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2 years ago

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An electron is accelerated through 2400 V from rest and then enters a region in which there is a uniform 1.70 T magnetic field.

What are (a) the maximum and (b) the minimum magnitudes for the magnetic force this charge can experience

1 answer:

aalyn [17]2 years ago

5 0

Answer:

Explanation:

Let v be the velocity acquired by electron in electric field

V q = 1/2 m v²

V is potential difference applied on charge q , m is mass of charge , v is velocity acquired

2400 x 1.6 x 10⁻¹⁹ = .5 x 9.1 x 10⁻³¹ x v²

v² = 844 x 10¹²

v = 29.05 x 10⁶ m /s

Maximum force will be exerted on moving electron when it moves perpendicular to magnetic field .

Maximum force = Bqv , where B is magnetic field , q is charge on electron and v is velocity of electron

= 1.7 x 1.6 x 10⁻¹⁹ x 29.05 x 10⁶

= 79.02 x 10⁻¹³ N .

Minimum force will be zero when electron moves along the direction of magnetic field .

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1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

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Explanation:

1)

The period of a simple pendulum is given by

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2)

The angular acceleration of the rotating disc is given by the equation

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$\omega_i$ is the initial angular velocity

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For the compact disc in this problem we have:

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For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

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x is the displacement

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For the oscillator in this problem, we have the following relationship

$a=-100 x$

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Also, the angular frequency is related to the frequency $f$ by

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Therefore, the frequency of this simple harmonic oscillator is

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4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

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x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

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According to the equation of continuity, the volume flow rate must remain constant, so we can write

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$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

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$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

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$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

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