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3 years ago

How fr do you think you would go on a car while sneezing for 2.5 seconds ?

Physics

2 answers:

Liula [17]3 years ago

7 0

the same speed. its only sneezing.

arsen [322]3 years ago

4 0

depends on the speed limit

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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

Which best describes the relationship between the direction of energy and wave motion in a transverse wave?

sammy [17]

I think the correct answer from the choices listed above is the second option. The relationship between the direction of energy and wave motion in a transverse wave would be the <span>energy direction is perpendicular to the motion of the wave. Hope this answers the question. Have a nice day.</span>

6 0

3 years ago

Read 2 more answers

What is the formula for silver nitrate

serious [3.7K]

Answer:

Explanation:

Agno3

8 0

4 years ago

The charges and coordinates of two charged particles held fixed in the xy plane are: q1 = +3.3 µc, x1 = 3.5 cm, y1 = 0.50 cm, an

Readme [11.4K]

1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is
$d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
substituting the coordinates of the two charges, we get
$d= \sqrt{(3.5+2)^2+(0.5-1.5)^2}=5.6~cm=0.056~m$

2) Then, we can calculate the electrostatic force between the two charges $q_1$ and $q_2$ , which is given by
$F=k_e \frac{q_1 q_2}{d^2}$
where $k_e=8.99\cdot10^{9} Nm^2C^{-2}$ is the Coulomb's constant.
Substituting numbers, we get
$F=8.99\cdot10^{9} Nm^2C^{-2} \frac{(3.3\cdot10^{-6}~C) (-4\cdot10^{-6}~C)}{(0.056~m)^2} =-37.8~N$
and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.

7 0

3 years ago

What is the magnitude of g at a height above Earth's surface where free-fall acceleration equals 6.5m/s^2?

prohojiy [21]

You've given the answer, right there in your question.

The "magnitude of gravity" is described in terms of the acceleration
due to it, and you just told us what that is.

We can also notice that the figure you gave is about 0.66 of the
acceleration due to gravity on the Earth's surface. That tells us that
the distance from the Earth's center at that height is about

(1 / √0.66) = 1.23 times

the Earth's radius, so the height is about 910 miles above the surface.

7 0

3 years ago