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2 years ago

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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?

1 answer:

antoniya [11.8K]2 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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Answer:

Total work done in expansion will be $3.60\times 10^5J$

Explanation:

We have given pressure P = 2.10 atm

We know that 1 atm $=1.01\times 10^5Pa$

So 2.10 atm $=2.10\times 1.01\times 10^5=2.121\times 10^5Pa$

Volume is increases from 3370 liter to 5.40 liter

So initial volume $V_1=3.70liter$

And final volume $V_2=5.40liter$

So change in volume $dV=5.40-3.70=1.70liter$

For isobaric process work done is equal to $W=PdV=2.121\times 10^5\times 1.70=3.60\times 10^5J$

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2 years ago

If a mass of .343 kg moves down a 5m ramp in 5.8 seconds, what is the velocity and KE developed by the moving mass

Soloha48 [4]

Velocity 0.86m/s

0.13J

Explanation:

Given parameters:

Mass = 0.343kg

distance = 5m

time taken = 5.8s

Unknown:

Velocity of mass = ?

Kinetic energy = ?

Solution:

Velocity is the rate of change of displacement with time. It is a vector quantity that shows magnitude and direction.

Mathematically;

Velocity = $\frac{displacement}{time}$

Velocity = $\frac{5}{5.8 }$ = 0.86m/s

Kinetic energy is the energy due to the motion of a body. It is expressed mathematically as:

Kinetic energy = $\frac{1}{2} m v^{2}$

m is the mass

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Kinetic energy = $\frac{1}{2 } x 0.343 x 0.86^{2}$ = 0.13J

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3 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

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3 years ago

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Anvisha [2.4K]

Friction? For example, like when a car's tires skid on rough concrete.

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Brrunno [24]

The correct answer that would complete the given statement above would be option B. UNIT (of measurement). <span>A standard quantity of a physical property used as a factor to compare other occurring quantities of that property is called a unit of measurement. Hope this answers your question. </span>

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