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MAVERICK [17]
3 years ago
14

A horseshoe magnet is held vertically with the north pole on the left and south pole on the right. A wire passing between the po

les, equidistant from them, carries a current directly away from you. In what direction is the force on the wire?
Physics
1 answer:
sergejj [24]3 years ago
8 0

Answer:

Force acts vertically downward

Explanation:

Solution:-

The force is downward. The field lines point from the north pole to the south pole, or left to right. We will apply the use of right-hand rule. Your fingers point in the direction of the current (away from you). Curl them in the direction of the field (to the right). Your thumb points in the direction of the force(downward).

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A rocket is fired with an initial VELOCITY OF 100m/s at an angle of 55° above the horizontal, It explodes On the mountain Side 1
GuDViN [60]

Answer

688.32m and 277.44m

Explanation :

⠀

\large{\maltese{\textsf{\underline{To find :-}}}}

The X and Y coordinates of the rocket relative of firing

⠀

⠀

\large{\maltese{\textsf{\underline{Given :-}}}} \\ \\ \sf velocity (v_i) = 100m{s}^{-1} \\ \sf angle ({\theta}_{1}) = 55.0{\degree} \\ \sf time (t) = 12s

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\Large{\maltese{\textsf{\underline{\underline{Step by Step Solution:-}}}}}

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<u>The</u><u> </u><u>horizontal</u><u> </u><u>range</u><u> </u><u>of</u><u> </u><u>projectile</u><u> </u><u>at</u><u> </u><u>x</u><u>.</u><u> </u>

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\sf \large{x = v_{xi}} \times t \\ \\ \sf \large{x = v_i \times \cos {\theta}_{i} \times t}

⠀

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\large\textsf{\underline{Now substituting the required values}}

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⠀

\sf x = 300 \times \cos 55{\degree} \times 12 \\ \\ \sf x = 100 \times 0.5756 \times 12 \\ \\ {\underline{\boxed{\bold{ x = 688.32m}}}}

⠀

⠀

The vertical position of projectile at y.

⠀

⠀

\sf \large  y = v_{yi} \times t -  (\frac{1}{2}  \times g  \times {t}^{2}) \\  \\   \sf  \large y = v_i \times  \cos \theta  \times t -  \frac{1}{2} g {t}^{2}

⠀

⠀

\textsf{ \large {\underline{Now substituting the required values}}  }

⠀

⠀

\sf y = 100 \times  \cos55{ \degree} \times 12 -  \frac{1}{2}   \times 9.80 \times  {12}^{2} \\  \\  \sf  y = 100 \times 0.8192 \times 12 - 0.5 \times 9.8 \times 144 \\  \\  \sf y = 983.04 - 705.6 \\  \\  \underline{ \boxed{ \bold{y = 277.44m}}}

⠀

⠀

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<h3><u>Henceforth</u><u>,</u><u> </u><u>the</u><u> </u><u>distance</u><u> </u><u>at</u><u> </u><u>horizon</u><u> </u><u>is</u><u> </u><u>6</u><u>8</u><u>8</u><u>.</u><u>3</u><u>2</u><u>m</u><u> </u><u>and</u><u> </u><u>at</u><u> </u><u>vertical</u><u> </u><u>is</u><u> </u><u>2</u><u>7</u><u>7</u><u>.</u><u>4</u><u>4</u><u>m</u><u>.</u></h3>

8 0
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Which of the following statements describes an anticyclone?
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B. 
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Answer:

a. The acceleration would increase.

Explanation:

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