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yarga [219]
2 years ago
10

What is speed of wave in terms of frequency and wavelength??​

Physics
1 answer:
ratelena [41]2 years ago
3 0

Answer:

In the case of a wave, the speed is the distance traveled by a given point on the wave (such as a crest) in a given interval of time. In equation form, If the crest of an ocean wave moves a distance of 20 meters in 10 seconds, then the speed of the ocean wave is 2.0 m/s.

Speed = Wavelength x Wave Frequency. In this equation, wavelength is measured in meters and frequency is measured in hertz (Hz), or number of waves per second. Therefore, wave speed is given in meters per second, which is the SI unit for speed.

Explanation:

can i get the crown please

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7 0
1 year ago
A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical ener
defon

Complete question:

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find

(a) the force constant of the spring and (b) the amplitude of the motion.

Answer:

(a) the force constant of the spring = 47 N/m

(b) the amplitude of the motion = 0.292 m

Explanation:

Given;

mass of the spring, m = 200g = 0.2 kg

period of oscillation, T = 0.410 s

total mechanical energy of the spring, E = 2 J

The angular speed is calculated as follows;

\omega = \frac{2\pi}{T} \\\\\omega = \frac{2\pi}{0.41} \\\\\omega = 15.33 \ rad/s

(a) the force constant of the spring

\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\k = m \omega^2\\\\k = (0.2)(15.33)^2\\\\k = 47 \ N/m

(b) the amplitude of the motion

E = ¹/₂kA²

2E = kA²

A² = 2E/k

A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2\times 2}{47} }\\\\A = 0.292 \ m

7 0
3 years ago
A solenoidal coil with 26 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 20.0 cm long
noname [10]

Answer:

Part a)

\phi = 2.76 \times 10^{-7} T m^2

Part B)

M = 5.52 \times 10^{-5} H

Part C)

EMF = 0.1 V/s

Explanation:

Part a)

Magnetic field due to a long ideal solenoid is given by

B = \mu_0 n i

n = number of turns per unit length

n = \frac{N}{L}

n = \frac{350}{0.20}

n = 1750 turn/m

now we know that magnetic field due to solenoid is

B = (4\pi \times 10^{-7})(1750)(0.100)

B = 2.2 \times 10^{-4} T

Now magnetic flux due to this magnetic field is given by

\phi = B.A

\phi = (2.2 \times 10^{-4})(\pi r^2)

\phi = (2.2 \times 10^{-4})(\pi(0.02)^2)

\phi = 2.76 \times 10^{-7} T m^2

Part B)

Now for mutual inductance we know that

\phi_{total} = M i

\phi_{total} = N\phi

\phi_{total} = 20(2.76 \times 10^{-4})

\phi_{total} = 5.52 \times 10^{-6}

now we have

M = \frac{5.52 \times 10^{-6}}{0.100}

M = 5.52 \times 10^{-5} H

Part C)

As we know that induced EMF is given as

EMF = M \frac{di}{dt}

EMF = 5.52 \times 10^{-5} (1800)

EMF = 0.1 V/s

3 0
3 years ago
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