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marusya05 [52]
3 years ago
14

The most soaring vocal melody is in Johann Sebastian Bach's Mass in B minor. In one section, the basses, tenors, altos, and sopr

anos carry the melody from a low D to a high A. In concert pitch, these notes are now assigned frequencies of 146.8 Hz and 880.0 Hz. (Use 343 m/s as the speed of sound, and 1.20 kg/m3 as the density of air.)
Find the wavelength of the initial note.
Physics
1 answer:
solmaris [256]3 years ago
3 0

Answer:

2.33651226158 m

Explanation:

From the question the required data is as follows

f = Frequency of the initial note = 146.8 Hz

v = Velocity of sound in air = 343 m/s

The wavelength of a wave is given by

\lambda=\dfrac{v}{f}

\Rightarrow \lambda=\dfrac{343}{146.8}

\Rightarrow \lambda=2.33651226158\ m

The wavelength of the initial note is 2.33651226158 m

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The radius of a sphere is 2.00 m. What’s the surface area
Svet_ta [14]

S.A of sphere = 4πr²

r = 2m

S.A = 4π *4

S.A = 50.26m²

5 0
3 years ago
A block with mass M = 3 kg is moving on a flat surface with constant speed v1 =
Alchen [17]

Answer:

this makes no since so i cant help you here sorry

5 0
2 years ago
oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student
12345 [234]

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

3 0
3 years ago
An iron anchor of density 7890.00 kg/m3 appears 299 N lighter in water than in air. (a) What is the volume of the anchor? (b) Ho
PtichkaEL [24]

Answer:

weigh is 2353.13 N

Explanation:

Given data

density = 7890.00 kg/m3

lighter =  299 N

to find out

the volume of the anchor and weigh in air

solution

from question we can say that

apparent weight = actual weight - buoyant force

we know weight = mg and buoyant force = water density × g

so volume of anchor is = actual weight - apparent weight / buoyant force

volume of anchor is = 299 / 1000 × 9.81

volume of anchor is = 0.0304791 m³

and

weight of anchor is mg

here mass m = density Fe g

density Fe = 7870 from table 14-1

so weight = 7870 × 0.0304791  × 9.81

weigh is 2353.13 N

7 0
4 years ago
White light (light containing all frequencies of light in the visible spectrum) passes through a diffraction grating with 1300 l
lara31 [8.8K]

Answer:

The distance between the 1st order blue fringe and the 1st order red fringe is 6.5 cm.

Explanation:

Given that,

Diffraction grating =1300 lines/cm

Distance of screen = 2 m

Order number = 1

We need to calculate the distance between the 1st order blue fringe and the 1st order red fringe

Using formula of distance

For first order red fringe

d\sin\theta=m\lambda

d\times\dfrac{y}{l}=m\times\lambda

y_{r}=\dfrac{m\times\lambda l}{d}

y_{r}=\dfrac{1\times700\times10^{-9}\times2}{\dfrac{1}{1300}\times10^{-2}}

y_{r}=0.182\ m

For first order blue fringe

y_{b}=\dfrac{1\times450\times10^{-9}\times2}{\dfrac{1}{1300}\times10^{-2}}

y_{b}=0.117\ m

We need to calculate the distance between blue fringe and red fringe

\Delta y=y_{r}-y_{b}

\Delta y=0.182-0.117

\Delta y=0.065\ m=6.5 cm

Hence, The distance between the 1st order blue fringe and the 1st order red fringe is 6.5 cm.

7 0
3 years ago
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