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Tcecarenko [31]
3 years ago
13

Problem 4: a long wire carries current towards east. a positive charge moves westward and just north from the wire. what is the

direction of the force experienced by this charge?
Physics
1 answer:
Alex787 [66]3 years ago
4 0
The direction of the force experienced by the positive charge is upward.

We can use the right-hand rule to understand the direction of the Lorentz force acting on the charge: let's put the thumb in the same direction of the current in the wire (eastward), while the other fingers "wrap themselves" around the wire. These other fingers give the direction of the Lorentz force in every point of the space around the wire. Since the charge is located north of the wire, in that point the fingers are directed upward, so the positive charge experiences a force directed upward.
(if it was a negative charge, we should have taken the opposite direction)
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Which terrestrial planet exhibits retrograde rotation?.
Sati [7]

Answer:

Planets that are farther from the sun than the earth (all but Mercury and Venus) will exhibit retrograde motion.

If the position of the planet is observed relative to the background stars, the planet will appear to move backward relative to the stars when the earth is moving in an Eastward direction faster than the planet, and the planet appears to move backwards relative to the stars

(The planet will be on the side of the earth that is opposite that of the sun)

3 0
2 years ago
Help on finding kinetic energy??
Jobisdone [24]
Trick question? In order to have kinetic energy, an object must be moving. Therefore, in this case, kinetic energy would be 0. If it were asking about potential energy, it would be a different story. 
8 0
3 years ago
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Two forces F1 and F2 act on an object at a point P in the indicated directions. The magnitude of F1 is 15 lb and the magnitude o
adoni [48]

Answer

given,

F₁ = 15 lb

F₂ = 8 lb

θ₁ = 45°

θ₂ = 25°

Assuming the question's diagram is attached below.

now,

computing the horizontal component of the forces.

F_h = F₁ cos θ₁ - F₂ cos θ₂

F_h = 15 cos 45° - 8 cos 25°

F_h = 3.36 lb

now, vertical component of the forces

F_v = F₁ sin θ₁ + F₂ sin θ₂

F_v = 15 sin 45° + 8 sin 25°

F_v = 13.98 lb

resultant force would be equal to

F = \sqrt{F_h^2+F_v^2}

F = \sqrt{3.36^2+13.98^2}

F = 14.38 lb

the magnitude of resultant force is equal to 14.38 lb

direction of forces

\theta =tan^{-1}(\dfrac{F_v}{F_h})

\theta =tan^{-1}(\dfrac{13.98}{3.36})

   θ = 76.48°

3 0
3 years ago
21. (a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500-μC charge and flies due west at a
beks73 [17]

Answer:

Explanation:

The picture attached shows the full explanation and i hope it helps. Thank you

6 0
3 years ago
A car is traveling on a level road at a constant rate of speed. What additional force is necessary to bring the car into equilib
iragen [17]

Answer: Zero.

Explanation:

By the first Newton's law, we know that:

every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.

Now, we know that the car is moving with constant speed, then there is no net force acting on the car, which means that the car is already in equilibrium.

Then if we add one force to the situation, the car will not be anymore in equilibrium.

The correct option is zero.

5 0
3 years ago
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