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skad [1K]
3 years ago
6

Two compact sources of sound oscillate in phase with a frequency of 100 Hz. At a point 5.00m from one source and 5.85 m from the

other, the amplitude of the sound from each source separately is A. (a) What is the phase difference of the two waves at that point? (b) What is the amplitude (in terms of A) of the resultant wave at that point?
Physics
1 answer:
antoniya [11.8K]3 years ago
5 0

Answer:

(a)1.557 radian (b) 1.424 A

Explanation:

Frequency of oscillation of sound = 100 Hz

\Delta r=5.85-5=0.85m

(a)The phase difference is given by \frac{2\pi \Delta r}{λ} where v is the velocity of sound in air

So phase difference \frac{2\pi \Delta rf}{v} as \lambda =\frac{v}{f}

So phase difference =\frac{2\times 3.14\times 0.85\times 100}{343}=1.557radian

(b) The  resultant amplitude is given by 2Acos\frac{\Phi }{2}=2\times A\times cos\frac{1.557}{2}=1.424A

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A student pushes against a wall with a force of 30N. The wall does not move. What amount of force does the wall exert on the stu
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A car travels 40 miles in 30 minutes.
lukranit [14]

Answer:

(a)Average velocity ,v =128.74 Km/hr

(b)Kinetic Energy , K=958546.875 Joule

(c)Distance, s=268.8m

(d)Acceleration, a= - 2.38 m/s^2

<u>Explanation</u>:

<u>Given</u>:

Distance travelled = 40 miles

Time taken = 30 minutes.

(A) The average velocity in kilometres/hour

Converting 40 miles into km ,

we know that,

1 mile = 1.60934

40 miles =  40 x 1.60934

so 40 miles  =  64.3738 Km

similarly converting 30 minutes into hours

1 minute = \frac{1}{60}hours

30 minute = \frac{30}{60}hours

30 minute = \frac{1}{2}hours

Now

Average velocity = \frac{Speed}{time}

Substituting Values,

Average velocity = \frac{64.3738}{\frac[1}{2}}

Average velocity = 64.3738 \times 2

Average velocity =128.74 Km/hr

(B) If the car weighs 1.5 tons, what is its If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)? in joules (Note: you will need to convert your velocity to m/s)?

Converting 1.5 tons into kg we get

1 ton = 1000 kg

so 1.5 ton =1500 kg

converting  velocity to m/s

128.74  \times \frac{5}{18}

=>35.75 m/s----------------------------------------------------------(1)

kinetic energy  K= \frac{1}{2}mv^2

Substituting the values,

K= \frac{1}{2}1500(35.75)^2

K= \frac{1}{2}1500(1278.06)

K= \frac{1500 \times (1278.06)}{2}

K= \frac{1917093.75}{2}

K=958546.875 Joule---------------------------------------------(2)

(c)When the driver applies the brake, it takes 15 seconds to stop. How far does the car travel (in meters) while stopping

Lets use Distance formula,

S= ut+\frac{1}{2}at^2

Substituting the known values,

s= ut+\frac{1}{2}at^2

s= (37.75)(15)+\frac{1}{2}a(15)^2

s=566.25+\frac{1}{2}a(225)

s=566.25+\frac{(225a)}{2}-------------------------------------(3)

(D) What is the average acceleration of the car (in m/s2) during braking?

Using the formula

v=u +at

re arranging the formula we get,

a = \frac{v - u}{t}

Substituting the values

a = \frac{0 - 35.75}{15}

a = \frac{- 35.75}{15}

a= - 2.38 m/s^2----------------------------------------(4)

Now substituting 4 in 3 we get

s=566.25+\frac{(225( - 2.38)}{2}

s=566.25+\frac{-535.5}{2}

s=536.25-267.75

s=268.8m--------------------------------------------------------------(5)

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