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3 years ago

Which species below is an ionic compound? A) NaF B) CO2 C) AlAs D) OF2

Physics

2 answers:

Svet_ta [14]3 years ago

8 0

A The answer to the question A)NaF

marysya [2.9K]3 years ago

6 0

A) NaF is the answer

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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

When air resistance balances the weight of an object that is falling, the velocity _____.

miskamm [114]

Decreases. Air resistance will slow a falling object to its terminal velocity, placing a limit on its acceleration.

5 0

3 years ago

The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?

Yuki888 [10]

Answer:

-.457 m/s^2

Explanation:

Actual weight = 60 .3 (9.81) = 591.54 N

Accel of lift changes this to 60.3 ( 9.81 - L) where L - accel of lift

60.3 ( 9.81 - L ) = 564

solve for L = .457 m/s^2 DOWNWARD

so L = - .457 m/s^2

4 0

1 year ago

A car travels up a hill at a constant speed of 38 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the a

mojhsa [17]

Answer:

Average speed will be 48.23 km/h

Explanation:

Let the distance up to hill is = d km

Speed when car goes to hill = 38 km/h

So time required $t=\frac{distance}{speed}=\frac{d}{38}hour$

Speed when car return from hill = 66 km/h

So time required to return fro hill $t=\frac{d}{66}h$

Total time $t_{total}=\frac{t}{38}+\frac{t}{66}$

Total distance = d+d =2d

So average speed $=\frac{total\ distance}{total\ time}=\frac{2d}{\frac{d}{38}+\frac{d}{66}}=48.23km/h$

8 0

3 years ago

Two separate masses on two separate springs undergo simple harmonic motion indefinitely (the surface is frictionless). In CASE 1

Artemon [7]

Answer:

$\frac{F_1}{F_2} =\frac{1}{2}$

Explanation:

Assuming the we have to find ratio maximum forces on the mass in each case

we know that in a spring mass system

F= Kx

K= spring constant

x= spring displacement

Case 1:

$F_1=2k\times d$

case 2:

$F_2=2k\times 2d$

therefore, $\frac{F_1}{F_2} = \frac{2K\times d}{2K\times 2d}$

$\frac{F_1}{F_2} =\frac{1}{2}$

4 0

3 years ago