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pochemuha
2 years ago
12

A proton is projected into a magnetic field that is directed along the positive x axis. Find the direction of the magnetic force

exerted on the proton for each of the following directions of the proton's velocity: (b) the negative y direction
Physics
1 answer:
bulgar [2K]2 years ago
4 0

Te direction of the magnetic force for the velocity of the proton in the

-ve  y direction will be +ve z direction.

As we know that the right-hand rule is based on the relation of magnetic fields and the forces that they exert on moving charges.When a charged particle moves under a magnetic field, it exerts a force on the particle, which is not in the same direction but different than the direction of the magnetic field.Under the right-hand rule,  if we point our pointer finger in the direction of the charged particle is moving and the middle finger is representing the direction of the magnetic field then our thumb depicts the direction of the magnetic force which is exerted on the charged particle.

So,  we are given that the direction of the velocity of the proton is in the negative y direction and the direction of the magnetic field is in the positive x  direction,  so the magnetic force is acting in the positive z direction.

To know  more about the right-hand rule refer to the link brainly.com/question/9750730?referrer=searchResults.

#SPJ4

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To practice Problem-Solving Strategy 21.1 Coulomb's Law. Three charged particles are placed at each of three corners of an equil
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Answer:

The force felt by charge 3 is F=(-5.6*10⁻⁶,3.36⁻⁵)N

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As the superposition principle applies to static charges, we can find the net electric force as the sum of the two forces felt by q3.

Looking at the drawing and knowing that they form an equilateral triangle of lenght 4 we can conclude that each internal angle is 60°.

So, the positions in our coordinate system are:

r_1=(0,0)\\r_2=(4\ cos(60\°),4\ sin(60\°))\\r_3=(4,0)\\

Now  using Coulomb's force:

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Where d=4, q1 = -7.8*10⁻⁹C, q2 = -15.6 *10⁻⁹C, q3 = 8.0 *10⁻⁹C, k=8.98*10⁹, e0=8.8*10¹⁰:

Replacing we get 2 equations:

F_{13}=\frac{-kq_1q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(-0.04\ cos(60\°),-0.04\ sin(60\°))\\\\F_{23}=\frac{-kq_2q_3}{d^2}(\vv{r}_1-\vv{r}_3)=\frac{-kq_1q_2}{d^2}(0.04-0.04\ cos(60\°),-0.04\ sin(60\°))\\

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=((3.5*10⁻⁴-7*10⁻⁴)*0.016,(3.5*10⁻⁴+7*10⁻⁴)*0.032)=

F=(-5.6*10⁻⁶,3.36⁻⁵)N

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3 years ago
A 93.5 kg snowboarder starts from rest and goes down a 60 degree slope with a 45.7 m height to a rough horizontal surface that i
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Answer:

a. 29.9 m/s, b. 29.6 m/s, c. 44.7 m

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v = √(2×9.8×45.7)

v = 29.9 m/s

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Sum of the forces in the y direction:

∑F = ma

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c) This is the same as part a, but this time, the weight component parallel to the incline is pointing left.

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h = 44.7 m

Using energy:

KE = PE

1/2 mv² = mgh

h = v² / (2g)

h = (29.6)² / (2×9.8)

h = 44.7 m

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