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Alexus [3.1K]
2 years ago
6

What is the wavelength of a 10 Hz wave that travels with a speed of 5 m/s?

Physics
1 answer:
garik1379 [7]2 years ago
8 0
B

V= f x lambda
V= 5m/s
F = 10hz
Lambda = ?
5 = 10 x lamba
5 /10 = lambda
Wavelength =0.5
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"A short-wave radio antenna is supported by two guy wires, 150 ft and 170 ft long. Each wire is attached to the top of the anten
Nonamiya [84]

Answer:

The anchor points are 78.37 ft and 111.99 ft

Explanation:

If you look at the attached (Fig 1) you will find that the union of antenna and its guy wires forms two right triangles. To solve problems that involve this kind of triangles, you can apply trigonometric functions (sine, cosine, etc) and Pythagoras Theorem. Trigonometric functions states the relation between angles, sides and hypotenuse of a right triangle. If you look Fig2, considering α angle, "b" is the opposite side, "a" the adjacent side and "c" the hypotenuse. Then

a) Sine (α) = b/c it means opposite side/hypotenuse

b) Cosine (α)= a/c, it means adjacent side/hypotenuse

c) Tangent (α) = b/a opposite side/adjacent side.

Pythagoras theorem states that if you called "a" and "b", the sides of the right triangle, and "c" the hypotenuse, then:

                                      a² + b² = c²    

As the problem states the lengths 150 ft and 170 ft represents the value of the hypotenuse of each triangle and 65° is one of the angles of the triangle with 150 ft hypotenuse. So you can solve this using sin (65°) to find the height of the antenna (h) and then the two distances (x and y,).

Sine (65°) = h/ 150 ft ⇒ Sine (65°) x 150ft = h ⇒ h = 127.9 ft.

To find x : Cosine (65°) = x/ 150 ft ⇒ Cosine (65°) x 150 ft = x

⇒ x = 78.37 ft.

And finally, to find y we can apply Pythagoras theorem

(170 ft)² = (127.9 ft)² + y² ⇒ y² = (170 ft)² - (127.9 ft)² ⇒ y = 111.99 ft

Summarizing, the anchor points are 78.37 ft and 111.99 ft

4 0
2 years ago
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Radius = 2.41 cm =0.0241 m

The magnetic field strength increases from 50.9 T to 96.3 T so change in magnetic field dB = 96.3-50.9=45.4 T

Time interval dt = 0.151 sec

We know that the induced emf e=-NA\frac{dB}{dt}

Area A=\pi r^2=3.14\times 0.0241^2=1.8237\times 10^{-3}m^2

Putting all these values in emf equation e=-123\times 1.8237\times 10^{-3}\times \frac{45.4}{0.151}=-67.44\ V here negative sign indicates that it opposes the cause due to which it is produced

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