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natita [175]
2 years ago
7

The equation of a transverse wave traveling along a string is given by y=0.3sin (0.5x-50t) find the maximum displacement of the

particles for the equilibrium position​
Physics
1 answer:
uysha [10]2 years ago
5 0

Answer:

0.3cm

Explanation:

Y = 0.3 sin(0.5x - 50t) compare with,

y = A sin(kx - wt)

A = 0.3m

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How much fuel dose it take to get a rocket to orbit the earth?
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Read 2 more answers
A police car is driving down the street with it's siren on. You are standing still on the sidewalk beside the street. If the fre
AleksandrR [38]

Answer:

A) 1568.60 Hz

B) 1437.15 Hz

Explanation:

This change is frequency happens due to doppler effect

The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source

f_(observed)=\frac{(c+-V_r)}{(C+-V_s)} *f_(emmited)\\

where

C = the propagation speed of waves in the medium;

Vr= is the speed of the receiver relative to the medium,(added to C, if the receiver is moving towards the source, subtracted if the receiver is moving away from the source;

Vs= the speed of the source relative to the medium, added to C, if the source is moving away from the receiver, subtracted if the source is moving towards the receiver.

A) Here the Source is moving towards the receiver(C-Vs)

and the receiver is standing still (Vr=0) therefore the observed frequency should get higher

f_(observed)=\frac{C}{C-V_s} *f_(emmited)\\=\frac{343}{343-15}*1500\\ =1568.60 Hz

B)Here the Source is moving away the receiver(C+Vs)

and the receiver is still not moving (Vr=0) therefore the observed frequency should be lesser

f_(observed)=\frac{C}{C+V_s} *f_(emmited)\\=\frac{343}{343+15}*1500\\ =1437.15 Hz

3 0
3 years ago
What is the minimum runway length that will serve? hint: you can solve this problem using ratios without having any additional i
ehidna [41]

There are many factors that determine if an aircraft can operate from a given airport. Of course the availability of certain services, such as fuel, access to air stairs and maintenance are all necessary. But before considering anything else, one must determine if the plane can physically land at an airport, and equally as important, take off.

What is the minimum runway length that will serve?

Looking at aerial views of runways can lead some to the assumption that they are all uniform, big and appropriate for any plane to land. This couldn’t be further from the truth.

A given aircraft type has its own individual set of requirements in regards to these dimensions. The classic 150’ wide runway that can handle a wide-body plane for a large group charter flight isn’t a guarantee at every airport. Knowing the width of available runways is important for a variety of reasons including runway illusion and crosswind condition.

Runways also have different approach categories based on width, and have universal threshold markings that indicate the actual width.

To learn more about runway

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6 0
2 years ago
What is the energy stored between 2 Carbon nuclei that are 1.00 nm apart from each other? HINT: Carbon nuclei have 6 protons and
Andrei [34K]

Answer:

A. 8.29\times 10^{-18}\ J

Explanation:

Given that:

p = magnitude of charge on a proton = 1.6\times 10^{-19}\ C

k = Boltzmann constant = 9\times 10^{9}\ Nm^2/C^2

r = distance between the two carbon nuclei = 1.00 nm = 1.00\times 10^{-9}\ m

Since a carbon nucleus contains 6 protons.

So, charge on a carbon nucleus is q = 6p=6\times 1.6\times 10^{-19}\ C=9.6\times 10^{-19}\ C

We know that the electric potential energy between two charges q and Q separated by a distance r is given by:

U = \dfrac{kQq}{r}

So, the potential energy between the two nuclei of carbon is as below:

U= \dfrac{kqq}{r}\\\Rightarrow U = \dfrac{kq^2}{r}\\\Rightarrow U = \dfrac{9\times10^9\times (9.6\times 10^{-19})^2}{1.0\times 10^{-9}}\\\Rightarrow U =8.29\times 10^{-18}\ J

Hence, the energy stored between two nuclei of carbon is 8.29\times 10^{-18}\ J.

8 0
3 years ago
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