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Burka [1]
3 years ago
14

A grating whose slits are 3.2x10^-4 cm apart produces a third-order fringe at a 25.°0 angle. What is the wavelength of light tha

t is used?
Physics
1 answer:
Ber [7]3 years ago
7 0

Answer:

The light used has a wavelenght of 4.51×10^-7 m.

Explanation:

let:

n be the order fringe

Ф be the angle that the light makes

d is the slit spacing of the grating

λ be the wavelength of the light

then, by Bragg's law:

n×λ = d×sin(Ф)

λ = d×sin(Ф)/n

λ = (3.2×10^-4 cm)×sin(25.0°)/3

  = 4.51×10^-5 cm

  ≈ 4.51×10^-7 m

Therefore, the light used has a wavelenght of 4.51×10^-7 m.

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1. The speed of sound in air is roughly 1,484 m/s. How far, in meters, would a sound wave travel in 3 seconds?
kobusy [5.1K]
Speed = distance/time taken

1484 = distance/3

Distance = 1484 x 3

Distance = 4452 m
7 0
3 years ago
An experimenter using a gas thermometer found the pressure at the triple point of water (0.01°C) to be 4.80 × 10⁴ Pa and the pre
irakobra [83]

Answer:

T = -282.33^o C

Explanation:

As we know that the relation between temperature and pressure is a linear relation

so we have

P - P_o = \frac{P_1 - P_o}{T_1 - T_o} (T - T_o)

here we know that

P_1 = 6.50 \times 10^4

P_o = 4.80 \times 10^4

T_1 = 100^o C

T_o = 0.01^o C

now we will have

P - 4.80 \times 10^4 = \frac{(6.50 - 4.80)\times 10^4}{100 - 0.01}(T - 0.01)

P = 4.80 \times 10^4 + 170.02(T - 0.01)

now if P = 0

then we will have

0 = 4.80 \times 10^4 + 170.02(T - 0.01)

T = -282.33^o C

7 0
3 years ago
Which of the following is NOT a kind of energy store?
snow_tiger [21]

Answer:

Option D is your answer ☺️☺️. If I'm right so,

Please mark me as brainliest. thanks!!!

6 0
3 years ago
A centrifuge in a medical laboratory rotates at an angular speed of 3,650 rev/min. when switched off, it rotates through 48.0 re
AlladinOne [14]
First of all we need to convert everything into SI units.

Let's start with the initial angular speed, \omega _i = 3650 rev/min. Keeping in mind that
1 rev = 2 \pi rad
1 min=60 s
we have
\omega _i = 3650  \frac{rev}{min} \cdot  \frac{2 \pi rad/rev}{60 s/min} =382.0 rad/s

And we should also convert the angle covered by the centrifuge:
\theta = 48.0 rev= 48.0 rev \cdot  2 \pi  \frac{rad}{rev}=301.4 rad

This is the angle covered by the centrifuge before it stops, so its final angular speed is \omega_f =0.

To solve the problem we can use the equivalent of
2aS = v_f^2 -v_i^2
of an uniformly accelerated motion but for a rotational motion. It will be
2 \alpha \theta = \omega_f^2-\omega_i^2
And by substituting the numbers, we can find the value of \alpha, the angular acceleration:
\alpha=- \frac{\omega_i^2}{2 \theta}=- \frac{(382 rad/s)^2}{2 \cdot 301.4 rad}=-242.1 rad/s^2
4 0
3 years ago
How do you solve #3 and #5?
julia-pushkina [17]

Answer:

your answers are correct i have done this many times

8 0
3 years ago
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