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Anna007 [38]
3 years ago
9

A wedge shaped air film is made between two sheets of glass using a spacer at one end of the glass sheets. If light of wavelengt

h 589 nm is used to illuminate the plates, how thick is the spacer if five dark fringes are observed between the edges of the glass
Physics
1 answer:
pychu [463]3 years ago
8 0

Answer:

1178 nm

Explanation:

We are given that

Wavelength of light=\lambda=589nm=589\times 10^{-9} m

1nm=10^{-9} m

We have to find the thickness of spacer if five dark fringes are observed between the edges of the glass.

Suppose that first dark fringe and fifth dark fringe near spacer, then the path length of light is 4 times the wavelength of light.

The light passes through air film is two times  then the change in air film thickness  from one edge to other is two times the wavelength of light.

Change in air film thickness  from one edge to other edge  is same as the thickness of spacer.

Therefore, thickness of spacer=2\lambda

Thickness of spacer=2\times 589\times 10^{-9}m

Thickness of spacer=1178 nm

Hence, the thickness of spacer=1178 nm

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6. An excited dog runs full-speed toward his owner who has just returned home from
mixas84 [53]

Answer:

v = 8.65 m/s

Explanation:

Given that,

Distance covered by the doge, d = 45 m

Time taken, t = 5.2 s

We need to find its average speed. The total distance covered divided by the total time taken is called the average speed of an object. So,

v=\dfrac{45\ m}{5.2\ s}\\\\=8.65\ m/s

So, the average speed is 8.65 m/s.

4 0
3 years ago
What Is a Sound Wave? Learning Goal: To understand the nature of a sound wave, including its properties: frequency wavelength, l
NISA [10]

Answer:

A)  Propagation of pressure fluctuations in a medium

B) air is the medium in which the wave is transported,

Explanation:

Part A.

A sound wave is a longitudinal oscillation of the molecules that forms in a material medium, they can be solid, liquid or gases, therefore the wave propagates in the same direction as the oscillation of the particles.

The most correct answer is:

* Propagation of pressure fluctuations in a medium

Part b

air is the medium in which the wave is transported, otherwise it cannot propagate

8 0
3 years ago
Is it proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinde
Gelneren [198K]

Answer:

No, it is not proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinder.

Explanation:

A cylinder is said to be infinitely long when is of a sufficient length. Also, when the diameter of the cylinder is relatively small compared to the length, it is called infinitely long cylinder.

Cylindrical rods can also be treated as infinitely long when dealing with heat transfers at locations far from the top or bottom surfaces. However, it not proper to treat the cylinder as being infinitely long when:

* When the diameter and length are comparable (i.e have the same measurement)

When finding the temperatures near the bottom or top of a cylinder, it is NOT PROPER TO USE AN INFINITELY LONG CYLINDER because heat transfer at those locations can be two-dimensional.

Therefore, the answer to the question is NO, since it is not proper to use an infinitely long cylinder when finding temperatures near the bottom or top of a cylinder.

8 0
3 years ago
a bus is moving at 22m/s [E] for 12s. Then the bus driver slows down at 1.2m/s2 [W] until the bus stops. Determine the total dis
KatRina [158]
The total displacement is equal to the total distance. For the east or E direction, the distance is determined using the equation:

d = vt = (22 m/s)(12 s) = 264 m

For the west or W direction, we use the equations:
a = (v - v₀)/t
d = v₀t + 0.5at²

Because the object slows down, the acceleration is negative. So,
-1.2 m/s² = (0 m/s - 22 m/s)/t
t = 18.33 seconds
d = (22 m/s)(18.33 s) + 0.5(-1.2 m/s²)(18.33 s)²
d = 201.67 m

Thus,
Total Displacement = 264 m +  201.67 m = 465.67 or  approximately 4.7×10² m.
7 0
3 years ago
You and a friend each hold a lump of wet clay. Each lump has a mass of 30 grams. You each toss your lump of clay into the air, w
Vesna [10]

Answer:

\ \text{m/s}

Explanation:

u_1 = Velocity of one lump = 3x+3y-3z

u_2 = Velocity of the other lump = -4x+0y-4z

m = Mass of each lump = 30\ \text{g}

The collision is perfectly inelastic as the lumps stick to each other so we have the relation

mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=\ \text{m/s}

The velocity of the stuck-together lump just after the collision is \ \text{m/s}.

4 0
3 years ago
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