Kamal said the distance from the top of the balloon to the ground in the Example is Ï···· 353 ft. What mistake might Kamal have
1 answer:
Complete question is;
Kamal said the distance from the top of the balloon to the ground in the Example image attached is √353 ft. What mistake might Kamal have made?
Answer:
the mistake Kamal made is that she probably used 17 ft as the perpendicular side of the triangle with b as the hypotenuse instead of using 17ft as the hypotenuse
Explanation:
From the image attached, we can see that the distance from the top of the balloon which is blue in color to the ground is denoted by "b".
Now the triangle is a right angle triangle with hypotenuse = 15ft + 2ft = 17 ft; the adjacent side = 8 ft, while the opposite side is "b".
Thus, we can use pythagoras theorem to solve this as;
b = √(17² - 8²)
b = √(289 - 64)
b = √225
b = 15ft
However,we are told Kamal got b as √353 ft.
From inspection of the calculations we just did, if we had used addition instead of subtraction, we would have gotten b = √353 ft.
Thus, we can under that the mistake Kamal made is that she probably used 17 ft as the perpendicular side of the triangle with b as the hypotenuse instead of using 17ft as the hypotenuse.
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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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