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3 years ago

HELP PLS

2 answers:

4 0

If the field is in a vacuum, the magnetic field is the dominant factor determining the motion. Since the magnetic force is perpendicular to the direction of travel, a charged particle follows a curved path in a magnetic field. The particle continues to follow this curved path until it forms a complete circle. Another way to look at this is that the magnetic force is always perpendicular to velocity, so that it does no work on the charged particle. The particle’s kinetic energy and speed thus remain constant. The direction of motion is affected but not the speed.

A negatively charged particle moves in the plane of the paper in a region where the magnetic field is perpendicular to the paper (represented by the small × ’s—like the tails of arrows). The magnetic force is perpendicular to the velocity, so velocity changes in direction but not magnitude. The result is uniform circular motion.

Free_Kalibri [48]3 years ago

4 0

I think it would be a electron
I’m not sure if my explanation is correct but I’m gonna say it’s bc they attract ik it had to be right it’s just my explanation that’s wrong

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A wind instrument that produces only odd-numbered standing wave modes has what configuration of its ends?

Luda [366]

Answer:

Open- closed

Explanation:

It has Open-closed configuration of its end

4 0

3 years ago

A certain capacitor, in series with a resistor, is being charged. At the end of 10 ms its charge is half the final value. The ti

vichka [17]

To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

$q(t) = e^{-\frac{t}{(R*C)}}$

Here,

q = Charge

t = Time

R = Resistance

C = Capacitance

When the charge reach its half value it has passed 10ms, then the equation is,

$\frac{1}{2}*q_{final} = e^{-\frac{0.01}{(R*C)}}$

$Ln(\frac{1}{2}) = -\frac{0.01}{RC}$

$- RC = \frac{0.01}{Ln(1/2)}$

$RC = 0.014s$

We know that RC is equal to the time constant, then

$T = RC = 0.014s = 14ms$

Therefore the time constant for the process is about 14ms

5 0

3 years ago

To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp

motikmotik

Answer:

0.79 s

Explanation:

We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:

$\sum F_x:f_{max}=ma_x\\\sum F_y:N-mg=0$

The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that $f_{max}=\mu N$

Now, we find the acceleration:

$\mu N=ma_x\\\mu mg=ma_x\\a_x=\mu g\\a_x=0.83(9.8\frac{m}{s^2})\\a_x=8.134\frac{m}{s^2}$

Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:

$x=v_0t+\frac{1}{2}a_xt^2\\2x=a_xt^2\\t=\sqrt{\frac{2x}{a}}\\t=\sqrt{\frac{2(3.2m)}{8.134\frac{m}{s^2}}}\\t=0.79 s$

8 0

4 years ago

A railroad train is traveling at a speed of 26.0 m/s in still air. The frequency of the note emitted by the locomotive whistle i

olga55 [171]

Answer: 0.757m; 0.881m; 432.70Hz; 371.89Hz

Explanation:

Give the following :

Velocity of train (Vt) = 26m/s

Frequency of sound (Fs) = 420Hz

Speed of sound (Vs) = 344m/s

1) wavelength = (Vs - Vt) / Fs

Wavelength = (344 - 26) / 420 = 318/420 = 0.757m

11) Wavelength = (Vs + Vt) / Fs

Wavelength = (344 + 26) / 420 = 370/420 = 0.881m

111) According to the doppler effect :

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = frequency of listener ; fs = frequency of sound source ; V = speed of sound ; Vl = Velocity of listener ; Vs = speed of sound source

Vs = - ve (train moving towards listener)

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = [(344 + 0) / (344 - 26)] * 400

Fl = (344 / 318) * 400 = 432.70Hz

1V) Vs = + ve (train moving away listener)

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = [(344 + 0) / (344 + 26)] * 400

Fl = (344 / 370) * 400 = 371.89Hz

6 0

3 years ago

Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the

astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

$B = \frac{\mu_oI}{2\pi R}$

Where $\mu_o$ is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

$B_1 = \frac{\mu_oI_1}{2\pi R}$

$B_2 = \frac{\mu_oI_2}{2\pi R}$

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T = $\frac{\mu_oI_1}{2\pi R}$ - $\frac{\mu_oI_2}{2\pi R}$

4.0×10⁻⁶T = $\frac{\mu_o}{2\pi R}\times (I_1-I_2 )$

also

$\frac{I_1}{I_2} = \frac{3}{1}$

⇒ $I_1 = 3\times I_2$

substituting the values in the above equation we get

4.0×10⁻⁶T = $\frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)$

⇒ $I_2 = 1A$

also

$I_1 = 3\times I_2$

⇒ $I_1 = 3\times 1A$

⇒ $I_1 = 3A$

Hence, the larger of the two currents is 3A

3 0

3 years ago