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mars1129 [50]
3 years ago
15

Two sound waves, from two different sources with the same frequency of 420 Hz travel in the same direction at 336 m/s. The sourc

es are in phase. What is the phase difference of the waves at a point that is 4.40 m from one source and 4.00 m from the other?
Physics
1 answer:
Tresset [83]3 years ago
8 0

Answer:

Pi(3.14) radians or 180º degrees

Explanation:

First of all, we need to obtain the wavelength of a wave traveling to the speed of sound and 420 Hz of frequency.

The formula is:

l=wavevelocity/frequency

where l = wavelength in meters

With current values:

l = 336 [m/s]/420[1/s] = 0.8 meters

Since a complete cycle (360º or 2pi radians) needs 0.8 meters to complete, 0.4 meters or 40 cm is just half of it, making a 180º degree phase or 3.14 radians.

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A loudspeaker diaphragm is vibrating in simple harmonic motion with a frequency of 760 Hz and a maximum displacement of 0.85 mm.
Alchen [17]

Answer:(a) 4775.2Hz (b) 4.06m/s (c) 19382.15m/s²

Explanation: Given that the frequency of oscilation f, is 760Hz and the maximum displacement x, is 0.85mm= 0.00085m

(a) Angular frequency w= 2πf

w= 2π × 760 = 4775.2Hz

(b) Maximum speed v is given as the product of angular frequency and maximum displacement

V=wx

V= 4775.2 × 0.00085

V= 4.06m/s

(c) The maximum acceleration a

= w²x

= (4775.2)² × (0.00085)

a= 19382.15m/s².

5 0
4 years ago
The U. S. Navy is helping these Congolese soldiers conduct an inspection of vehicles, looking mostly for car bombs. They can ins
Mashutka [201]
A flat mirror; reflection
6 0
3 years ago
Read 2 more answers
What happens if magnesium is added to hydrochloric acid (hci) in a single replacement reaction
STatiana [176]
It produces hydrogen gas
3 0
3 years ago
Un the way to the moon, the Apollo astro-
kherson [118]

Answer:

Distance =  345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2}  } \\

F_{m} =G*\frac{m_{m}*m_{a}  }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]

When we match these equations the masses cancel out as the universal gravitational constant

G*\frac{m_{e} *m_{a} }{r_{e}^{2}  } = G*\frac{m_{m} *m_{a} }{r_{m}^{2}  }\\\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2}  }

To solve this equation we have to replace the first equation of related with the distances.

\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m}  )^{2}  } = \frac{7.36*10^{22}  }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17}  \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c }  }{2*a}\\  where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) }  }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

<u />

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2}  } \\a=3.33*10^{19} [m/s^2]

6 0
3 years ago
5. 3 women push a stalled car. Each woman pushes with a 400N force. What is the mass of the car if the car accelerates at 0.85 m
iris [78.8K]

Answer:

<h2>470.59 kg</h2>

Explanation:

The the mass of the car can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{400}{0.85}  \\  = 470.58823...

We have the final answer as

<h3>470.59 kg</h3>

Hope this helps you

7 0
3 years ago
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