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Alona [7]
3 years ago
15

Light from a laser of wavelength λ1 shines normally on a pair of narrow slits separated by distance D. This results in a differe

nce in angle between the central and first intensity maxima of 0.02 radians. A second laser of wavelength λ2=2λ1 then shines on the same slit system. The new angular difference between the central first intensity maxima is closest to:
Physics
1 answer:
lana66690 [7]3 years ago
6 0

Answer:

Explanation:

The angular position of first intensity maxima is given by the expression

= λ / d where λ is wave length of light and d is slit separation

putting the values given in the problem. ,

.02 = λ₁ / D

Now wavelength has been increased to 2λ₁

angular separation of first maxima

=  2λ₁ /D

= 2 x .02

= .04 radians.

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Water boiling<br><br> Which one is shown? (Look at pic)
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Answer:

conduction.

Explanation:

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3 years ago
How long would it take a drag racer to increase her speed from 10.m/s to 20 m/s if her car accelerates at a uniform rate of 15 m
garik1379 [7]
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3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
You use 35 J of energy to move an object 5 m. What is the weight of the object
Vlad [161]

Explanation:

35 ÷ 5 = 7kg.

=======================

7 0
2 years ago
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