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maria [59]
4 years ago
5

What is the frequency heard by a person driving at 15 m/s toward a factory whistle emitting a

Physics
1 answer:
zmey [24]4 years ago
4 0

Answer:

835.29 Hz

Explanation:

When moving towards the source of sound, frequency will be given by

f*=f(vd+v)/v

Where f is the freqiency of the source, vd is the driving speed, v is the speed of sound in air, f* is the inkown frequency when moving forward.

Substituting 800 Hz for f, 340 m/s for v and 15 m/s for vd then

f*=800(15+340)/340=835.29411764704 Hz

Rounded off, the frequency is approximately 835.29 Hz

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Explanation:

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3 years ago
If a nucleus was as big as a nonpareil, an atom would be ____
Mamont248 [21]

Answer:

small marble  

Explanation:

A nonpareil are confectionery tiny ball items that are made up of starch and sugar. It is very small of the size of sugar crystal or sand grains. They were the miniature version of the comfits. They are generally opaque white but bow available in all colors.

In the context, if the nucleus is compared to the size of a nonpareil then its atom would be of the size of small size marble. An atom is bigger in size than that of nucleus as the nucleus is located inside the atoms.

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I’m steel, the solvent is And the solute is. .
sleet_krkn [62]
Iron is the solvent and carbon is the solute.
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A sphere of radius 1.59 cm and a spherical shell of radius 7.72 cm are rolling without slipping along the same floor. The two ob
valina [46]

Answer: 4.86

Explanation:

sphere moment of Inertia Iₑ = (2/5)mrₑ²

Let the sphere of radius 1.59 cm be x

Let the spherical shell of radius 7.72 cm be y, so that

Iₑ(x) = 2/5 * m * 1.59²

Iₑ(x) = 2/5 * m * 2.5281

Iₑ(x) = 1.011m

Iₑ(y) = 2/5 * m * 7.72²

Iₑ(y) = 2/5 * m * 59.5984

Iₑ(y) = 23.84m

Also, the angular speed of the sphere's would be ωₑ(x) and ωₑ(y)

total k.e = rotational k.e + linear k.e

for sphere = ½Iₑωₑ² + ½mωₑ²rₑ²

For sphere x

{ωₑ²[ 1.011 + 1.59²]} =

ωₑ²(1.011 + 2.5281) =

ωₑ²(3.5391)

For sphere y

{ωₑ²[ 23.84 + 7.72²]} =

ωₑ²(23.84 + 59.5984) =

ωₑ²(83.4384)

If the ratio of x/y = 1, then

ωₑ(x)²(3.5391) / ωₑ(y)²(83.4384) = 1

ωₑ(x)²(3.5391) = ωₑ(y)²(83.4384)

[ωₔ(x)/ωₑ(y)]² = [83.4384] / [3.5391] ~= 23.5762

[ωₔ(x)/ωₑ(y)] = √(23.5762)

[ωₔ(x)/ωₑ(y)] = 4.86

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3 years ago
What is the potential difference across lamp 1
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