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Sedbober [7]
3 years ago
8

How can the distance between the first bright band and the central band be increased in a double-slit experiment?

Physics
1 answer:
Strike441 [17]3 years ago
8 0

Answer:

Decrease the slit separation, increase the distance of the screen from the slits, and increase the wavelength.

Explanation:

The distance y from the central band to the first bright band is given by

y = \dfrac{ \lambda D}{d}

where \lambda is the wavelength of light (or any particle), D is the distance to the screen, and d is the slit separation.

From this equation we see that, by increasing the wavelength \lambda , increasing the distance from the screen D, and decreasing the slit separation d, we increase the distance between the first bright band and the central band.

Therefore, the 2nd choice "<em>Decrease the slit separation, increase the distance of the screen from the slits, and increase the wavelength.</em>" is correct.

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Answer:C

Explanation:

5 0
3 years ago
Assume that the driver begins to brake the car when the distance to the wall is d=107m, and take the car's mass as m-1400kg, its
Evgen [1.6K]

Answer:

Explanation:

a ) Let let the frictional force needed be F

Work done by frictional force = kinetic energy of car

F x 107 = 1/2 x 1400 x 35²

F = 8014 N

b )

maximum possible static friction

= μ mg

where μ is coefficient of static friction

= .5 x 1400 x 9.8

= 6860 N

c )

work done by friction for μ = .4

= .4 x 1400 x 9.8 x 107

= 587216 J

Initial Kinetic energy

= .5 x 1400 x 35 x 35

= 857500 J

Kinetic energy at the at of collision

= 857500 - 587216

= 270284 J

So , if v be the velocity at the time of collision

1/2 mv² = 270284

v = 19.65 m /s

d ) centripetal force required

= mv₀² / d which will be provided by frictional force

= (1400 x 35 x 35) / 107

= 16028 N

Maximum frictional force possible

= μmg

= .5 x 1400 x 9.8

= 6860 N

So this is not possible.

4 0
3 years ago
Metals can be described as?
luda_lava [24]

They tend to be lustrous, ductile, malleable, and good conductors of electricity, while nonmetals are generally brittle (for solid nonmetals), lack lustre, and are insulators. Use google.

6 0
3 years ago
Snail 3, trying to keep up with Snail 2, managed to get to
Kipish [7]

Answer:

Acceleration is the change in velocity over the change in time = Δv/Δt.  To do these problems, you need to find out how much the speed changed and over what period of time it changed.

Snail 1 changes from 4 cm/min to 7 cm/min in 3 minutes.  Subtract the starting velocity (4 cm/min) from the ending velocity (7 cm/min) then divide by the time (3 min):

Snail 1 = (7 cm/min. - 4 cm/min)/(3 minutes) = ?    (remember to put down the units)

Snail 2 changed from 7 cm/min. down to 1 cm/min. in 3 minutes

Snail 2 = (1 cm/min. - 7 cm/min.)/(3 min.) = ?        (note that the acceleration is negative when you slow down)

I hope this helps you

6 0
3 years ago
A 15 kilogram mass is traveling with a velocity of 2.7m/s. What is the object's kinetic energy?
Usimov [2.4K]

Answer:

KE=1/2*m*v^2

KE=1/2*15*2.7^2

KE=54.675 J

Explanation:

4 0
3 years ago
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