Answer:
Diameter decreases by the diameter of 0.0312 m.
Explanation:
Given that,
Bulk modulus = 14.0 × 10¹⁰ N/m²
Diameter d = 2.20 m
Depth = 2.40 km
Pressure = ρ g h = 1030 × 9.81 × 2.4 × 1000
= 24.25 × 10⁶ N/m²
Volume =
Bulk modulus is equal to
now
Δ r = -0.0156 m
change in diameter
Δ d = -2 × 0.0156
Δ d = -0.0312 m
Diameter decreases by the diameter of 0.0312 m.
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 and Explanation:
1. Evaluate the function x(t) at t=0.5
2. The period of motion T can be calculated as:
Where:
So:
3. The angular frequency can be expressed as:
Solving for k:
4. Find the derivate of x(t):
Now, the sine function reach its maximum value at π/2 so:
Solving for t:
Evaluating v(t) for 0.6603981634:
So the maximum speed of the block is:
In the negative direction of x-axis
5. The force is given by:
The cosine function reach its maximum value at 2π so:
Solving for t:
Evaluating x(t) for 3.016592654:
Therefore the the maximum force on the block is: