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4 years ago

PLEASE HELP URGENT!!!!!!!!!!!!!

Physics

1 answer:

Nimfa-mama [501]4 years ago

4 0

c i think sorry if wrong

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Please help on this one somebody?

coldgirl [10]

7.5 x 10⁻¹¹m. An electromagnetic wave of frecuency 4.0 x 10¹⁸Hz has a wavelength of 7.5 x 10⁻¹¹m.

Wavelength is the distance traveled by a periodic disturbance that propagates through a medium in a certain time interval. The wavelength, also known as the space period, is the inverse of the frequency. The wavelength is usually represented by the Greek letter λ.

λ = v/f. Where v is the speed of propagation of the wave, and "f" is the frequency.

An electromagnetic wave has a frecuency of 4.0 x 10 ¹⁸Hz and the speed of light is 3.0 x 10⁸ m/s. So:

λ = (3.0 x 10⁸ m/s)/(4.0 x 10¹⁸ Hz)

λ = 7.5 x 10⁻¹¹m

8 0

4 years ago

Why are most objects not magnetic?

Vedmedyk [2.9K]

Because there’s like no metal stuff idk

6 0

3 years ago

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A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tunin

JulijaS [17]

Answer:

Explanation:

given,

tuning fork vibration = 508 Hz

accelerates = 9.80 m/s²

speed of sound = 343 m/s

observed frequency = 490 Hz

$f_s = f(\dfrac{v}{v-(-v_s)})$

$f_s = f(\dfrac{v}{v+v_s})$

$v_s = v[\dfrac{f_s}{f_o}-1]$

$= 343[\dfrac{508}{490}-1]$

$v_s=12.6 m/s$

distance the tunning fork has fallen

$y_1=\dfrac{v^2}{2a_y}$

$=\dfrac{12.6^2}{2\times 9.8}$

=8.1 m

now, time required for the observed will be

$t = \dfrac{8.1}{343} = 0.023 s$

now, for the distance calculation

$y_2 = u\ t + \dfrac{1}{2}at^2$

$= 12.6\times 0.023 +\dfrac{1}{2}\times 9.8 \times 0.023^2$

=0.293 m

total distance

= 8.1 + 0.293 = 8.392 m

3 0

3 years ago

If an atomic nucleus were the size of a dime how far away might one of its electrons be?

mihalych1998 [28]

The radius of a nucleus of hydrogen is approximately $r_{n1}=1\cdot 10^{-15}m$ , while we can use the Borh radius as the distance of an electron from the nucleus in a hydrogen atom: $r_{e1}=5.3 \cdot 10^{-11}m$

The radius of a dime is approximately $r_{n2} = 9\cdot 10^{-3}m$ : if we assume that the radius of the nucleus is exactly this value, then we can find how far is the electron by using the proportion
$r_{n1}:r_{e1}=r_{n2}:r_{e2}$
from which we find
$r_{e2}= \frac{r_{e1} r_{n2}}{r_{n1}}= \frac{(5.3 \cdot 10^{-11}m)(9\cdot 10^{-3}m)}{1 \cdot 10^{-15}m}=477 m$

So, if the nucleus had the size of a dime, we would find the electron approximately 500 meters away.

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3 years ago

Work is done if you carry a plant across a room at constant velocity. True False

BlackZzzverrR [31]

it is false hope this helps

5 0

4 years ago

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