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ryzh [129]
3 years ago
5

Middle C also known as "do" on the fixed do-solfege scale, is a note used when playing or singing music. The frequency of middle

C is 261.63 HZ, and the wavelength is 131.87 cm. What is the speed of sound for middle c ?
Physics
1 answer:
saveliy_v [14]3 years ago
8 0

Answer:

198.2m/s

Explanation:

Speed of a wave(v) is the product of the frequency of the wave (f) and its wavelength(¶).

Mathematically, v = f/¶

Given frequency of the middle C = 261.63Hz

Wavelength = 131.87cm

Converting this to meters we have;

131.87/100 = 1.32m

Speed of the sound = 261.63/1.32

Speed of the sound = 198.20m/s

Therefore the speed of sound for middle C is 198.2m/s

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Q|C The speed of a one-dimensional compressional wave traveling along a thin copper rod is 3.56 km/s . The rod is given a sharp
Gekata [30.6K]

The length of rod in terms v(r) and t is L = [  Δt / (1/343) - (1/v(r)) ].

3.56 km/s is the speed of a one-dimensional compressional wave moving along a thin copper rod.

At one end of the rod, a hard hammer strike is delivered. With a time interval of Δt between the two pulses, a listener at the other end of the rod hears the sound twice as it travels through the metal and the air.

The time interval is given by t = L/v.

The delay between pulses arrivals is:

Δt = L [(1/v(air)) - (1/v(copper))]

Now,

When the copper rod is swapped out for a different substance and the sound speed is measured as v(r).

The speed of air, v(air) = 343 m/s

Then,

L = [  Δt / (1/v(air)) - (1/v(r)) ]

L = [  Δt / (1/343) - (1/v(r)) ]

Here L is the length of the rod, Δt is in seconds and v(r) is the speed of sound in the rod.

Learn more about speed here:

brainly.com/question/13943409

#SPJ4

8 0
1 year ago
If your front lawn is 21.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1350 new snowflakes every minu
ziro4ka [17]

Answer:

snow is 64.638 kg / hr

Explanation:

Given data

wide w = 21 feet

long L = 20 ft

area A = 1350 square foot

mass of snow m  = 1.90 mg

to find out

snow in kilograms / hour

solution

we will find snow in kg

so we apply formula that is

snow kg / hour  = w × L ×A ×  m × 60/10^6

put all value we get  snow

snow =  21 × 20 × 1350 ×  1.90 × 60/10^6

snow =  420 × 1350 ×  1.90 × 60/10^6

snow =  1077300 × 60/10^6

snow =  64.638

hence snow is 64.638 kg / hr

7 0
3 years ago
What is sap?
Paladinen [302]
Its A Sap is basically a fluid which is transported in to the xylem tubes.
4 0
3 years ago
Read 2 more answers
A cannonball is fired vertically upwards at 100.0 m/s a) How long will it take to return to the cannon? b) what is it's maximum
V125BC [204]
Answer:
a) 20s
b) 500m

Explanation:
Given the initial velocity = 100 m/s, acceleration = -10m/s^2 (since it is moving up, acceleration is negative), and at the maximum height, the ball is not moving so final velocity = 0 m/s.

To find time, we apply the UARM formula:

v final = (a x t) + v initial

Replacing the values gives us:

0 = (-10 x t) + 100

-100 = -10t

t = 10s

It takes 10s for the the ball to reach its max height, but it must also go down so it takes 2 trips, once going up and then another one going down, both of which take the same time to occur

So 10s going up and another 10s going down:

10x2 = 20s

b) Now that we have v final = 0, v initial = 100, a = -10, t = 10s (10s because maximum displacement means the displacement from the ground to the max height) we can easily find the displacement by applying the second formula of UARM:

Δy = (1/2)(a)(t^2) + (v initial)(t)

Replacing the values gives us:

Δy = (1/2)(-10)(10^2) + (100)(10)

= (-5)(100) + 1000

= -500 + 1000

= 500 m

Hope this helps, brainliest would be appreciated :)
7 0
3 years ago
Which describes Einstein’s idea of space and time?
Sholpan [36]
Time and space are both relative
8 0
3 years ago
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