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ale4655 [162]
3 years ago
11

There is a uniform magnetic field of magnitude B, pervading all space, perpendicular to the plane of rod and rails. The rod is r

eleased from rest, and it is observed that it accelerates to the left. In what direction does the magnetic field point?
Physics
1 answer:
Charra [1.4K]3 years ago
6 0

The right hand rule to find the direction of the magnetic field for a falling bar is:

  • The charge is positive the magnetic field is outgoing, horizontally and towards us.
  • The charge of the bar is negative, the magnetic field is incoming, that is horizontal away from us.

The magnetic force is given by the vector product of the velocity and the magnetic field.

        F = q v x B

Where the bolds indicate vectors, F is the force, q the charge on the particle, v the velocity and B the magnetic field.

In the vector product, the vectors are perpendicular, which is why the right-hand rule has been established, see attached:

  • The thumb points in the direction of speed.
  • Fingers extended in the direction of the magnetic field.
  • The palm is in the direction of the force if the charge is positive and in the opposite direction if the charge is negative.

They indicate that the bar is dropped, therefore its speed is vertical and downwards, it moves to the left therefore this is the direction of the force, we use the right hand rule, the magnetic field must be horizontal, we have two possibilities:

  • If the charge is positive the magnetic field is outgoing, horizontally and towards us.
  • If the charge of the bar is negative, the magnetic field is incoming, that is, horizontal away from us

In conclusion using the right hand rule we can find the direction of the magnetic field for a falling bar is:

  • The charge of the bar is negative, the magnetic field is incoming, that is horizontal away from us.
  • The charge is positive the magnetic field is outgoing, horizontally and towards us.

Learn more about the right hand rule here:  brainly.com/question/12847190

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Which of the following scientists taught physics in the United States?
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In the figure, a weightlifter's barbell consists of two identical small but dense spherical weights, each of mass 50 kg. These w
kondaur [170]

The moment of inertia is 24.8 kg m^2

Explanation:

The total moment of inertia of the system is the sum of the moment of inertia of the rod + the moment of inertia of the two balls.

The moment of inertia of the rod about its centre is given by

I_r = \frac{1}{12}ML^2

where

M = 24 kg is the mass of the rod

L = 0.96 m is the length of the rod

Substituting,

I_r = \frac{1}{12}(24)(0.96)^2=1.84 kg m^2

The moment of inertia of one ball is given by

I_b = mr^2

where

m = 50 kg is the mass of the ball

r=\frac{L}{2}=\frac{0.96}{2}=0.48 m is the distance of each ball from the axis of rotation

So we have

I_b = (50)(0.48)^2=11.5 kg m^2

Therefore, the total moment of inertia of the system is

I=I_r + 2I_b = 1.84+ 2(11.5)=24.8 kg m^2

Learn more about inertia:

brainly.com/question/2286502

brainly.com/question/691705

#LearnwithBrainly

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3 years ago
If you push any floating object down from equilibrium and release it, it bobs up and down. That looks like an oscillation, so le
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Answer:

  F_{y} = ( ρ_fluid g A) y

Explanation:

This exercise can be solved in two parts, the first finding the equilibrium force and the second finding the oscillating force

for the first part, let's write Newton's equilibrium equation

        B₀ - W = 0

        B₀ = W

        ρ_fluid g V_fluid = W

the volume of the fluid is the area of ​​the cube times the height it is submerged

      V_fluid = A y  

For the second part, the body introduces a quantity and below this equilibrium point, the equation is

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        ρ_fluid g A (y₀ + y) - W = m a

        ρ_fluid g A y + (ρ_fluid g A y₀ -W) = m a

       ρ_fluid g A y + (B₀-W) = ma

the part in parentheses is zero since it is the force when it is in equilibrium

      ρ_fluid g A y = m a

      this equation the net force is

      F_{y} = ( ρ_fluid g A) y

we can see that this force varies linearly the distance and measured from the equilibrium position

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Ideal mechanical advantage is equal to the displacement of the effort force divided by the displacement of the load.
ELEN [110]
This would be false. hope this helps. good luck :)
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