Why echos are undesirable in a big hall
2 answers:
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To solve this problem, it is necessary to apply the concepts related to the constructive interference caused by the wavelengths of the sound traveling in the air.
From the definition of constructive interference we know that
Where
Distance between speakers
n = Integer which represent he number of repetition of the spectrum
Wavelength
At the same time the frequency is subject to the form,
Where
Velocity (of the sound at this case)
From our given values we have to is
The wavelength would be subject to the sound spectrum therefore for n = 1,
Then the frequency would be,
For the value of n = 2,
Then the frequency would be,
For n = 3,
Then the frequency would be,
From the frequencies obtained we can identify that the two lowest frequencies that can be heard due to constructive interference are 28Hz and 42Hz
Answer:
35.14°C
Explanation:
The equation for linear thermal expansion is , which means that a bar of length
with a thermal expansion coefficient
under a temperature variation
will experiment a length variation
.
We have then = 0.481 foot,
= 1671 feet and
= 0.000013 per centigrade degree (this is just the linear thermal expansion of steel that you must find in a table), which means from the equation for linear thermal expansion that we have a
= 22.14°. As said before, these degrees are centigrades (Celsius or Kelvin, it does not matter since it is only a variation), and the foot units cancel on the equation, showing no further conversion was needed.
Since our temperature on a cool spring day was 13.0°C, our new temperature must be = 35.14°C
The catcher can catch the ball at a height of 0.96 m from the ground.
The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.
Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.
The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.
Therefore,
Calculate the time t by substituting 18.44 m for x and 40 m/s for u.
The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.
Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.
Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,
Substitute 9.8 m/s² for g and 0.461 s for t.
The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.
Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by,
The catcher needs to hold his glove at a height of <u>0,96 m from the ground.</u>