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Veronika [31]
3 years ago
8

A small compass is placed near a current-carrying wire. The wire is vertical, and the compass is on a horizontal surface. The re

d end of the compass needle would point to geographic north if it weren’t near the wire. What direction is the current flowing in the wire, up or down? Explain your reasoning

Physics
1 answer:
azamat3 years ago
5 0

Answer:

The direction of magnetic field produced by a current carrying wire is given by the right hand thumb rule. If the thumb points in the direction of current the fingers curl along the direction of magnetic field.

further, The red end of the compass needle points in the direction of external magnetic field. As the red end of the needle is pointing away from us, the external magnetic field at its location should also be directed away from us. Using the right hand thumb rule, we can see that this is only possible if the current in the wire is flowing upwards.

Explanation:

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rosijanka [135]
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3 0
2 years ago
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Cool facts about satellites ???
svetoff [14.1K]
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7 0
2 years ago
Having aced your Physics 2111 class, you get a sweet summer-job working in the International Space Station. Your room-mate, Cosm
Sphinxa [80]

Answer:

a

The speed is   s =  5.857 m/s

b

The distance is  D = 22.4  \  m

Explanation:

From the question we are told that

     The speed of the banana is  v =  16 \ m/s

   The distance from my  location is  d =  8.2  \ m  

     The time taken is  t = 1.4 \ s

The speed of the ice cream is

          s =  \frac{d}{t}

substituting values

        s =  \frac{8.4}{1.4}

        s =  5.857 m/s

The distance of separation between i and Valdimir is the same as the distance covered by the banana

   So  

          D =  v * t

substituting values

        D = 16 *  1.4

        D = 22.4  \  m

     

3 0
2 years ago
The submarine emits a pulse of sound to detect other objects in the sea. The sped of sound in sea water is 1500m/s. An echo is r
r-ruslan [8.4K]

Answer:

d = 375 m

Explanation:

The speed of sound is constant in any medium, therefore we can use the uniform motion relationships

          v = x / t

          x = v t

In this case it indicates that the time since the sound is emitted and received is t = 0.50 s, in this time the sound traveled a round trip distance

           x = 2d

          2d = v t

          d = v t/2

     

let's calculate

          d = 1500 0.5 / 2

          d = 375 m

3 0
3 years ago
Two objects carry initial charges that are q1 and q2, respectively, where |q2| &gt; |q1|. They are located 0.160 m apart and beh
mart [117]

Answer:

\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force between two static point charges  \rm q_1 and \rm q_1, separated by a distance \rm r, is given by

\rm F = \dfrac{kq_1q_2}{r^2}.

where k is the Coulomb's constant.

Initially,

\rm r = 0.160\ m\\F_i = -1.30\ N.\\\\and \ \ |q_2|>|q_1|.

The negative sign is taken with force F because the force is attractive.

Therefore, the initial electrostatic force between the charges is given by

\rm F_i = \dfrac{kq_1q_2}{r^2}.\\-1.30=\dfrac{kq_1q_2}{0.160^2}\\\rm\Rightarrow q_2 = \dfrac{-1.30\times 0.160^2}{q_1k}\ \ \ ..............\ (1).

Now, the objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions.

The force is now repulsive, therefore, \rm F_f = +1.30\ N.

The new charges on the two objects are

\rm q_1'=q_2' = \dfrac{q_1+q_2}{2}.

The new force is given by

\rm F_f = \dfrac{kq_1'q_2'}{r^2}\\+1.30=\dfrac{k\left (\dfrac{q_1+q_2}{2}\right )\left (\dfrac{q_1+q_2}{2}\right )}{0.160^2}\\\Rightarrow \left (\dfrac{q_1+q_2}{2}\right )^2=\dfrac{+1.30\times 0.160^2}{k}\\(q_1+q_2)^2=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+q_2^2+2q_1q_2=\dfrac{4\times 1.30\times 0.160^2}{k}\\\\

Using (1),

\rm q_1^2+\left ( \dfrac{-1.30\times 0.160^2}{q_1k}\right )^2+2\left (\dfrac{-1.30\times 0.160^2}{k} \right )=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+\dfrac 1{q_1^2}\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0

\rm q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )=0\\q_1^4-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2=0

\rm q_1^4-q_1^2\left (2.22\times 10^{-11} \right )+\left ( 1.37\times 10^{-23}\right ) =0\\\Rightarrow q_1^2 = \dfrac{-(-2.22\times 10^{-11})\pm \sqrt{(-2.22\times 10^{-11})^2-4\cdot (1)\cdot (1.37\times 10^{-23})}}{2}\\=1.11\times 10^{-11}\pm 1.046\times 10^{-11}.\\=6.4\times 10^{-13}\ \ \ or\ \ \ 2.156\times 10^{-11}\\\Rightarrow q_1 = \pm 8.00\times 10^{-7}\ C\ \ \ or\ \ \ \pm 4.64\times 10^{-6}\ C.

Using (1),

When \rm q_1 = \pm 8.00\times 10^{-7}\ C,

\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 8.00\times 10^{-7}\times 9\times 10^9}=\mp4.6\times 10^{-6}\ C.

When \rm q_1=\pm 4.6\times 10^{-6}\ C,

\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 4.64\times 10^{-6}\times 9\times 10^9}=\mp7.97\times 10^{-7}\ C\approx 8.0\times 10^{-7}\ C.

Since, \rm |q_2|>|q_1|

Therefore, \rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.

7 0
2 years ago
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