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1 year ago

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A + 8.42 nC point charge and a - 3.75 nC point charge are 2.73 cm apart. What is the electric field strength at the midpoint bet

ween the two charges?

1 answer:

aev [14]1 year ago

4 0

Given:

The charge q1 = 8.42 nC

The charge q2 = -3.75 nC

The distance between the charges is d = 2.73 cm

To find the electric field strength at the midpoint between the two charges.

Explanation:

The distance between the charges and the midpoint is d' =d/2

The electric field strength can be calculated by the formula

$E\text{ = }\frac{k|q|}{d^{\prime2}}$

The electric field strength at the midpoint due to the charge q1 will be

$\begin{gathered} E_1=\frac{9\times10^9\times8.42\times10^{-9}\text{ C}}{(\frac{2.73}{2}\times10^{-2})^2} \\ =\text{ 4.07}\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint due to the charge q2 will be

$\begin{gathered} E_2=\frac{9\times10^9\times3.75\times10^{-9}}{(\frac{2.73}{2}\times10^{-2})^2} \\ =1.8\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint between the charges will be

$\begin{gathered} E=E_1+E_2 \\ =4.07\times10^5+1.8\times10^5 \\ =\text{ 5.87}\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint between the two charges is 5.87 x 10^(5) m.

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Elena L [17]

The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

$\theta =37.01^{\circ}$

We can apply the first Newton's law in x and y-direction.

If we do a free body diagram of the system we will have:

x-direction

All the forces acting in this direction are:

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Where:

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T(2) is the tension due to the rope 2

Here we just conclude that T(1) = T(2)

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The forces in this direction are:

$T_{1}cos(\theta)+T_{2}cos(\theta)-W=0$ (2)

Here W is the weight of the steel beam.

We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.

Knowing that T(1) = T(2) and W = mg, we have:

$T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0$

$2T_{1}cos(\theta)-m_{steel}g=0$

$2T_{1}cos(\theta)=m_{steel}g$

T(1) must be equal to 5479 N, so we have:

$cos(\theta)=\frac{m_{steel}g}{2T_{1}}$

$cos(\theta)=\frac{892*9.81}{2*5479}$

$cos(\theta)=\frac{892*9.81}{2*5479}$

$cos(\theta)=0.80$

Therefore, the maximum angle allowed is θ = 37.01°.

You can learn more about tension here:

brainly.com/question/12797227

I hope it helps you!

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an empty boat floats in water with 10% of its volume submerged. and when it is loaded with 1200kg , the volume submerged will in

Lostsunrise [7]

Let volume of empty boat be = 100% = 1V

and mass of boat be M

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boat submerging.

M + 1200kg leads to 0.7V boat submerging.

This is 60%, 0.6 V increase

By comparison

(M+1200kg) * 0.1V = 0.7V * M

0.1M + 120kg = 0.7M

120kg = 0.7M - 0.1M

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Answer:

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Our values can be defined like this,

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