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Dafna1 [17]
3 years ago
5

A particular HeNe laser beam at 633 nm has a spot size of 0.8 mm. Assuming a

Physics
1 answer:
Kobotan [32]3 years ago
8 0

Answer:

Explanation:

A

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Two loudspeakers (A and B) are 3.20m apart and emitting a sound with a frequency of 400Hz. An observer is 2.10m directly in fron
TiliK225 [7]

Answer:

The observer hears a loud sound

Explanation:

In order to know if the observer hears a loud or a quiet sound, you need to know if there is a constructive or destructive interference between the sound waves of the loudspeakers.

You first calculate the distance between the observer and the loudspeakers.

The distances are given by:

d1: distance to loudspeaker A = 2.10m

d2: distance to loudspeaker B

d_2=\sqrt{(3.20m)^2+(2.10m)^2}=3.827m

Next, you calculate the wavelength of the sound waves by using the following formula:

\lambda=\frac{v_s}{f}

vs: speed of sound =  343 m/s

f: frequency of the waves = 400Hz

λ: wavelength

\lambda=\frac{343m/s}{400Hz}=0.8575m

Next, you calculate the path difference between the distance from the observer to the loudspeakers:

\Delta d=3.827m-2.10m=1.727m

You obtain a constructive interference (loud sound) if the quotient between the wavelength of the sound and the difference path is an integer:

\frac{\Delta d}{\lambda}=\frac{1.727m}{0.857}\approx2

Then, there will be a constructive interference, and the sound who the observer hears is loud.

5 0
3 years ago
Why are people so rude???
DanielleElmas [232]

Answer:

I.D.K but the same measure they use to judge will be used to judge them

Explanation:

6 0
3 years ago
Read 2 more answers
Why is it safe to watch an eclipse of the Moon but not an eclipse of the Sun
gavmur [86]
A solar eclipse occurs when the moon crosses in front of the Sun, blocking some or all of its rays. A lunar eclipse happens when the moon is directly behind the earth, blocking the moon from receiving light. The only light comes from the light on earth's reflected shadow.

You can look at a lunar eclipse because there is very little light or none at all. You can't look at a solar eclipse because you are looking directly at the sun unless it is complete. Before totality, only some of the Sun is blocked, causing your pupils dilate to let in more light. Since they do this, more of the Sun's rays can be let in to the eye, which effectively allows your eyes to burn. 

Some doctors and eye care specialists say that after someone complains of blindness after looking at a solar eclipse unaided, they can see what the Sun and moon looked like at the time that they looked at it, as it is burned onto their retinas. 
8 0
3 years ago
A copper telephone wire has essentially
Lunna [17]

Answer:

128.21 m

Explanation:

The following data were obtained from the question:

Initial temperature (θ₁) = 4 °C

Final temperature (θ₂) = 43 °C

Change in length (ΔL) = 8.5 cm

Coefficient of linear expansion (α) = 17×10¯⁶ K¯¹)

Original length (L₁) =.?

The original length can be obtained as follow:

α = ΔL / L₁(θ₂ – θ₁)

17×10¯⁶ = 8.5 / L₁(43 – 4)

17×10¯⁶ = 8.5 / L₁(39)

17×10¯⁶ = 8.5 / 39L₁

Cross multiply

17×10¯⁶ × 39L₁ = 8.5

6.63×10¯⁴ L₁ = 8.5

Divide both side by 6.63×10¯⁴

L₁ = 8.5 / 6.63×10¯⁴

L₁ = 12820.51 cm

Finally, we shall convert 12820.51 cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

12820.51 cm = 12820.51 cm × 1 m / 100 cm

12820.51 cm = 128.21 m

Thus, the original length of the wire is 128.21 m

5 0
3 years ago
onsider two projectiles that can be shot upward by spring guns. Object A is made of solid aluminum and has a mass of 50 grams. O
BARSIC [14]

Answer:

B

Explanation:

From Newton's law of motion, we have:

V^2 = U^2 + 2gH

Where V and U are final and initial velocity respectively.

H is the height.

For the object to have a sustain a maximum height it means the final velocity of the object is zero.

By computing the height of the object sustain by A, we have:

0^2 = 2^2 -2×10×H

0= 4 -20H

4 = 20H;

H= 0.2m

For object B we have;

0^2 = 1^2 -2×10×H

0 = 1 -20H

H = 1/20= 0.05m

From computing the height sustain by both objects, we see object B is projected at a shorter height into atmosphere than A.

Hence object B will return to the ground first.

8 0
3 years ago
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