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allochka39001 [22]

4 years ago

12

These four photos show features of earth that change with time, some fast and some slow. which of these features could not exist

if earth did not have a magnetic field?

1 answer:

allochka39001 [22]4 years ago

5 0

If magnetic field on earth didnt exist then there are horrible things will happen to us. the compass wont work, compass only work due to the magnetic pole of the earth. cosmic rays can reach the surface of the earth, and it is very dangerous to us. the atmosphere will be gone

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Stellar objects moving away from earth________.

Zanzabum

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4 years ago

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A student is experimenting with some insulated copper wire and a power supply. She winds a single layer of the wire on a tube wi

OverLord2011 [107]

Answer:

$P=214.7187\,W$

Explanation:

Given that:

Diameter of the solenoid, $D=10\,cm=0.1\,m$

length of the solenoid, $L=90\,cm=0.9\,m$

diameter of the wire, $d=0.1\,cm=10^{-3}\,m$

magnetic field at the center of the solenoid, $B=7.4\times 10^{-3}\,T$

<u>Now we need the no. of turns incorporated in the length of 90 cm:</u>

$N=\frac{Length\,\,of\,\,solenoid}{diameter\,\,of\,\, wire}$

$N=\frac{L}{d}$

$N=\frac{0.9}{10^{-3}}$

$N=900\,\,turns$

For solenoids we have:

$B=\mu.n.I$ ...............................(1)

where:

$\mu=$ permeability of the medium

n = no. of turns per unit length

I = current in the coil

So,

$n=\frac{900}{0.9}$

$n=1000\,turns\,.\,m^{-1}$

Now putting the respective values in the eq. (1)

$7.4\times 10^{-3}=4\pi\times10^{-7}\times 1000\times I$

$I=5.8887\,A$

For copper we have resistivity:

$\rho=1.72\times 10^{-8}\, \Omega.m$

We know that resistance is given by:

$R=\rho.\frac{l}{a}$ .....................................(2)

where:

l = length of the conducting wire

a = cross sectional area of the conducting wire

<u>Now we need the length (l) of the wire:</u>

Circumference of the solenoid,

$C=\pi.D$

$C=0.1\pi\,m$

$\therefore l=C\times N$

$l=90\pi\,m$

&

<u>Cross-sectional area of wire:</u>

$a=\pi.\frac{d^2}{4}$

$a=\pi. \frac{(10^{-3})^2}{4}\,m^2$

<u>Resistance from eq. (2):</u>

$R=1.72\times 10^{-8}\times \frac{90\pi}{\pi. \frac{(10^{-3})^2}{4}}$

$R=6.192 \,\Omega$

For power we have:

$P=I^2.R$

$P=5.8887^2 \times 6.192$

$P=214.7187\,W$

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3 years ago

Study the graph above. What can you conclude about ticket prices and concert attendance, based on this data?

iren2701 [21]

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2 years ago

As the wave interacts with a wall, which kind of wave interaction is shown?

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3 years ago

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A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of ma

tresset_1 [31]

Answer:

Explanation:

a. Given that:

$m-$ mass of first & second piece, $3m-$ mass of 3rd piece, $\bar v_1$ -velocity of first piece( $-23\dot i \ m/s$ ) and $\bar v_2$ as velocity of 2nd piece ( $-23\dot j \ m/s$ )

Let $\bar v_3$ be velocity of 3rd piece=?

#Vessel is at rest before explosion. Considering conservation of linear momentum:

$m\bar v_1 +m\bar v_2 +3m \bar v_3=0$ #Dividing both sides by $m, m\neq 0$

$\bar v_1+ \bar v_2 +3\bar v_3=0\\3\bar v_3=-(\bar v_1 +\bar v_2)\\\bar v_3=-\frac{1}{3}(\bar v_1 +\bar v_2)$

#Plug the $\bar v_1 ,\bar v_2$ values:

$\bar v_3=-\frac{1}{3}(-23\dot i -23\dot j)=\frac{1}{3}(23\dot i +23\dot j)$

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$|\bar v_3|=\sqrt{(23/3)^2+(23/3)^2}\\=10.84m/s$

Magnitude of the 3rd piece is 10.84 m/s

b. To find direction of the magnitude (as an angle relative to the $x$ -axis), we find $\angle \theta$ . The angle is obtained by getting the tan inverse as:

$\theta=tan^-^1(\frac{23/3}{23/3})\\=45\textdegree$

-The direction of the magnitude (angle relative to the x-axis) is 45°

6 0

3 years ago