Can y’all please help
1 answer:
Answer:
<em>The cat's speed when it slid off the table was 2.53 m/s</em>
Explanation:
<u>Horizontal Motion</u>
When an object is thrown horizontally at a speed v, from a height h, it describes a curved path ruled by gravity, until it hits the ground.
the range or maximum horizontal distance traveled by the object can be calculated as follows:
The situation described in the problem fits into the concept of a horizontal launch when the cat slides off the table and gets to the ground some distance ahead.
It's given the range as d=1.3m and the height h=1.3 m, thus we can find the cat's speed by solving the equation for v:
Where
Substituting values:
Calculating:
v = 2.53 m/s
The cat's speed when it slid off the table was 2.53 m/s
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Given that,
Frequency emitted by the bat, f = 47.6 kHz
The speed off sound in air, v = 413 m/s
We need to find the wavelength detected by the bat. The speed of a wave is given by formula as follows :
or
So, the bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes i.e. 8.67 mm.
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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