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3 years ago

A sound wave travels through water at 1,500 m/s. How far would it travel in 5 s?

Physics

2 answers:

hichkok12 [17]3 years ago

8 0

7500m because 1,500 in 1 sec so 1500x5 7500

Andrews [41]3 years ago

7 0

Answer:

7,500 miles

Explanation:

lets see it gos 1,500 per second so if it gos for five seconds it will go 7500, (just 1500 times 5, its that simple).

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What is the difference between homophones and homographs.

Natali5045456 [20]

Answer:

*Homophones are words that sound the same but are different in meaning or spelling.

*Homographs are spelled the same, but differ in meaning or pronunciation.

Explanation:

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3 years ago

If you are sitting in a bus that is traveling along a straight, level road at 100 km/hr., you are traveling at 100 km/hr too. (a

aleksandr82 [10.1K]

Answer:

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3 years ago

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

Given:

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$

Let $\Delta U$ be the change in potential Energy given by

$\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J$

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

$\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s$

So the electron will be moving with $v=1.37\times10^7\ \rm m/s$

5 0

3 years ago

a stone is dropped from top of a tower of 50m high ,simltaneously another stone is thrown upward with a speed of 20m/s find the

aleksklad [387]

At the time that I'll call ' Q ', the height of the stone that was
dropped from the tower is

 H = 50 - (1/2 G Q²) ,

and the height of the stone that was tossed straight up
from the ground is

 H = 20Q - (1/2 G Q²) .

The stones meet when them's heights are equal,
so that's the time when

 50 - (1/2 G Q²) = 20Q - (1/2 G Q²) .

This is looking like it's going to be easy.

Add (1/2 G Q²) to each side.
Then it says
 50 = 20Q

Divide each side by 20: 2.5 = Q .

And there we are. The stones pass each other

 2.5 seconds

after they are simultaneously launched.

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3 years ago

What does the pupil of the eye controls

Alja [10]

The correct answer would be D

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4 years ago