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Vinil7 [7]
2 years ago
7

What is the distance between lines on a diffraction grating that produces a second-order maximum for 760-nm red light at an angl

e of 60.0º
Physics
1 answer:
Nataly [62]2 years ago
8 0
The condition for maximum intensity is the same as that for a double slit. However, angular separation of the maxima is generally much greater because the slit spacing is so small for a diffraction grating. For a diffraction grating with lines/mm = lines/inch, the slit separation is d = micrometers = x10^ m.
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ahrayia [7]
1. 12m/s and here’s how. The equation for calculating velocity of a wave is lambda x frequency. So 2m x 6Hz. Hz can be 1/s so the only unit of measurement we get is meters and seconds. So the answer will be 12m/s.

2. Frequency=2Hz period= 0.5 seconds
Equation for frequency is velocity/wavelength. 10m/s divides by 5m = 2Hz.
Equation for period is 1 / frequency. 1 divided by 2 = 0.5 seconds

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7 0
3 years ago
Two charged particles exert an electric force of 27 N on each other. What will the magnitude of the force be if the distance bet
Elena-2011 [213]

Answer:

243 N

Explanation:

The formula for electromagnetic force is F= Kq1q2/r^2

where r is the distance between the charges, if the distance between the charges is reduced by 1/3 then F will increase by 9 [(1/3r)^2 becomes 1/9r which is 9F] so 27*9 is 243N

4 0
3 years ago
What is the most commonly used method of food preservation today? refrigeration drying foods smoking raw meats salt curing
dmitriy555 [2]
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6 0
3 years ago
Read 2 more answers
An iron ball and an aluminum ball of mass 100 g each are heated to the same temperature and then cooled to a temperature of 20°
MrRa [10]

Answer:

The specific heat of aluminum is greater.

Explanation:

It lost the most heat.

7 0
3 years ago
Read 2 more answers
The driver of a car slams on the brakes, causing the car to slow down at a rate of
sdas [7]

Answer:

A. The time taken for the car to stop is 3.14 secs

B. The initial velocity is 81.64 ft/s

Explanation:

Data obtained from the question include:

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Final velocity (V) = 0

Time (t) =?

Initial velocity (U) =?

A. Determination of the time taken for the car to stop.

Let us obtain an express for time (t)

Acceleration (a) = Velocity (V)/time(t)

a = V/t

Velocity (V) = distance (s) /time (t)

V = s/t

a = s/t^2

Cross multiply

a x t^2 = s

Divide both side by a

t^2 = s/a

Take the square root of both side

t = √(s/a)

Now we can obtain the time as follow

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Time (t) =..?

t = √(s/a)

t = √(256/26)

t = 3.14 secs

Therefore, the time taken for the car to stop is 3.14 secs

B. Determination of the initial speed of the car.

V = U + at

Final velocity (V) = 0

Deceleration (a) = –26ft/s2

Time (t) = 3.14 sec

Initial velocity (U) =.?

0 = U – 26x3.14

0 = U – 81.64

Collect like terms

U = 81.64 ft/s

Therefore, the initial velocity is 81.64 ft/s

7 0
3 years ago
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