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ycow [4]
3 years ago
13

Sound with frequency 1300 Hz leaves a room through a doorway with a width of 1.03 m . At what minimum angle relative to the cent

erline perpendicular to the doorway will someone outside the room hear no sound
Physics
1 answer:
AlexFokin [52]3 years ago
6 0

Answer:

  about 14.7°

Explanation:

The formula for the angle of the first minimum is ...

  sin(θ) = λ/a

where θ is the angle relative to the door centerline, λ is the wavelength of the sound, and "a" is the width of the door.

The wavelength of the sound is the speed of sound divided by the frequency:

  λ = (340 m/s)/(1300 Hz) ≈ 0.261538 m

Then the angle of interest is ...

  θ = arcsin(0.261538/1.03) ≈ 14.7°

At an angle of about 14.7°, someone outside the room will hear no sound.

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Estimate the mass of the Great Pyramid of Giza, in tons. You make may use of the following information: the Great Pyramid is in
postnew [5]

Answer:

6005803.83105 short tons

Explanation:

The definition of density is \rho = \frac{m}{V}, and the volume of a pyramid is (confusingly written on the proposal) V=\frac{1}{3} Ah, so we can write:

m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h

Where s is the side of the base, being s^2 the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that \frac{1m}{100cm}=1, and we can use this factor to convert anything written in cm to anything written in m. Example:

500cm=500cm\frac{1m}{100cm}=5m

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so <em>it's not doing anything but changing the units</em>.

We can use this tool like this:

2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3

Where we have used the fact that 1^3=1 (<u>we can elevate any conversion factor to any number and they still will be 1</u>) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

Substituting then our values:

m=\rho V \frac{1}{3} s^2h=(2100Kg/m^3)\frac{1}{3} (230.34m)^2(146.7m)=5448373586.96Kg

And now we will convert to short tons using two conversion factors at the same time:

m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons

Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.

8 0
3 years ago
At what angle of projectile range of maximum​
Agata [3.3K]

Answer:

45°

Explanation:

The textbooks say that the maximum range for projectile motion (with no air resistance) is 45 degrees.

7 0
3 years ago
A 4.0kg block is sliding with a constant velocity of 3.0m/s on a frictionless table that is 0.5m high. If all of the block’s ene
MArishka [77]

Answer:

Velocity = 4.33[m/s]

Explanation:

The total energy or mechanical energy is the sum of the potential energy plus the kinetic energy, as it is known the velocity and the height, we can determine the total energy.

E_{M}=E_{p}  + E_{k} \\E_{p} = potential energy [J]\\E_{k} = kinetic energy [J]\\where:\\E_{p} =m*g*h\\E_{p} = 4*9.81*0.5=19.62[J]\\E_{k}=\frac{1}{2} *m*v^{2}  \\E_{k}=\frac{1}{2} *4*(3)^{2} \\E_{k}=18[J]\\Therefore\\E_{M} =18+19.62\\E_{M}=37.62[J]

All this energy will become kinetic energy and we can find the velocity.

37.62=\frac{1}{2} *m*v^{2} \\v=\sqrt{\frac{37.62*2}{4} } \\v=4.33[m/s]

8 0
3 years ago
This company is run by kids right
GREYUIT [131]

Answer:

i d k about that but I know it`s a Polish thing

4 0
3 years ago
Read 2 more answers
How can we maximise the rate of energy transfer to keep things cool?
faltersainse [42]
To do this we may use things that are good conductors - are painted dull black -
Have a air flow around them Maximised.


6 0
3 years ago
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