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Phoenix [80]
3 years ago
10

A rock is thrown from a bridge at an angle 30∘ below horizontal.immediately after the rock is released, is the magnitude of its

acceleration greater than, less than, or equal to g?
Physics
1 answer:
Wittaler [7]3 years ago
4 0
<span>The magnitude of the rock is equal to g. After the rock is released, there are no more forces acting on it, yet gravity remains. The initial inputs, on a bridge, at an angle of 30 deg below horizontal do not matter after the release.</span>
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An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r
bixtya [17]

1) Linear charge density of the shell:  -2.6\mu C/m

2)  x-component of the electric field at r = 8.7 cm: 1.16\cdot 10^6 N/C 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: 1.28\cdot 10^7 N/C 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

\lambda_2 = \rho A

where

\rho = -567\mu C/m^3 is charge volumetric density

A is the area of the cylindrical shell, which can be written as

A=\pi(b^2-a^2)

where

b=4.7 cm=0.047 m is the outer radius

a=2.7 cm=0.027 m is the inner radius

Therefore, we have :

\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m

 

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:

E=E_1+E_2

where:

E_1=\frac{\lambda_1}{2\pi r \epsilon_0}

where

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m 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

E_2=\frac{\lambda_2}{2\pi r \epsilon_0}

where

\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m

And this field points radially inward, because the charge is negative.

Therefore, the net field is

E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C

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 dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

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,

E_y=0

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:

E=\frac{\lambda_1}{2\pi \epsilon_0 r}

where:

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m 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,

E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C

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 dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago
To balance the forces on the box what direction must you push?
ss7ja [257]
The correct answer would be left
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What is the mathematical relationship among voltage current and resistance​
loris [4]

Answer:

The relationship between voltage, current, and resistance is described by Ohm's law. This equation, i = v/r, tells us that the current, i, flowing through a circuit is directly proportional to the voltage, v, and inversely proportional to the resistance, r.

5 0
3 years ago
According to Bode’s Law a planet is missing between Jupiter and Saturn. True False
defon
Bode law, a planet<span> was believed to exist </span>between<span> .... An Astronomer's Account of the </span>Missing Planet Between<span> Mars and </span>Jupiter<span> as Interpreted </span>Jupiter<span> ·</span>Saturn<span> · Uranus · Neptune.</span>
6 0
3 years ago
Hey help me plzzzzz i will mark brainliest​
vodomira [7]

Answer:

The answer to your question is given below.

Explanation:

Mechanical advantage (MA) = Load (L)/Effort (E)

MA = L/E

Velocity ratio (VR) = Distance moved by load (l) / Distance moved by effort (e)

VR = l/e

Efficiency = work done by machine (Wd) /work put into the machine (Wp) x 100

Efficiency = Wd/Wp x100

Recall:

Work = Force x distance

Therefore,

Work done by machine (wd) = load (L) x distance (l)

Wd = L x l

Work put into the machine (Wp) = effort (E) x distance (e)

Wp = E x e

Note: the load and effort are measured in Newton (N), while the distance is measured in metre (m)

Efficiency = Wd/Wp x100

Efficiency = (L x l) / (E x e) x 100

Rearrange

Efficiency = L/E ÷ l/e x 100

But:

MA = L/E

VR = l/e

Therefore,

Efficiency = L/E ÷ l/e x 100

Efficiency = MA ÷ VR x 100

Efficiency = MA / VR x 100

7 0
2 years ago
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