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gogolik [260]
3 years ago
7

A tuba creates a 4th harmonic of

Physics
1 answer:
nikklg [1K]3 years ago
6 0

Answer:

92.33Hz

Explanation:

A tuba is considered as having one open end and the other end is closed. Then:

f_n = n×v / 4L

so here

Where n = 4

f_4 = 116.5 Hz

116.5 Hz = nv/4L

116.5 Hz = 4v/4L

4×v / 4L = v / L

116.5 Hz = v/L............1

Let's assume the speed of sound is 343 m/s. Then substituting 343 m/s for v in equation 1

116.5 Hz = v/L

116.5 Hz = 343 m/s/L

Making L the subject

116.5 Hz × L = 343 m/s

L = 343m/s / 116.5Hz

= 2.944 m

Add 0.721 m to the length, and

f_4 = 343m/s / (2.944+0.721)m

f_4 = 343m/s ÷ 3.715

= 92.328 Hz

Approximately = 92.33Hz

Hence the new frequency for the 4th harmonic is 92.33Hz

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The big bang theory has finally answered one of the biggest questions of science—the origin of the universe.
ss7ja [257]

Answer:

False

Explanation:

According to the big bang theory, matter was an infinitely small and very high density point which at one point exploded and expanded in all directions, creating what we know as our Universe, which also includes space and time . This happened about 13.8 billion years ago. Theoretical physicists have managed to reconstruct this chronology of events from 1/100 of a second after the Big Bang. After the explosion, while the Universe expanded, it cooled sufficiently and the first subatomic particles were formed: Electrons, Positrons, Mesons, Barions, Neutrinos, Photons among others. Today more than 90 particles are known. This theory solves many unknowns and is very well received by the scientific community, however there is still much to solve, for example, one of the great unsolved scientific problems in the expanding Universe model is whether the Universe is open or closed.

An attempt to solve this problem is to determine if the average density of matter in the Universe is greater than the critical value in Friedmann's model. The mass of a galaxy can be measured by observing the movement of its stars; multiplying the mass of each galaxy by the number of galaxies, it is seen that the density is only 5 to 10% of the critical value.

6 0
3 years ago
What is the time lapse between seeing a lightning strike and hearing the thunder if the lightning flash is 47 km away? The speed
Natali5045456 [20]

Answer:

141.14098 secs

Explanation:

Time taken to see the lightning flash can be gotten from:

Velocity = distance/time

Time = distance/velocity

Time = (47 * 1000)/(3 * 10^8)

Time = 0.0001567 secs

Time taken to hear the thunder can be gotten from:

Velocity = distance/time

Time = distance/velocity

Time = (47 * 1000)/(333)

Time = 141.14114 secs

The time lapse between the lightning flash and the thunder will be:

141.14114 - 0. 0001567

= 141.14098 secs

6 0
3 years ago
A meteor moving 468 km per minute traveling in a south-to-north direction passed near Earth in 2013. Does this statement describ
Thepotemich [5.8K]

Answer:

this statement describes meteor's velocity,

because velocity is a vector quantity which has both magnitude as well as a specific direction and here the meteor's direction is specified in the statement hence we conclude that this statement describes meteor's velocity as well as speed too.

3 0
3 years ago
what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe
ArbitrLikvidat [17]

Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{7.4}\\\\v=67.56\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s

Acceleration,

a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2

Hence, this is the required solution.

8 0
3 years ago
Read 2 more answers
Josh did an experiment recording the changes in temperature in sand and water when exposed to a light source, and then when the
Marrrta [24]

Before going to solve this question first we have to understand specific heat capacity of a substance .

The specific heat of a substance is defined as amount of heat required to raise the temperature of 1 gram of substance through one degree Celsius. Let us consider a substance whose mass is m.Let Q amount of heat is given to it as a result of which its temperature is raised  from T to T'.

Hence specific heat  of a substance is calculated as-

                                              c= \frac{Q}{m[T'-T]}

Here c is the specific heat capacity.

The substance whose specific heat capacity is more will take more time to be heated up to a certain temperature as compared to a substance having low specific heat which is to be heated up to the same temperature.

As per the question John is experimenting on sand and water.Between sand and water,water has the specific heat 1 cal/gram per degree centigrade which is larger as compared to sand.Hence sand will be heated faster as compared to water.The substance which is heated faster will also cools faster.

From this experiment John concludes that water has more specific heat as compared to sand.

7 0
2 years ago
Read 2 more answers
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