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3 years ago

8

The velocity of sound in air at STP (Standard Temperature and Pressure) is approximately 344 m/s. If the frequency of the sound

wave is ν = 5.2 kHz, what is the wavelength in cm?

1 answer:

exis [7]3 years ago

5 0

Answer:

0.066m

Explanation:

use the formula

velocity= frequency × wave length

344m/s = 5200Hz× wavelength

wavelength = 344m/s/5200Hz

wavelength = 0.066m(to 2 sig fig)

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The resistance of 3A and 12V

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Resistance = V / I
= 12 / 3 = 4

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3 years ago

A light, flexible cable is wrapped around a solid cylinder with mass 3.3 kg and a radius of 0.8 meters. The cylinder rotates on

kari74 [83]

Answer:

$9.16rad/s^2$

Explanation:

We are given that

Mass, $m_1=3.3 kg$

Radius,r=0.8 m

$m_2=4.9 kg$

Height,h=2.9 m

We have to find the angular acceleration of the cylinder.

According to question

$4.9g-T=4.9a$

$Tr=I\alpha$

Where

$\alpha=\frac{a}{r}$

$Tr=\frac{1}{2}m_1ra$

$T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a$

Substitute the value

$4.9g-\frac{1}{2}(3.3a)=4.9a$

$4.9\times 9.8=4.9a+\frac{3.3a}{2}$

Where $g=9.8 m/s^2$

$48.02=a(4.9+1.65)=6.55a$

$a=\frac{48.02}{6.55}=7.33m/s^2$

Angular acceleration, $\alpha=\frac{a}{r}=\frac{7.33}{0.8}=9.16rad/s^2$

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3 years ago

HOW MANY KILOMETERS IS EARTH FROM THE SUN?

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Earth is 150 million kilometers away for the sun

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4 years ago

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A sensational 130 kg midfielder has 695 J of

antoniya [11.8K]

Answer:

v = 3.27 m/s

Explanation:

KE = 1/2 mv^2

695 J = 1/2 (130kg)(v^2)

695 J / (1/2 x 130kg) = v^2

v^2 = square root of 10.69

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3 years ago

In an experiment replicating Millikan’s oil drop experiment, a pair of parallel plates are placed 0.0200 m apart and the top pla

lianna [129]

Answer: See below

Explanation:

<u>Given:</u>

The potential between plates, V = 240 V

Distance between plates, d = 0.02 m

The mass of drop, m = 2x10^-11

Charge on electron, e = 1.6x10^-19

Part (a)

The free-body diagram is attached below

Part (b)

The electric field is given by,

$E=\frac{V}{d}$

On applying force balance, the force on oil drop is equal to the weight of the oil,

$\begin{aligned}F_{E} &=m g \\q E &=m g \\q \frac{V}{d} &=m g \\q &=\frac{m g d}{V}\end{aligned}$

Substituting the given values in the above equation,

$\begin{aligned}&q=\frac{2 \times 10^{-11} \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2} \times \frac{1 \mathrm{~N}}{1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}} \times 0.02 \mathrm{~m}}{240 \mathrm{~V} \times \frac{1 \mathrm{~N} \cdot \mathrm{m} / \mathrm{C}}{1 \mathrm{~V}}} \\&q=1.63 \times 10^{-14} \mathrm{C}\end{aligned}$

Therefore, the charge on the oil drop is 1.63x10^-14 C

Part (c)

There will be an excess of electrons on the oil drop.

The number of electrons on oil drop can be calculated as,

$\begin{aligned}q &=n e \\1.63 \times 10^{-14} \mathrm{C} &=n \times 1.6 \times 10^{-19} \mathrm{C} \\n &=1.01 \times 10^{5}\end{aligned}$

Therefore, the number of excess electrons is 1.01x10^5

3 0

2 years ago