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RUDIKE [14]
3 years ago
14

A wire is carrying current vertically downward. What is the direction of the force due to Earth's magnetic field on the wire?

Physics
1 answer:
AnnyKZ [126]3 years ago
6 0

Answer:

The direction of the force will be towards the east

Explanation:

From the question we are told that  

    The direction of the  downward

Generally according to Fleming's right-hand rule(

          Thumb -  direction of force

           Middle finger -  direction of current

           Index finger -  direction of the magnetic field

) and the fact that the earth magnetic field acts  from south to north with respect to the four cardinal points then the direction of the  force will be toward the east with respect to the four cardinal point on the earth

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The rare earth elements include
irga5000 [103]
~Hello there!

Your questions: The rare Earth elements include ___________________.

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2) Holmium 

3) Yttrium 

4) Promethium 

5) Samarium.

Note: These are some of the rare Earth elements, there are MANY rare elements that are found in Earth.

Any questions ^?

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3 years ago
Whay is the other name of negative acceleration.​
sladkih [1.3K]

Answer:

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3 years ago
Two equally charged, 2.807 g spheres are placed with 3.711 cm between their centers. When released, each begins to accelerate at
statuscvo [17]
\begin{gathered} m=\text{ 2.807 g} \\ d=\text{ 3.711cm} \\ a=260.125\text{ m/s}^2 \\ m=\text{ mass of  both the spheres} \\ d=\text{ distance between the centers of sphere.} \\ a=\text{ acceleration of spheres.} \end{gathered}\begin{gathered} force\text{ due to the sphere having charge q, outside its surface is given by } \\ \vec{F}=\frac{1}{4\pi\epsilon_o}\frac{q_1q_2}{r^2}\hat{r} \\ q_1=charge\text{ on the source object.} \\ q_2=charge\text{ of the object in which we are observing the force.} \\ F=\text{ the force on the charged particle outside the sphere} \\ r=\text{ distance of the charged particle from the center of the sphere} \\ \hat{r}\text{= direction of the force acting on the charged particle} \end{gathered}\begin{gathered} from\text{ Newton's second law} \\ F=ma \\ F=\text{ force acting on the particle.} \\ m=\text{ mass of the object.} \\ a=\text{ acceleration of the object.} \end{gathered}\begin{gathered} from\text{ both the equation } \\ ma=\frac{1}{4\pi\epsilon_o}\frac{q_1q_{\frac{2}{}}}{r^2}\hat{r} \\ here\text{ q}_1\text{ and q}_2\text{ are the same, according to the question.} \end{gathered}\begin{gathered} converting\text{ all the values in s.i. unit} \\ m=2.807*10^{-3}kg \\ d=3.711*10^{-2}m \\ according\text{ to the question q}_1=\text{ q}_2 \\ value\text{ of }\frac{1}{4\pi\epsilon_o}=9*10^9\text{ Nm}^2\text{/C}^2 \\ now\text{ put all the values in the above equation } \\ 2.807*10^{-3}kg*260.125\text{ m/s\textasciicircum2}=9*10^9Nm^2\text{/C}^2*\frac{q^2}{3.711*10^{-2}m} \\  \end{gathered}\begin{gathered} by\text{ trasformation} \\ q=\sqrt{\frac{2.807*10^{-3}kg*260.125m/s^2*3.711*^10^{-2}m}{9*10^9Nm^2\text{/C}^2}} \\ by\text{ solving this we get } \\ q=17.3514*10^{-7}C \\ q=1.73514\text{ micro coulombs.} \end{gathered}Hence the correct answer is q= 1.73514 micro coulombs.
5 0
1 year ago
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melomori [17]

Notice that

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7 0
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kirza4 [7]

Explanation:

Bonjour , pourriez-vous m'aider à faire cet exercice svp ??

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