A ball is kicked from the top of a building with a velocity of 50 m/s and lands 165 m away from the base of the building.
1 answer:
Answer:
3.1 s
Explanation:
we can assume that this motion has a horizontal projectile motion.
projectile motion happens when an object is subjected to only gravity. Gravity acts to influence the vertical motion of the projectile, it makes a vertical acceleration.
The horizontal motion of the projectile is the result of the constant velocity of the object.We assume no acceleration in the horizontal direction.
applying motion equations
where
s - distance
u - initial velocity
a-acceleration
t - time
s = ut + (1/2)at²
→ 165 = 50t + 0 (as no acceleration is in horizontal direction)
t = 3.1 s
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The correct option is C) The angle between the vectors is 120°.
Why?
We can solve the problem and find the correct option using the Law of Cosine.
Let A and B, the given two sides and R the resultant (sum),
Then,
So, using the law of cosines, we have:
Hence, we have that the angle between the vectors is 120°. The correct option is C) The angle between the vectors is 120°
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1) Linear charge density of the shell:
2) x-component of the electric field at r = 8.7 cm: outward
3) y-component of the electric field at r =8.7 cm: 0
4) x-component of the electric field at r = 1.15 cm: outward
5) y-component of the electric field at r = 1.15 cm: 0
Explanation:
1)
The linear charge density of the cylindrical insulating shell can be found by using
where
is charge volumetric density
A is the area of the cylindrical shell, which can be written as
where
is the outer radius
is the inner radius
Therefore, we have :
2)
Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.
The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:
where:
where
is the linear charge density of the wire
r = 8.7 cm = 0.087 m is the distance from the axis
And this field points radially outward, since the charge is positive .
And
where
And this field points radially inward, because the charge is negative.
Therefore, the net field is
in the outward direction.
3)
To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.
However, we notice that since the wire is infinite, for the element of electric field produced by a certain amount of charge
along the wire there exist always another piece of charge
on the opposite side of the wire that produce an element of electric field
, equal and opposite to
.
Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.
We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,
4)
Here we want to find the x-component of the electric field at a point at
r = 1.15 cm
from the central axis.
We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.
Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:
where:
is the linear charge density of the wire
r = 1.15 cm = 0.0115 m is the distance from the axis
This field points radially outward, since the charge is positive . Therefore,
5)
For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field produced by a certain amount of charge
along the wire there exist always another piece of charge
on the opposite side of the wire that produce an element of electric field
, equal and opposite to
.
Therefore, the y-component of the electric field is zero.
Learn more about electric field:
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Answer:
No, the volume don't affect the potential energy.
Explanation:
The volume does not affect the potential energy, as this energy depends on the mass and elevation of the body relative to the reference point. This analysis can be easily seen in the equation expressing potential energy