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Complete Question </u></h3>
Q)Two loudspeakers, A and B, are driven by the same amplifier as shown and emit sinusoidal waves in phase. Speaker B is 2.00 m to the right of speaker A. Consider point Q along the extension of the line connecting the speakers, 1.00 m to the right of speaker B. Both speakers emit sound waves that travel directly from the speaker to point Q. a) What is the lowest frequency for which constructive interference occurs at point Q? b) What is the lowest frequency for which destructive interference occurs at point Q? The speed of sound wave is 344 m/s.
Answer:
lowest frequency is 172 Hz. for n = 1
lowest frequency is 86 Hz for n = 0
Explanation:
(a) Constructive interference takes place when the path difference is n where n=1,2,3 ...
frequency f = v /λn
= v n / d
= n ( 344 / 2 )
= n ( 172 ) Hz.
lowest frequency is 172 Hz
b)Destructive interference takes place when the path difference is n/2 x where n=1,3,5 ...
λ = 4/2n+1
for n = 0
λ = 4m
f = 344/4
f = 86 Hz.
lowest frequency is 86 Hz
Explanation:
Given:
Acceleration of car = 10 m/s
Distance travelled in 5 sec = 150 m
To find:
Distance travelled in the next 5 seconds
Concept:
There are 2 ways to app this kind of questions .
Either , we can find the total distance travelled in 10 seconds and then subtract 150 m from it.
Whereas , you can find out the final velocity at the end of the 5th seconds. Using equation of motion, then get the distance travelled in next 5 seconds.
Calculation:
v² = u² + 2as
=> ( u + at)² = u² + 2as
=> u² + 2uat + a²t² = u² + 2as
=> 2uat + (at)² = 2as
=> 2u (10)(5) + (10 × 5)² = 2 (10)(150)
=> 100u = 3000 - 2500
=> 100u = 500
=> u = 5 m/s
Distance travelled in 10 seconds :
s = ut + ½at²
=> s = (5 × 10) + ½(10)(10)²
=> s = 50 + 500
=> s = 550 m
Distance travelled in the 2nd half will be :
d = 550 - 150
=> d = 400 m
So final answer is :
The appropriate answer is c. silty clay loam. This is the most likely soil that was present in the garden before sand was added to balance it. This type of soil contains an even mix of silt and clay. This type of soil does not drain well and tends to hold water. This would not be suitable for most garden variety plants. Adding sand to the soil ensures better drainage and removes moisture that would rot roots or create conditions for fungi to develop.
Answer:
the major difference between the rate of diffusion between urea and albumin is that urea diffused more slowly because it is larger than albumin.
Explanation:
As described in activity 1, the major difference between the rate of diffusion between urea and albumin is that urea diffused more slowly because it is larger than albumin. Urea basically is an heavy compound, thus it is at the base of the chain reactions which which major purpose is to break down the amino acids that make up proteins. Albumin on the other hand is a form of protein that simply helps in the transmission of blood and prevent blood from flowing through to other tissue in the body. The liver produces the albumin the body. So going back to the question, the major difference between the rate of diffusion between urea and albumin is that urea diffused more slowly because it is larger than albumin.
Answer:
I think the answer will be water ,sorry if ik wrong
