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Ronch [10]
3 years ago
7

A particle accelerator fires a proton into a region with a magnetic field that points in the +x-direction (a) If the proton is m

oving in the ty-direction, what is the direction of the magnetic force on the proton? +x-direction -x-direction +y-direction -y-direction +z-direction -z-direction zero force What is the formula used to find the vector magnetic force? What is the right-hand rule for a cross prod (b) If the proton is moving in the -y-direction, what is the direction of the magnetic force on the proton? +x-direction -x-direction +y-direction -y-direction +z-direction -z-direction zero force (c) If the proton is moving in the x-direction, what is the direction of the magnetic force on the proton?
Physics
1 answer:
pochemuha3 years ago
4 0

Answer:

a) -z direction, b) +z direction, c) F=0  

Explanation:

The magnetic force is given by the expression

        F = q v x B

the bold indicate vectors, this equation can be separated in its module

        F = a v B sin θ

and where θ is the angles between the speed and the magnetic field.

The direction of the force can be found with the right-hand rule. For a positive charge, the thumb goes in the direction of speed, the fingers extended in the direction of the magnetic field and the palm points in the direction of the force, if the charge is negative the force is in the opposite direction.

a) Let's apply this to our case

the proton is positively charged

moves in the direction of + x

The magnetized field goes in the direction of y

therefore applying the right hand rule the force must be in the direction of the negative part of the z-axis (-z)

The right-hand rule is used to find this address.

b)  in this case it indicates that the proton moves in the recode of -y

again we apply the right hand rule and the force is in the direction of + z

c)   The proton moves in the x direction

In this case the force is zero because the angle between the field and the speed is zero and the sine is zero, therefore the force is zero

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Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

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Explanation:

Given;

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intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

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P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

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F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

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A marble rolling at a speed of 2.71 m/s falls of the end of a table that is 1.25 m high. How far from the base of the table does
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Answer:

The marble lands at a distance of 1.36 m from the base of the table.

Explanation:

The motion of the marble falling off the table is a projectile with initial velocity in the horizontal direction only.

The motion can be solved in two directions, the horizontal and vertical direction.

Along the vertical direction, the initial velocity is 0 m/s as it has only horizontal component initially. Also acceleration in the vertical direction is acceleration due to gravity.

Let us use equation of motion in vertical direction.

y-y_0=v_{0y}t+\frac{1}{2}at^2

Where,

v_{0y}\rightarrow \textrm{vertical component of the initial velocity}.\\y\rightarrow \textrm{final position of the marble}\\y_0\rightarrow \textrm{initial vertical position}\\a_y\rightarrow \textrm{acceleration in the vertical direction}\\t\rightarrow \textrm{time taken to reach bottom}

Now, plug in 0 for y, 1.25 for y_0, 0 for v_{0y}, -9.8 for a_y. This gives,

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Now, consider the horizontal motion. There is no acceleration in the horizontal direction. So, distance is given as the product of horizontal velocity and time taken.

Horizontal distance covered by the marble is given as:

x=v_{0x}\times t=2.71\times 0.501=1.36 \textrm{ m}

Therefore, the marble lands at a distance of 1.36 m from the base of the table.

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