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Morgarella [4.7K]
3 years ago
8

Suppose a uniform electric field of 4N/C is in the positive x direction. When a charge is placed and at a fixed to the origin, t

he resulting electric field on the x axis at x =2m becomes zero. What is the magnitude of the electric field at x =4m on the x axis at this time?
Physics
1 answer:
Levart [38]3 years ago
4 0

Answer:

3N/C

Explanation:

Thinking process:

Let, electric field of 4N/C is in the positive x direction, that is + 4 NC

And the the resulting electric field on the x axis at x =2m

Therefore, based on the data above, the net charge must be in the positive direction. Thus, the magnitude is given by the equation:

q = \frac{Er^{2} }{k} \\   = \frac{4 (4)}{9*10^{4} } \\   = \frac{16}{9}NC

Now considering that at x = + 4 NC, the net charge will be:

E_{net} = E_{0} + E_{q}   \\             = 4 -(\frac{9*10^{} }{16}*\frac{16}{9} * 10^{-90})\\             = 3 (N/C)

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Two adjacent natural frequencies of an organ pipe are AMT determined to be 550 Hz and 650 Hz. Calculate (a) the M fundamental fr
allochka39001 [22]

Answer:

The fundamental frequency and length of the pipe are 100 Hz and 1.7 m.

Explanation:

Given that,

Frequency f = 550 Hz

Frequency f' = 650 Hz

We know that,

AMT pipe is open pipe.

(b). We need to calculate the length of the pipe

Using formula of organ pipe

f=\dfrac{nv}{2L}

For 550 Hz,

550=\dfrac{n\times340}{2L}...(I)

For 650 Hz,

650=\dfrac{(n+1)\times340}{2L}...(II)

From equation (I) and (II)

550-650=\dfrac{340}{2L}-\dfrac{340}{L}

L=\dfrac{340}{2\times100}

L=1.7\ m

(a). We need to calculate the fundamental frequency for n = 1

Using formula of  fundamental frequency

=f=\dfrac{n\lambda}{2L}

put the value of L

f=\dfrac{1\times340}{2\times1.7}

f=100\ Hz

Hence, The fundamental frequency and length of the pipe are 100 Hz and 1.7 m.

4 0
3 years ago
b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul
Nat2105 [25]

Answer:

b) Betelgeuse would be \approx 1.43 \cdot 10^{6} times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is \approx 1.37 \cdot 10^{-5} } the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

B = \displaystyle{\frac{L}{4\pi d^2}} (1)

where B is the brightness, L is the star luminosity and d, the distance from the star to the point where the brightness is calculated (measured). Thus:

b) B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}} and B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}} where L_{Sun} is the Sun luminosity (3.9 x 10^{26} W) but we don't need to know this value for solving the problem. ly is light years.

Finding the ratio between the two brightness we get:

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 1.581 \cdot 10^{-5}\ ly. Then

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0
3 years ago
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