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Svetradugi [14.3K]
3 years ago
10

When a charged particle moves along a helical path in a uniform magnetic field, which component determines the pitch of the path

? the velocity component perpendicular to the magnetic field vector the velocity component parallel to the magnetic field vector the acceleration component perpendicular to the magnetic field vector the acceleration component parallel to the magnetic field vector the acceleration component radially inward the acceleration component radially outward
Physics
1 answer:
Llana [10]3 years ago
3 0

Answer:

the velocity component parallel to the magnetic field vector

Explanation:

When a charged particle moves in a helical path, we can decompose its velocity into two parts v_parallel and v_perpendicular to the magnetic field.

Let's analyze which component receives a force

            F = q vxB

the bold letters indicate vectors, in the vector product if the two vectors are parallel the angle is zero and the sin 0 = 0 for which there is no force. therefore the velocity parallel to the field remains constant

If the two vectors are perpendicular, the angle is 90º and the sin 90 = 1, for which there is a force, which has a radial direction and consequently a centripetal acceleration that gives a circular path that does not remove the particle from the magnetic field

When checking the different answers, the correct one is: the velocity component parallel to the magnetic field vector

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Which of these examples displays CONDUCTION based on the definition?
dangina [55]
A most probably because conduction transfer heat by movement of collisions of particles and movement of electrons within a body
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3 years ago
c) If the ice block (no penguins) is pressed down even with the surface and then released, it will bounce up and down, until fri
eduard

Answer:

y = 20.99 V / A

there is no friction    y = 20.99 h

Explanation:

Let's solve this exercise in parts: first find the thrust on the block when it is submerged and then use the conservation of energy

when the block of ice is submerged it is subjected to two forces its weight  hydrostatic thrust

         

              F_net= ∑F = B-W

the expression stop pushing is

              B = ρ_water g V_ice

where rho_water is the density of pure water that we take as 1 10³ kg / m³ and V is the volume d of the submerged ice

We can write the weight of the body as a function of its density rho_hielo = 0.913 10³ kg / m³

             W = ρ-ice g V

              F_net = (ρ_water - ρ_ ice) g V

this is the net force directed upwards, we can find the potential energy with the expression

            F = -dU / dy

            ΔU = - ∫ F dy

            ΔU = - (ρ_water - ρ_ ice) g ∫ (A dy) dy

            ΔU = - (ρ_water - ρ_ ice) g A y² / 2

we evaluate between the limits y = 0,  U = 0, that is, the potential energy is zero at the surface

             U_ice = (ρ_water - ρ_ ice) g A y² / 2

now we can use the conservation of mechanical energy

starting point. Ice depth point

             Em₀ = U_ice = (ρ_water - ρ_ ice) g A y² / 2

final point. Highest point of the block

             Em_{f} = U = m g y

as there is no friction, energy is conserved

            Em₀ = Em_{f}

            (ρ_water - ρ_ ice) g A y² / 2 = mg y

let's write the weight of the block as a function of its density

            ρ_ice = m / V

            m = ρ_ice V

we substitute

             (ρ_water - ρ_ ice) g A y² / 2 = ρ_ice V g y

              y = ρ_ice / (ρ_water - ρ_ ice) 2 V / A

let's substitute the values

             y = 0.913 / (1 - 0.913) 2 V / A

             y = 20.99 V / A

This is the height that the lower part of the block rises in the air, we see that it depends on the relationship between volume and area, which gives great influence if there is friction, as in this case it is indicated that there is no friction

                V / A = h

where h is the height of the block

                 y = 20.99 h

7 0
4 years ago
A big league hitter attacks a fastball! The ball has a mass of 0.16 kg. It is pitched at 38 m/s. After the player hits the ball,
Strike441 [17]
Momentum = (mass) x (velocity)
Original momentum before the hit = 
                   (0.16 kg) x (38 m/s) this way <==
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Momentum after the hit = 
                   (0.16) x (44 m/s) that way  ==>
               =           7.04 kg-m/s  that way ==>
Change in momentum = (6.08 + 7.04) =  13.12 kg-m/s  that way ==> .-----------------------------------------------
Change in momentum = impulse.
                                   Impulse = (force) x (time the force lasted)
                          13.12 kg-m/s  = (force) x (0.002 sec)
  (13.12 kg-m/s) / (0.002 sec)  =  Force
             6,560 kg-m/s² = 6,560 Newtons  =  Force    
                            ( about 1,475 pounds  ! ! ! )
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