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Nezavi [6.7K]
3 years ago
11

A graph of a wave is shown if the frequency f is 8.0 Hz what is the waves speed

Physics
1 answer:
defon3 years ago
6 0

Answer:

C

6 =  \frac{3}{2} \lambda \\  wavelength :  \:  \lambda = 4 \: m \\ from \: wave \: equation \\ v = f\lambda  \\ v = 8 \times 4 \\ v = 24 \:  {ms}^{ - 1}

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A ray of light traveling in air strikes the surface of a liquid. if the angle of incidence is 29.7◦ and the angle of refraction
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When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted ray, and it is given by:
\theta_c = \arcsin ( \frac{n_r}{n_i} ) (2)
where n_r is the refractive index of the second medium and n_i is the refractive index of the first medium.

We can find the ratio n_r / n_i by using Snell's law:
n_i \sin \theta_i = n_r \sin \theta_r (1)
where
\theta_i is the angle of incidence
\theta_r is the angle of refraction

By using the data of the problem and re-arranging (1), we find
\frac{n_r}{n_i} =  \frac{\sin \theta_i}{\sin \theta_r} = \frac{\sin 16.3^{\circ}}{\sin 29.7^{\circ}} =0.566

and if we use eq.(2) we can now find the value of the critical angle:
\theta_c = \arcsin ( \frac{n_r}{n_i} ) = \arcsin (0.566) = 34.5^{\circ}
3 0
3 years ago
Can an object with constant acceleration reverse its direction of travel? Can it reverse its direction twice? Explain.
Neko [114]

Answer:yes

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6 0
3 years ago
Un pintor de 75.0 kg sube por una escalera de 2.75 m que está inclinada contra una pared vertical. La escalera forma un ángulo d
dezoksy [38]

Answer:

Work done, W = 1786.17J

Explanation:

The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "

Mass of a painter, m = 75 kg

He climbs 2.75-m ladder that is leaning against a vertical wall.

The ladder makes an angle of 30 degrees with the wall.

We need to find the work done by the gravity on the painter.

The angle between the weight of the painter and the displacement is :

θ = 180 - 30

= 150°

The work done by the gravity is given by :

W=Fd\cos\theta\\\\=75\times 10\times 2.75\times \cos30\\\\=1786.17\ J

Hence, the required work done is 1786.17 J.

6 0
2 years ago
true or false: a image created by an object located inside the focal point is virtual, enlarged, and upright
crimeas [40]

true or false: a image created by an object located inside the focal point is virtual, enlarged, and upright

True :D

4 0
3 years ago
How far should the second antenna be moved in order to receive a minimum signal from a station that broadcasts at 99.4 MHz?
raketka [301]

Answer:

So the distance of the antenna from the station will be 3.018 m

Explanation:

We have given the frequency of the broadcast f=99.4MHz=99.4\times 10^4Hz

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The distance of the antenna to receive a minimum signal from the station is given by d=\frac{v}{f}=\frac{3\times 10^{8}}{99.4\times 10^6}=3.018m

So the distance of the antenna from the station will be 3.018 m

5 0
3 years ago
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