What are two ways in which radio waves and X-rays are different?

I NEED PHYSICS HELP ASAP! DO NOT GUESS! I WILL GIE BRAINLEIST IF RIGHT! THANKS! :D

ale4655 [162]

<span>The repelling of the support magnet decreases friction. is the answer you're looking for . :)

hope i helped - beanz</span>

6 0

3 years ago

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The Millennium Falcon was constructed 43.6 m long. a) How fast is it traveling past Luke when he measures its length to be 30.1

Zinaida [17]

Answer:

a)0.5564c

b)43.6 m

Explanation:

Given proper length of falcrum L₀= 43.6 m

improper length L=30.1 m (when viewed from moving frame)

we know that $L=L_{0}\sqrt{1-\frac{v^2}{c^2}}$

$30.1=43.6\sqrt{1-\frac{v^2}{c^2}$

⇒ $\{1-\frac{v^2}{c^2}}$ = $\frac{30.1}{43.6}$

${1-\frac{v^2}{c^2}}=0.6904$

$\frac{v^2}{c^2}=0.3096$

$v^{2}=3096c^{2}$

v=0.5564c

this is the required speed of falcon when it passes luke

b). Since Han solo is on the Falcon its reference frame will be falcon itself hence there wont be any change in the length of Falcon that its length will be

43.6 m

8 0

3 years ago

The possible state for an electron of an atom that is proportional to its distance to the nucleus.

aksik [14]

Answer:

Energy Level..........................................................................................

5 0

4 years ago

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The kinetic energy of an object depends upon its _____ *

spin [16.1K]

The kinetic energy of an object depends upon its- C. speed and mass

7 0

3 years ago

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Completenlo por favor

Ainat [17]

La velocidad $\mathbf v$ del objeto al tiempo $t$ es descrito por

$\mathbf v(t)=v_x(t)\,\mathbf i+v_y(t)\,\mathbf j$

donde

$\begin{cases}v_x(t)={v_x}_0+a_xt\\v_y(t)={v_y}_0+a_yt\end{cases}$

El objeto no tiene aceleración horizontal, pues $a_x=0$ y $v_x$ está determinado exactamente por su velocidad inicial en la dirección horizontal. En la dirección vertical, la gravedad es la única fuerza que actúa en el objeto, pues $a_y=-9.81\,\dfrac{\mathrm m}{\mathrm s^2}$ . Entonces, la velocidad después de $3\,\mathrm s$ satisface

${v_x}_0\,\mathbf i+\left({v_y}_0+\left(-9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)(3\,\mathrm s)\right)\,\mathbf j=20\,\mathbf i-4\,\mathbf j$

Inmediatamente, vemos que ${v_x}_0=20\,\dfrac{\mathrm m}{\mathrm s}$ y podemos encontrar que ${v_y}_0=25.43\,\dfrac{\mathrm m}{\mathrm s}$ .

Su posicíon es descrita al tiempo $t$ por

$\mathbr r(t)=r_x\,\mathbf i+r_y\,\mathbf j$

con

$\begin{cases}r_x={r_0}_x+{v_0}_xt+\dfrac12a_xt^2\\\\r_y={r_0}_y+{v_0}_yt+\dfrac12a_yt^2\end{cases}$

Si la posición inicial del objeto es el origen, suponemos que $\mathbr r(0)=\mathbf 0$ , y además tenemos

$\mathbf r(t)=\left(20\,\dfrac{\mathrm m}{\mathrm s}\right)t\,\mathbf i+\left(\left(25.43\,\dfrac{\mathrm m}{\mathrm s}\right)t+\dfrac12\left(-9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\right)\,\mathbf j$

Queremos determinar el máximo de $r_y$ . Encontramos que ${r_y}_{\mathrm{max}}\approx32.9\,\mathrm m$ cuando $t\approx2.59\,\mathrm s$ .

3 0

3 years ago