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musickatia [10]
3 years ago
11

if a force of 50N is used to pull a box along the ground, at a distance of 8m. The box moves in the same direction as the force.

Calculate the work done.​
Physics
1 answer:
ruslelena [56]3 years ago
5 0

Answer:

<h2>400 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 50 × 8

We have the final answer as

<h3>400 J</h3>

Hope this helps you

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There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is
Mademuasel [1]

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹  = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

4 0
3 years ago
The coordinates of a bird flying in the xy-plane are given by x(t)=αt and y(t)=3.0m−βt2, where α=2.4m/s and β=1.2m/s2.part a:Cal
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x(t)=at

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vectorV=[α;-2βt]

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ay(t)=-2βt

vector a [0;-2βt]


3 0
3 years ago
If the light wave has a wavelength of 10m what would be its velocity
s2008m [1.1K]

If this case could ever happen, the speed would follow from this formula:

v = f \cdot \lambda

with f the frequency and lambda the wavelength. We are give a wavelength of 10m. The frequencies of the visible light can range between 400 to about 790 Terahertz, so let us pick a middle point of 600 THz ("green-ish") as a "representative."

v = 600THz\cdot 10m = 6\cdot 10^{14} \frac{1}{s}\cdot 10 m = 6\cdot10^{15}\frac{m}{s}

The speed of such a wave would have to be 6e+15 m/s (which would be 7 orders of magnitude higher than the universal speed of light constant)

7 0
3 years ago
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the clown can be put on a sealion and the other clown can hold a lot of peole  he help a clown that was tall and heivy

Explanation:

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A car accelerates from 10 km/hr to 50 km/hr in 8 seconds. What is the acceleration?
nikklg [1K]
The answer is 5 km/hr

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