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FinnZ [79.3K]
3 years ago
5

A proton travels through a region of uniform magnetic field at an angle \thetaθ relative to the magnetic field. The magnitude of

the magnetic field in this region is 1.17 T, and the proton's velocity is 500,000 m/s when it experiences an acceleration whose magnitude is 1.50 \times 10^{13}~\frac{m}{s^2}1.50×10 ​13 ​​ ​s ​2 ​​ ​ ​m ​​ . Calculate the angle \thetaθ
Physics
1 answer:
MArishka [77]3 years ago
7 0

Answer:

\theta=15.52^{\circ}

Explanation:

The magnitude of the force of a moving cahrge, in our case a proton, trought a magentic field is given by:

F=|q||vB|sin(\theta) (1)

where:

q is the proton charge (q=1.6*10^{-19} C)

v is the proton velocity (v=5*10^{5} m/s)

B is the magnetic field (B = 1.17 T)

Now, we just need to solve the equaton (1) for \theta.

\theta=sin^{-1}\left(\frac{F}{qvB}\right)

But the force F = ma, then:

m is the mass of proton  (m=1.67*10^{-27} kg)

a is the acceleration (a=1.5*10^{13} m/s^{2})

\theta=sin^{-1}\left(\frac{m_{p}a}{q_{p}vB}\right)

\theta=sin^{-1}\left(\frac{1.67*10^{-27}*1.5*10^{13}}{1.6*10^{-19}*5*10^{5}*1.17}\right)

\theta=15.52^{\circ}

I hope it helps you!

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user100 [1]

Answer:

a = - 50 [m/s²]

Explanation:

To solve this problem we simply have to replace the values supplied in the given equation.

Vf = final velocity = 0.5 [m/s]

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s = distance = 100 [m]

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Now replacing we have:

(0.5)^{2}-(10)^{2} = 2*a*(100)\\0.25-10000=200*a\\200*a=-9999.75\\a =-50 [m/s^{2} ]

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2 years ago
An inductor in an LC circuit has a maximum current of 2.4 A and a maximum energy of 56 mJ.
Harrizon [31]

Answer:

The energy stored in the capacitor, when the current in the inductor is 1.2 A, is 41.6 mJ.

Explanation:

In a LC oscillating circuit, the energy is stored in the electric field (between the plates of the capacitor) and in the magnetic field (surrounding the wires of the inductor).

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When the capacitor is completely discharged, all the energy is stored in the magnetic field, and at that time, the current is maximum.

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