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oksian1 [2.3K]
3 years ago
14

Hurryyyyyy When using the right-hand rule to determine the direction of the magnetic force on a charge, which part of the hand p

oints in the direction that the charge is moving? palm
fingers
back
thumb
Physics
2 answers:
masya89 [10]3 years ago
6 0
When using the right-hand rule to determine the direction of the magnetic force on a charge, which part of the hand points in the direction that the charge is moving? The answer is <span>thumb.

</span>One way to remember this is that there is one velocity, represented accordingly by the thumb. There are many field lines, represented accordingly by the fingers. The force is in the direction you would push with your palm. The force on a negative charge is in exactly the opposite direction to that on a positive charge. Because the force is always perpendicular to the velocity vector, a pure magnetic field will not accelerate a charged particle in a single direction, however will produce circular or helical motion (a concept explored in more detail in future sections). It is important to note that magnetic field will not exert a force on a static electric charge. These two observations are in keeping with the rule that <span>magnetic fields do no </span>work<span>.</span>
Ronch [10]3 years ago
3 0

Answer: The fingers.

Explanation: The magnetic force on a charge is given by the equation:

F = q(VxB)

where q is the charge of the particle, V is the velocity of the particle and B is the magnetic field.

If you use the right-hand method, you need to point with the tip of your fingers to the direction of the velocity, face the palm of your hand in the direction of the magnetic field, and now your thumb will point in the direction of the magnetic force.

So the part of the hand that points in the direction that the charge is moving is the second option, "fingers"

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Answer:

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Explanation:

<em>Here </em><em>it </em><em>is </em><em>given </em>

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<em>Time </em><em>(</em><em>t) </em><em> </em><em>=</em><em> </em><em>5</em><em>5</em><em> </em><em>sec</em>

<em>power </em><em>(</em><em>P) </em><em> </em><em>=</em><em> </em><em>?</em>

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<em>p  = \frac{w}{t}</em>

<em>P </em><em>=</em><em> </em><em>3</em><em>5</em><em>6</em><em>0</em><em>/</em><em>5</em><em>5</em>

<em>P </em><em>=</em><em> </em><em>6</em><em>4</em><em>.</em><em>7</em><em>3</em><em> </em><em>watt</em>

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If you push an 5 N object 2 m and then push a 10 N object 2 m. Which is TRUE?
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So you do more work when you move the 10N object because the Force (F) has doubled.

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A Texas rancher wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in
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The answer:
the full question is as follow:
 <span>A Texas rancher wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure below , where A = 4.90 km and θC = 15°. He then correctly calculates the length and orientation of the fourth side D. What is the magnitude and direction of vector D? 

As shown in the figure, 
A + B + C + D = 0, so to find the </span>magnitude and direction of vector D, we should follow the following method:
 D = 0 - (A + B + C) , 
let  W = - (A + B + C), so the magnitude and direction of vector D is the same of the vector W characteristics

Magnitude
 A + B + C = <span> (4.90cos7.5 - 2.48sin16 - 3.02cos15)I</span>
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lapo4ka [179]

Answer:

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g - Gravitational acceleration, measured in meters per square second.

h - Maximum height of the ball, measured in meters.

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v = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (5\,m)}

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The minimum speed needed at the top of the circle so that the cord remains tensioned and the ball's path remains circular is approximately is 9.903 meters per second.

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