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pantera1 [17]
3 years ago
6

Which of the following statements incorrectly describes a scenario where the source of the sound waves and the person hearing th

e sound are remaining stationary? A. The frequency of the sound waves is not affected by the nature of the medium. B. The frequency of the sound waves is not altering as a result of the distance between the sound source and the listener. C. The frequency of the sound waves will change for the listener as a result of the Doppler effect. D. The frequency of the sound waves will remain the same to the listener
Physics
1 answer:
iris [78.8K]3 years ago
8 0
The answer to this question is C.  The reason why this is an incorrect statement is that the Doppler effect takes into account two moving objects.  In this scenario, both the source and the listener are stationary to the frequency of the sound waves will not change due to the Doppler effect.  For the other answers, the frequency of sound waves is not affected by nature of the medium.  The distance between a sound source and listener does not alter the frequency of the sound wave.  The frequency of the sounds waves remains the same to the listener.
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With a frequency of 500 hz, what is the period of a wave
Pavlova-9 [17]
<span>500 hz means 500 times in a second therefore its 2.</span>
5 0
3 years ago
A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is
alexgriva [62]

Answer:

R=4.22*10⁴km

Explanation:

The tangential speed v of the geosynchronous satellite is given by:

v=\frac{2\pi R}{T}

Because 2\pi R is the circumference length (the distance traveled) and T is the period (the interval of time).

Now, we know that the centripetal force of an object undergoing uniform circular motion is given by:

F_c=\frac{mv^{2} }{R}

If we substitute the expression for v in this formula, we get:

F_c=\frac{m(\frac{2\pi R}{T})^{2}}{R}=\frac{4m\pi ^{2}R}{T^{2}}

Since the centripetal force is the gravitational force F_g between the satellite and the Earth, we know that:

F_g=\frac{GMm}{R^{2}}\\\\\implies \frac{GMm}{R^{2}}=\frac{4m\pi ^{2}R}{T^{2}}\\\\R^{3}=\frac{GMT^{2}}{4\pi^{2}} \\\\R=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}} }

Where G is the gravitational constant (G=6.67*10^{-11} Nm^{2}/kg^{2}) and M is the mass of the Earth (M=5.97*10^{24}kg). Since the period of the geosynchronous satellite is 24 hours (equivalent to 86400 seconds), we finally can compute the radius of the satellite:

R=\sqrt[3]{\frac{(6.67*10^{-11}Nm^{2}/kg^{2})(5.97*10^{24}kg)(86400s)^{2}}{4\pi^{2}}}\\\\R=4.22*10^{7}m=4.22*10^{4}km

This means that the radius of the orbit of a geosynchronous satellite that circles the earth is 4.22*10⁴km.

5 0
3 years ago
In nuclear reaction 5 kg of reactants give 2kg of products
dlinn [17]
Choice A is correct.======Kinetic energy equation:   KE = (1/2)(m)(v²)This tells us that KE is directly proportional to mass and the square of velocity. In other words, the more mass and more velocity an object has, the more kinetic energy.If an object is sitting at the top of a ramp, there is no velocity and therefore no kinetic energy.    Choices B and D are wrong.A golf ball has more mass than a ping-pong ball, so a ping-pong ball would have less kinetic energy than a golf ball rolling off the end of a ramp.    Choice C is wrong.Choice A is correct.


6 0
3 years ago
A 12,000kg. railroad car is traveling at +2m/s when it
ivann1987 [24]

<u>Answer:</u>

The final velocity of the two  railroad cars is 1.09 m/s

<u>Explanation:</u>

Since we are given that the two cars lock together it shows that the collision is inelastic in nature. The final velocity due to inelastic collision is given by  

\mathrm{V}=\frac{V 1 M 1+V 2 M 2}{M 1+M 2}

where

V= Final velocity

M1= mass of the first object in kgs = 12000

M2= mas of the second object in kgs = 10000

V1= initial velocity of the first object in m/s = 2m/s

V2= initial velocity of the second object in m/s = 0 (given at rest)

Substituting the given values in the formula we get

V = 2×12000 + 0x100012000 + 10000= 2400022000= 1.09 m/s  

\mathrm{V}=\frac{2 \times 1200+0 \times 1000}{12000+10000}=\frac{24000}{22000}=1.09 \mathrm{m} / \mathrm{s}

Which is the final velocity of the two  railroad cars

8 0
3 years ago
I need help to figure out how to solve this problem and solve it!!!
wariber [46]

well it looks like the walk at a constant increasing pace then at a constant pace then increaseing pace then constant pace then they slow down then walk at a constant pace then walk at a constantly increasing pace

plz rate me  brainliest

4 0
3 years ago
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