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3 years ago

9

You connect three resistors with resistances R, 2R, and 3R in parallel. The equivalent resistance of the three resistors will ha

ve a value that is

1 answer:

belka [17]3 years ago

7 0

Answer:

<h2>The equivalent is 6R/11</h2>

Explanation:

We know that the equivalent resistance of resistors connected in parallel is expressed as

$\frac{1}{Re} =\frac{1}{R1} +\frac{1}{R2}+\frac{1}{R3}\\\\\frac{1}{Re} =\frac{1}{R} +\frac{1}{2R}+\frac{1}{3R}\\$

the L.C.M is 6R

$\frac{1}{Re} =\frac{6+3+2}{6R} = \frac{11}{6R} \\\\Re= \frac{6R}{11}$

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When a burning stick of increase is moved fast in a circle a circle of red light is seen.

anzhelika [568]

Answer:

The impression of the image on the retina lasts for about 1/16th of a second after the removal of the object. If a burning stick of incense is revolved at a rate of more than sixteen revolutions per second, we see a circle of red light due to persistence of vision.

Explanation:

7 0

2 years ago

An athlete is running a 400m race around a 400m track. On the backstretch the athlete's velocity is 8m/s but he is running into

Aleksandr-060686 [28]

Answer:

33 N

Explanation:

v = Velocity of fluid = 8+2 = 10 m/s

$\rho$ = Density of fluid = 1.2 kg/m³

C = Coefficient of drag = 1.1

A = Cross sectional area = 0.5 m²

Drag force is given by

$F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2\times 1.1\times 0.5\times (8+2)^2\\\Rightarrow F=33\ N$

The drag force on the athlete is 33 N

3 0

2 years ago

What is the wavelength and frequency of a photon emitted by transition of an electron from a n- orbit to a n-1 orbit'?

PolarNik [594]

Answer:

$\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m$

$\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}$

Explanation:

$E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$n_i=n\ and\ n_f=n-1$

Thus solving it, we get:

$\Delta E=2.179\times 10^{-18}(\frac{1}{n^2} - \dfrac{1}{{(n-1)}^2})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{{(n-1)}^2-n^2}{{{(n-1)}^2}\times n^2}})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{n^2+1-2n-n^2}{{{(n-1)}^2}\times n^2}})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J$

Also, $\Delta E=\frac {h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

So,

$\frac {h\times c}{\lambda}=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J$

$\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2.179\times 10^{-18}}\times \frac {{{{(n-1)}^2}\times n^2}}{{1-2n}}\ m$

So,

$\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m$

Also, $\Delta E=h\times \nu$

So,

$h\times \nu=2.179\times 10^{-18}\frac{1-2n}{{{(n-1)}^2}\times n^2}}$

$\nu=\frac {2.179\times 10^{-18}}{6.626\times 10^{-34}}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}$

$\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}$

8 0

3 years ago

6. A student travels 4.0 m east, 8.0 m south, 4.0 m west, and finally 8.0 m north. What is her

Helen [10]

The student's path traces out a rectangle and she ends up at the same spot, so her <u>displacement</u> is 0 m.

On the other hand, she travels a <u>total distance</u> equal to the perimeter of that rectangle,

4.0 m + 8.0 m + 4.0 m + 8.0 m = 24 m

3 0

1 year ago

Add these two velocity vectors to find the magnitude of their resultant vector

a_sh-v [17]

<u>Answer</u>

4.6 m/s to the right.

<u>Explanation</u>

Velocity is a vector quantity. It has both magnitude and direction.

Taking the 5.8 m/s moving to the right as positive, then the 1.2 m/s moving to the left will be negative.

∴ 5.8 + -1.2 = 5.8 - 1.2

= 4.6 m/s.

The resultant velocity vector is 4.6 m/s to the right.

7 0

3 years ago