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2 years ago

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Light of wavelength 560 nm passes through a slit of width 0. 170 mm. (a) the width of the central maximum on a screen is 8. 00 m

m. how far is the screen from the slit?

1 answer:

faltersainse [42]2 years ago

3 0

The distance between slit and the screen is 1.214m.

To find the answer, we have to know about the width of the central maximum.

<h3>How to find the distance between slit and the screen?</h3>

It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.

We have the expression for slit width w as,

$w=\frac{2*wavelength*d}{a}$

where, d is the distance between slit and the screen, and a is the slit width.

Thus, distance between slit and the screen is,

$d=\frac{w*a}{2*wavelength} =\frac{8*10^{-3}*0.17*10^{-3}}{560*10^{-9}*2} \\\\d=1.214m$

Thus, we can conclude that, the distance between slit and the screen is 1.214m.

Learn more about the width of the central maximum here:

brainly.com/question/13088191

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which planet should punch travel to if his goal is to weigh in at 118 lb? refer to the table of planetary masses and radii given

Harrizon [31]

The planet that Punch should travel to in order to weigh 118 lb is Pentune.

<h3 /><h3 /><h3>The given parameters:</h3>

Weight of Punch on Earth = 236 lb
Desired weight = 118 lb

The mass of Punch will be constant in every planet;

$W = mg\\\\m = \frac{W}{g}\\\\m = \frac{236}{g}$

The acceleration due to gravity of each planet with respect to Earth is calculated by using the following relationship;

$F = mg = \frac{GmM}{R^2} \\\\g = \frac{GM}{R^2}$

where;

M is the mass of Earth = 5.972 x 10²⁴ kg
R is the Radius of Earth = 6,371 km

For Planet Tehar;

$g_T =\frac{G \times 2.1M}{(0.8R)^2} \\\\g_T = 3.28(\frac{GM}{R^2} )\\\\g_T = 3.28 g$

For planet Loput:

$g_L =\frac{G \times 5.6M}{(1.7R)^2} \\\\g_L = 1.94(\frac{GM}{R^2} )\\\\g_L = 1.94g$

For planet Cremury:

$g_C =\frac{G \times 0.36M}{(0.3R)^2} \\\\g_C = 4(\frac{GM}{R^2} )\\\\g_C = 4 g$

For Planet Suven:

$g_s =\frac{G \times 12M}{(2.8R)^2} \\\\g_s = 1.53(\frac{GM}{R^2} )\\\\g_s = 1.53 g$

For Planet Pentune;

$g_P =\frac{G \times 8.3 }{(4.1R)^2} \\\\g_P = 0.5(\frac{GM}{R^2} )\\\\g_P = 0.5 g$

For Planet Rams;

$g_R =\frac{G \times 9.3M}{(4R)^2} \\\\g_R = 0.58(\frac{GM}{R^2} )\\\\g_R = 0.58 g$

The weight Punch on Each Planet at a constant mass is calculated as follows;

$W = mg\\\\W_T = mg_T\\\\W_T = \frac{236}{g} \times 3.28g = 774.08 \ lb\\\\W_L = \frac{236}{g} \times 1.94g =457.84 \ lb\\\\ W_C = \frac{236}{g}\times 4g = 944 \ lb \\\\ W_S = \frac{236}{g} \times 1.53g = 361.08 \ lb\\\\W_P = \frac{236}{g} \times 0.5 g = 118 \ lb\\\\W_R = \frac{236}{g} \times 0.58 g = 136.88 \ lb$

Thus, the planet that Punch should travel to in order to weigh 118 lb is Pentune.

<u>The </u><u>complete question</u><u> is below</u>:

Which planet should Punch travel to if his goal is to weigh in at 118 lb? Refer to the table of planetary masses and radii given to determine your answer.

Punch Taut is a down-on-his-luck heavyweight boxer. One day, he steps on the bathroom scale and "weighs in" at 236 lb. Unhappy with his recent bouts, Punch decides to go to a different planet where he would weigh in at 118 lb so that he can compete with the bantamweights who are not allowed to exceed 118 lb. His plan is to travel to Xobing, a newly discovered star with a planetary system. Here is a table listing the planets in that system (<em>find the image attached</em>).

<em>In the table, the mass and the radius of each planet are given in terms of the corresponding properties of the earth. For instance, Tehar has a mass equal to 2.1 earth masses and a radius equal to 0.80 earth radii.</em>

Learn more about effect of gravity on weight here: brainly.com/question/3908593

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Answer:

The initial velocity is 50 m/s.

(C) is correct option.

Explanation:

Given that,

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$h =\dfrac{v^2}{2g}$ ....(I)

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$h =ut+\dfrac{1}{2}gt_{2}^2$

$h=0+\dfrac{1}{2}gt_{2}^2$

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$t_{2}=\dfrac{v}{g}$

According to question,

$t_{1}+t_{2}=10$

$t_{1}=t_{2}$

Put the value of t₁ and t₂

$\dfrac{v}{g}+\dfrac{v}{g}=10$

$\dfrac{2v}{g}=10$

$v=\dfrac{10\times g}{2}$

Here, g = 10

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$v=\dfrac{10\times10}{2}$

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Positive feedback interactions in earth’s systems are always a result of human action is a FALSE statement.

<u>Explanation:</u>

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Human can alter, or can modify a very small part of the events that occur on earth’s surface. But they don’t have any control on what’s going inside earth’s core. So positive feedback interactions are not only the results of human interaction, but also different other factors.

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