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Kryger [21]
3 years ago
10

When you hear an echo you are hearing a ________of the original sound wave

Physics
1 answer:
krek1111 [17]3 years ago
7 0

Answer:

reflection

Choice D is correct

Explanation:

An echo is simply a reflection of the original sound wave which arrives at the listener with a delay after the original sound.  Sound waves can be reflected, that is bouncing off objects in the same manner a solid object can bounce off the ground. The echo sounds entirely the same as the original sound but the direction of the sound is altered.

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You and your friend live in different states and plant identical hydrangea bulbs in your garden. When they bloom, you send each
Whitepunk [10]

Answer:

sample. rewrite in your own words if you'd like.

Explanation:

Since the hydrangeas are changing color depending on whether the soil is acidic or basic, they must have an indicator. This means that any hydrangea planted in acidic soil will have blue flowers and any hydrangea planted in basic soil will have pink. One common indicator present in many plants is anthocyanin, so maybe hydrangeas contain that chemical.

8 0
3 years ago
Read 2 more answers
An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between t
viktelen [127]

Answer:

a)    F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} ),   b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

         F =k \frac{q_2q_2}{r^2}

         

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

          F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron       r₁ = x

right side -electron     r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

          F_net = k q Q  ( \frac{1}{r_1^2}+ \frac{1}{r_2^2})

          F _net = kqQ  ( \frac{1}{x^2} + \frac{1}{(d-x)^2} )

         

let's substitute the values

          F_net = 9 10⁹  1.6 10⁻¹⁹ 4.50 10⁻⁹ ( \frac{1}{x^2} + \frac{1}{(0.30-x)^2} )

          F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values ​​of the distance (x) and the net force are given

x (m)        F (N)

0.05        27.0 10-16

0.10          8.10 10-16

0.15          5.76 10-16

0.20         8.10 10-16

0.25        27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

4 0
3 years ago
Since the universe is infinite, the probability of events and reoccurrence is infinite. So that means that there is a chance, if
il63 [147K]

Answer:

That insane it might be true because  a planet sometimes quoted to be an Earth 2.0 or Earth's Cousin based on its characteristics; also known by its Kepler Object of Interest designation KOI-7016.01) is an exoplanet orbiting the Sun-like star Kepler-452 about 1,402 light-years (430 pc) from Earth in the constellation Cygnus.

Explanation:

7 0
3 years ago
If a Stone with an of original velocity of zero is falling from a ledge and takes eight seconds to hit the ground what is the fi
DaniilM [7]
Let t=time to reach the ground=8 secs, g= acceleration of gravity. The speed v on reaching the ground is gt=8g=78.4 m/s where g=9.8 m/s/s approx.
3 0
3 years ago
The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5T, collide with molecules of t
ollegr [7]

Answer:

8.79*10^6 rad/s

Explanation:

To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:

r=\frac{mv}{qB}         (1)

r: radius of the trajectory

m: mass of the electron = 9.1*10^-31 kg

v: speed of the electron = 1.0*10^6 m/s

q: charge of the electron = 1.6*10^-19 C

B: magnitude of the magnetic field = 5.0*10^-5 T

You use the fact that the angular frequency in a circular motion is given by:

\omega=\frac{v}{r}

Then, you solve the equation (1) in order to obtain v/r:

\frac{v}{r}=\omega=\frac{qB}{m}

Finally, you replace the values of the parameters:

\omega=\frac{(1.6*10^{-19}C)(5.0*10^{-5}T)}{9.1*10^{-31}kg}\\\\\omega=8.79*10^6\frac{rad}{s}

hence, the angular frequency is 8.79*10^6 rad/s

The frequency is:

f=2\pi \omega=5.5*10^7Hz

5 0
3 years ago
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