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3 years ago

What is the speed of sound at 33 °C (m/s)? For a frequency of 5 kHz, how large do you expect the wavelength to be (m)?

Physics

1 answer:

Vedmedyk [2.9K]3 years ago

7 0

Answer:

-The speed of sound at 33°C is 362.8 m/s.

-The wavelength at a frequency at 5 kHz is 0.07256 m .

Explanation:

let v = 343 m/s be the speed of sound.

let T be the temperature.

then the speed of sound V, at 33°C is given by:

V = v + 0.6×T

= 343 + 0.6×33

= 362.8 m/s

Therefore, the speed of sound at 33°C is 362.8 m/s.

the wavelength at a frequency of f = 5kHz = 5000 Hz is given by:

λ = V/f

= (362.8)/(5000)

= 0.07256 m

Therefore, the wavelength at a frequency at 5 kHz is 0.07256 m .

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What is the average speed for the entire graph?

Maru [420]

Answer:

Explanation:

Divide the total distance traveled by the total time spent traveling. This will give you your average speed. . So if Ben traveled 150 miles in 3 hours, 120 miles in 2 hours, and 70 miles in 1 hour, his average speed was about 57 mph.

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2 years ago

On the surface of the earth the weight of an object is 200 lb. Determine the height of the

siniylev [52]

Answer:

The height of the object is 5007.4 miles.

Explanation:

Given that,

Weight of object = 200 lb

We need to calculate the value of $Gmm_{e}$

Using formula of gravitational force

$F=\dfrac{Gmm_{e}}{r^2}$

Put the value into the formula

$200=\dfrac{Gmm_{e}}{(3958.756)^2}$

$200\times(3958.756)^2=Gmm_{e}$

$Gmm_{e}=3.134\times10^{9}$

We need to calculate the height of the object

Using formula of gravitational force

$F=\dfrac{Gmm_{e}}{r^2}$

Put the value into the formula

$125=\dfrac{200\times(3958.756)^2}{r^2}$

$r^2=\dfrac{200\times(3958.756)^2}{125}$

$r^2=25074798.5$

$r=\sqrt{25074798.5}$

$r=5007.4\ miles$

Hence. The height of the object is 5007.4 miles.

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3 years ago

The energy goes from _____ to the _____ above it.

Artist 52 [7]

Answer:

The energy goes from the ground state to the excited states above it.

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3 years ago

PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer this correctly!!!! (60pts)

Alona [7]

Answer:

Weight

Explanation:

"An object will float if the buoyancy force exerted on it by the fluid balances its weight, i.e. if FB=mg F B = mg . But the Archimedes principle states that the buoyant force is the weight of the fluid displaced. So, for a floating object on a liquid, the weight of the displaced liquid is the weight of the object."

Hope this helps! :)

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3 years ago

The axle of an automobile is acted upon by the forces and couple shown. knowing that the diameter of the solid axle is 32 mm, de

saul85 [17]

First we need to convert the mm to inches to make our computation easier.

1mm = 0.0393701

32mm * 0.0393701 = 1.25 in

Solution:

C = 1/2d = ½ (1.25) = 0.625 in^4

Tension: tension = Te/J = 2T/ piC^3

= (2)(2500)/pi (0.0625)^3 = 6.519 x 10^3 psi = 6.519 ksi

Bending:

I = pi/4 * c^4 = 119.842 x 10^-3 in^4

M = (5)(600) = 3600 lb in

G = My/I = (3600)(0.625)/119.842 x 10^-3 = -18.775 x 10^2 psi = -18.775ksi

Gx = -18.775 ksi

Gy = 0

Txy = 6.519 ksi

G ave – ½ (Gx + Gy) = -9.387 ksi
R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2 + (6.519)^2 = 11.429 ksi

1. G1 = Gave + R = -9/387 + 11.429 = 2.04 ksi

G2 = Gave - R = -9/387 - 11.429 = -20.8

Tan 2ϴp = 2txy/Gx – Gy = 2(6.519)/-9.387 = -1.3889

ϴp = -27.1 degrees and 62.9 degrees

2. Tmax = R = 11.43 ksi

R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2 + (6.519)^2 = 11.429 ksi

3 0

3 years ago