravitational force and electric force:
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Answer:
Explanation:
Given that
The mass of the body is 0.04kg
M=0.04kg
The radius of the paths is 0.6m
r=0.6m
The normal force exerted at A is 3.9N
Fa=3.9N
The normal force exerted at B is 0.69N
Fb=0.69N
Then work done by friction from point A to B will be the change in K.E
W=∆K.E+P.E
So we need to know the velocity at both point A and B
Then since the centripetal force is given as
Ft=mv²/r
Then,
For point A
Fa=mv²/r
3.9=0.04v²/0.6
3.9=0.0667v²
v²=3.9/0.0667
v²=58.5
v=√58.5
v=7.65m/s
Va=7.65m/s
Now at point B
Fb=mv²/r
0.69=0.04v²/0.6
0.69=0.0667v²
v²=0.69/0.0667
v²=10.35
v=√10.35
v=3.22m/s
Vb=3.22m/s
Then, the work done is
W=∆K.E+P.E
P.E is given as mgh
The height will be 2R =1.2m
P.E=mgh
P.E=0.04×9.81×1.2
P.E=0.471J
Final kinetic energy at B minus initial kinetic energy at A
W=K.Eb-K.Ea
K.E is given as 1/2mv²
W=1/2m(Vb²-Va²) +P.E
W=0.5×0.04(3.22²-7.65²) +0.471
W=0.5×0.04×(-48.1541) +0.471
W=-0.96+0.471
W=-0.49J
work was done on the block by friction during the motion of the block from point A to point B is 0.49J.
Friction opposes motions and that is why the work done is negative
The answer is C. 23.5°. <span>It's because of this </span>tilt<span> that the </span>Earth<span> experiences seasons as it orbits around the Sun. Imagine the Sun is at the center of a spinning record.</span>
1 Watt = 1 joule/second
650 watts = 650 joules/second
(650 J/sec) x (3,600 seconds/1 hour) = <em>2,340,000 Joules/hour</em>
