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3 years ago

Choose correct answer

Physics

2 answers:

dsp733 years ago

8 0

Answer: The answer would be the primary purpose is to construct a logical argument.

rodikova [14]3 years ago

7 0

C is the correct answer urwelcom

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How can you tell the direction a river flows

NARA [144]

Answer: Regarding the image posted, a unicellular paramecium relies on mitosis for reproduction.

I hope this helps!

4 0

3 years ago

Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’2) mm apart and then falls on a screen 7

IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

$sin \theta=\frac{n \lambda}{d}$

where

n is the order of the maximum

$\lambda$ is the wavelength

$a$ is the distance between the slits

In this problem,

n = 5

$\lambda=525 nm =5.25\cdot 10^{-7} m$

$a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m$

So we find

$\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}$

And given the distance of the screen from the slits,

$D=75.0 cm = 0.75 m$

The distance of the 5th bright fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm$

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

$sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}$

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

$\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}$

And the distance of the 8th dark fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm$

5 0

3 years ago

Being a living example of a tobacco-free lifestyle can __________.

anygoal [31]

Help you live longer.

3 0

4 years ago

Read 2 more answers

A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti

Ugo [173]

Use Newton's second law to determine the acceleration being applied to the sled. There are three forces at work on the sled (its weight, the force normal to the ground, and friction) but two of them cancel, leaving friction as the only effective force. This vector is pointed in the opposite direction of the sled's movement, so if we take the direction of its movement to be the positive axis, we would find the acceleration due to the friction to be

$\vec F_G+\vec F_N+\vec F_F=m\vec a\iff-12\,\mathrm N=(20\,\mathrm{kg})a\implies a=-0.6\,\dfrac{\rm m}{\mathrm s^2}$

Now we use the formula

${v_f}^2-{v_i}^2=2a(x_f-x_i)$

to find the distance it travels. The sled comes to a rest, so $v_f=0$ , and let's take the starting position $x_i=0$ to be the origin. Then the distance traveled $x_f-x_i=x_f$ is

$-\left(4.5\,\dfrac{\rm m}{\rm s}\right)^2=2\left(-0.6\,\dfrac{\rm m}{\mathrm s^2}\right)x_f\implies x_f\approx17\,\mathrm m$

6 0

3 years ago

The potential energy of an object attached to a spring is 2.60 J at a location where the kinetic energy is 1.40 J. If the amplit

anyanavicka [17]

Answer:

This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant.

U = (1/2)kx^2

U = (1/2)(5.3)(3.62-2.60)^2

U = 2.75706 J

Read more on Brainstorm - httpd://brainstorm/question/7981441#read more

Explanation:

6 0

3 years ago