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3 years ago

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A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It en

ters a magnetic field of ( ) mT. What is the acceleration of the particle?

1 answer:

Virty [35]3 years ago

8 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The acceleration would be $a = 0.003* 6 =0.018\ m/s^2$

Explanation:

The objective of this solution is to obtain the acceleration of the particle

Now looking at Newton law which is mathematically represented as

$F = ma$

Where F is the force experience by a particle

m is the mass of the particle

a is the acceleration of the particle

And also this force is equivalent to magnetic force in a magnetic field which is mathematically represented as

$F = qvB$

Where q is the charge of the particle

v is the velocity of the charge

B is the magnetic field the charge is under it influence

Now equating this two formulas

$ma = qvB$

Making a the subject we have

$a = \frac{qvB}{m}$

In the question the direction of the is in the positive x-axis which is $i$ hence the direction would be in the $i$ direction

So substituting $(2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T$ for B

$a = \frac{q}{m} * v (2.0i +3.0j +4.0k)*10^{-3}$

Substituting $3.0 Km /s = 3.0*10^{3}\ m/s$ for v and $-6.0 \muC = -6.0*10^{-6} C$ for q

$a = \frac{-6.0*10^{-6}}{2.0*10^{-3}} * 3.0*10^{3} *(2i+3j+4k) *10^{-3}$

$a = 0.003 * 3i(2i+3j+4k)$

$a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j \ + (3*4)i \ \cdot \ k)$

According to vector multiplication

$i \cdot i = j \cdot j = k\cdot k = 1\\\\and \ i\cdot j = i\cdot k = 0$

So

$a = 0.003* 6 =0.018\ m/s^2$

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<span>f(x) = 5.05*sin(x*pi/12) + 5.15

First, you need to determine the period of the function. The period will be the time interval between identical points on the sinusoidal function. For this problem, the tide is rising and at 5.15 at midnight for two consecutive days. So the period is 24 hours. Over that 24 hour period, we want the parameter passed to sine to range from 0 to 2*pi. So the scale factor for x will be 2*pi/24 = pi/12 which is approximately 0.261799388. The next thing to note is the magnitude of the wave. That will simply be the difference between the maximum and minimum values. So 10.2 ft - 0.1 ft = 10.1 ft. And since the value of sine ranges from -1 to 1, we need to divide that magnitude by 2, so 10.1 ft / 2 = 5.05 ft.

So our function at this point looks like f(x) = 5.05*sin(x*pi/12) But the above function ranges in value from -5.05 to 5.05. So we need to add a bias to it in order to make the low value equal to 0.1. So 0.1 = X - 5.05, 0.1 + 5.05 = X, 5.15 = X. So our function now looks like:
f(x) = 5.05*sin(x*pi/12) + 5.15

The final thing that might have been needed would have been a phase correction. With this problem, we don't need a phase correction since at X = 0 (midnight), the value of X*pi/12 = 0, and the sine of 0 is 0, so the value of the equation is 5.15 which matches the given value of 5.15. But if the problem had been slightly different and the height of the tide at midnight has been something like 7 feet, then we would have had to calculate a phase shift value for the function and add that constant to the parameter being passed into sine, making the function look like:
f(x) = 5.05*sin(x*pi/12 + C) + 5.15
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But we don't need it for this problem, so the answer is:
f(x) = 5.05*sin(x*pi/12) + 5.15

Note: The above solution assumes that angles are being measured in radians. If you're using degrees, then instead of multiplying x by 2*pi/24 = pi/12, you need to multiply by 360/24 = 15 instead, giving f(x) = 5.05*sin(x*15) + 5.15</span>

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$x_1 = \frac{2 + 5.2}{2.4}$

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