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Deffense [45]
3 years ago
14

A proton traveling due north enters a region that contains both a magnetic field and an electric field. The electric field lines

point due west. It is observed that the proton continues to travel in a straight line due north. In which direction must the magnetic field lines point
Physics
1 answer:
Bess [88]3 years ago
3 0

Answer:

upward

Explanation:

In the electromagnetic system of force if the direction of motion of proton does not changes it means that the electric and magnetic forces are such a ways that they are cancelling each other's effect.

Since, electric field lines will exert a force on the proton to the west, hence, the magnetic force must force it to the east. It is well known that magnetic force acts perpendicular to the direction of magnetic field.  magnetic field should point upward direction.

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Two 1.0 cm * 2.0 cm rectangular electrodes are 1.0 mm apart. What charge must be placed on each electrode to create a uniform el
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Answer:

The number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.

Explanation:

<u>Step 1:</u> calculate the charge on each electrode

Given;

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The distance between the electrode = 1mm = 1 X 10⁻³ m

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E =\frac{Kq}{r^2}

where;

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K is coulomb's constant = 8.99 X 10⁹ Nm²/C²

r is the distance between the electrodes = 1 X 10⁻³ m

q is the charge in each electrode = ?

q = \frac{Er^2}{K} = \frac{(2X10^6)(1X10^{-3})^2}{8.99 X10 ^9} = 0.2225 X 10⁻⁹ C

The charge on each electrode is 0.2225 X 10⁻⁹ C

<u>Step 2:</u> calculate the number of electrons to be moved from one electrode to the other.

1 electron contains 1.602 X 10⁻¹⁹ C

So, 0.2225 X 10⁻⁹ C will contain how many electrons ?

= (0.2225 X 10⁻⁹)/(1.602 X 10⁻¹⁹)

= 1.4 X 10⁹ electrons

Therefore, the number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.

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A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 20.0 m/s. Then the truck trav
aniked [119]
<span>You can use the equation
V_xf = V_xi + a_x(t)

V_xf = 20.0m/s
V_xi = 0m/s
ax = 2.0 t

Thus, solve for t and get 10seconds and then take 5 seconds to break after 20 seconds of driving so for

a) 10 + 20 + 5 = 35 seconds

</span><span>for part b)
You can use the formula
Delta x/Delta t = average velocity
 
Need to find xf, knowing xi = 0

Thus, use the formula
 x_f = x_i + V_xi(t) + (1/2)a_x(t)^(2)
x_f = 0 + 0(10) + (1/2)(2.0)(10)^(2)
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so for the first 10 seconds the truck traveled 100ms At a speed of 20m/s

20m/s = xm/20s 20*20 = x
x = 400
 
thus we have 100+400 = 500m then it slows down from 500m to x_f
 
thus I use the equation
x_f = x_i + (1/2)(V_xf + V_xi)t
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therefore the total distance traveled is 550m
</span>
<span>to calculate average velocity
550/35 = 16m/s

thus V_xavg = 16m/s</span>
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3 years ago
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