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Vlad1618 [11]
3 years ago
5

What is the speed of a wave with a frequency of 100 hz and a wave length of .5 m?

Physics
2 answers:
____ [38]3 years ago
7 0

Answer:

Explanation:

There are 2 ways to help with this. Explain the details, which are fairly simple in this topic, or give the formula. My hope is that an explanation will last longer than memorizing the formula. I give you both.

If a wave has frequency, f, of 3 Hz, its period, T, is

1

3

s

. The wavelength,

λ

, is 5 meters. That means that in the time of one period, the wave travels 5 m.

In general,

S

p

e

e

d

=

distance

time

In applying this general definition of speed

↑

to a wave, we have

speed of the wave

=

wavelength

period

Note: we generally use v for speed of a wave. Using the variable names, then that last formula is written

v

=

λ

T

Since

T

=

1

f

, we can also say that

v

=

λ

⋅

f

So, using that last formula

v

=

5

m

⋅

3

H

z

=

15

m

s

Note: the unit Hz is equivalent to what it was called 100 years ago,

cycles

second

(

also cps

)

. Cycles is not a true unit, so the Hz contributed only the "per second" to the result

15

m

s

.

ElenaW [278]3 years ago
4 0

Answer: the speed is =30ms^-1 The speed of a wave is given by "speed" (ms^-1)= "frequency(Hz)" xx "wavelength (m)" The frequency is f=100Hz The wavelength is lambda=0.3m The speed is v=lambdaf=0.3*100=30ms^-1

Explanation:

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Therefore, the correct option is C.

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A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on th
ira [324]

Answer:

(a) q = 2.357 x 10⁻⁵ C

(b) Φ = 2.66 x 10⁶ N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a)  Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻⁵ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

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Explanation:

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The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from
grigory [225]

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

Number of turns, N = 78

Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

\epsilon=-N\dfrac{d\phi}{dt}

Where

\dfrac{d\phi}{dt} is the rate of change of magnetic flux,

And \phi=BA

\epsilon=-NA\dfrac{dB}{dt}

\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}

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Using Ohm's law, \epsilon=I\times R

Induced current, I=\dfrac{\epsilon}{R}

I=\dfrac{28.32}{1.5}

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So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.

5 0
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