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Elza [17]
3 years ago
8

Back 8 8. 9. 10. 11. 12. 13. 13

Physics
1 answer:
deff fn [24]3 years ago
3 0

Answer:

i dont know the answer for this question sorry for the misunderstanding

Explanation:

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***PLEASE HELP WITH ANSWER AND EXPLANATION: Imagine the current in a current-carrying wire is flowing into the screen. What is t
aalyn [17]

Magnetic field direction is given by right hand thumb rule.

If we put our thumb in the direction of current then curl of fingers will show the magnetic field direction around the wire.

Now here since current is going into the screen so we will put our thumb into the screen and then the curl of fingers is clockwise around it.

The magnetic field is clockwise.

So this would be the direction of magnetic field

5 0
3 years ago
Read 2 more answers
A thin metallic spherical shell of radius 0.357 m has a total charge of 5.03 times 10^-6 C placed on it. At the center of the sh
Artyom0805 [142]

Answer:

The electric field is 5.623\times10^{4}\ N/C

Explanation:

Given that,

Radius = 0.357 m

Charge Q=5.03\times10^{-6}\ C

Point charge q=4.15\times10^{-6}\ C

Distance = 0.815 m

We need to calculate the total electric field

Using formula of electric field

E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{r^2}

Where, q = point charge

r = distance

Put the value into the formula

E=\dfrac{9\times10^{9}\times4.15\times10^{-6}}{(0.815)^2}

E=5.623\times10^{4}\ N/C

Hence, The electric field is 5.623\times10^{4}\ N/C

4 0
3 years ago
South pole
Nata [24]

Side A: north pole

Side C: south pole

7 0
3 years ago
How hot is the sun in degrees Celsius?
Vlada [557]

Answer:

The temperature of the mantle varies greatly, from 1000° Celsius near its boundary with the crust, to 3700° Celsius

hope this helps you!  

8 0
3 years ago
A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?
dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

I = MR^2

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

\Omega = \frac{Mgd}{I\omega}

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

\omega= Angular velocity

Replacing the value for moment of inertia

\Omega= \frac{MgR}{MR^2 \omega}

\Omega = \frac{g}{R\omega}

The value for our angular velocity is not in SI, then

\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})

\omega = 104.7rad/s

Replacing our values we have that

\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}

\Omega = 1.17rad/s

The precession frequency is

\Omega = \frac{2\pi rad}{T}

T = \frac{2\pi rad}{\Omega}

T = \frac{2\pi}{1.17}

T = 5.4 s

Therefore the precession period is 5.4s

7 0
3 years ago
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