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3 years ago

If the period of a certain wave (wavelength = 4.5 m) is 2 seconds, what is the speed of the wave?

Physics

1 answer:

taurus [48]3 years ago

5 0

Answer:

2.25m/s

Explanation:

wavelength=4.5m

period=2sec

frequency= 1/period

=1/2

speed= wavelength×wave frequency

=4.5×1/2

=2.25 m/s

so, none of the options are correct

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3 years ago

For an object to be classified as a ____, it must meet certain definite criteria: It must be massive enough to pull itself into

Orlov [11]

Answer:

a planet

Explanation:

a planet is one which exerts these properties and therefore is the answer

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4 years ago

You have a stopped pipe of adjustable length close to a taut 62.0 cmcm, 7.25 gg wire under a tension of 4710 NN. You want to adj

Oksana_A [137]

Answer:

6 cm long

Explanation:

F = 4110N

Vo(speed of sound) = 344m/s

Mass = 7.25g = 0.00725kg

L = 62.0cm = 0.62m

Speed of a wave in string is

V = √(F / μ)

V = speed of the wave

F = force of tension acting on the string

μ = mass per unit density

F(n) = n (v / 2L)

L = string length

μ = mass / length

μ = 0.00725 / 0.62

μ = 0.0116 ≅ 0.0117kg/m

V = √(F / μ)

V = √(4110 / 0.0117)

v = 592.69m/s

Second overtone n = 3 since it's the third harmonic

F(n) = n * (v / 2L)

F₃ = 3 * [592.69 / (2 * 0.62)

F₃ = 1778.07 / 1.24 = 1433.927Hz

The frequency for standing wave in a stopped pipe

f = n (v / 4L)

Since it's the first fundamental, n = 1

1433.93 = 344 / 4L

4L = 344 / 1433.93

4L = 0.2399

L = 0.0599

L = 0.06cm

L = 6cm

The pipe should be 6 cm long

6 0

3 years ago

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drek231 [11]

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3 years ago

What is the repulsive force between two pith balls that are 10.0 cm apart and have equal charges of -40.0 NC?

galina1969 [7]

Answer:

So the repulsive force between the pith ball will be $1.44\times 10^{-3}N$

Explanation:

We have given that the pith ball have the equal charge q = -40 nC = $-40\times 10^{-9}C$

Distance between the charges = 10 cm =0.1 m

According to coulombs law $F=\frac{KQ_1Q_2}{R*2}$

$F=\frac{9\times 10^9\times -40\times 10^{-9}\times -40\times 10^{-9}}{0.1^2}=1440000\times 10^{-9}=1.44\times 10^{-3}N$

So the repulsive force between the pith ball will be $1.44\times 10^{-3}N$

5 0

3 years ago