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Scorpion4ik [409]
3 years ago
12

Importance of choke coil?​

Physics
1 answer:
Verdich [7]3 years ago
8 0

Answer:The choke coil works because it can act as an inductor. When the current pass through will change as AC currents creates a magnetic field in the coil that works against that current. This is known as inductance and blocks most of the AC current from passing through.

Explanation:

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5 0
3 years ago
Consider an electron and a proton separated by a distance of 4.5 nm. (a) what is the magnitude of the gravitational force betwee
Dmitry_Shevchenko [17]
A.) For letter a, we use the law of universal gravitation using the constant G = 6.674×10−<span>11 m3</span>⋅kg−1⋅s−<span>2

Grav. F = G*m1*m2*(1/d^2)

m1 is mass of electron = </span>9.11 × 10-31<span> kg
m2 is mass of proton = </span>1.67 × 10<span>-27 kg
d = 4.5 nm = 4.5 x 10^-9 m

Grav F = 5.01 x 10^-51 N

b.) </span>For letter b, we use the Coulomb's using the constant k = 9×10^9 N

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6 0
3 years ago
A deuteron particle consists of one proton and one neutron and has a mass of 3.34x10^-27 kg. A deuteron particle moving horizont
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_dThe radius of curvature of a subatomic particle under a magnetic field is given by the following formula:

r=\frac{mv}{qB}

Where:

\begin{gathered} r=\text{ radius} \\ v=\text{ velocity} \\ q=\text{ charge} \\ B=\text{ magnetic field} \end{gathered}

We can determine the quotient between the velocity and the charge of the deuteron particle from the formula. First, we divide both sides by the mass:

\frac{r_d}{m_d}=\frac{v}{q_B_}

Now, we multiply both sides by the magnetic field "B":

\frac{Br_d}{m_d}=\frac{v}{q}

Since the charge of the deuterion is the same as the charge of the proton and the velocity we are considering are the same this means that the quotient between velocity and charge is the same for both particles. Therefore, we can apply the formula for the radius again, this time for the proton:

r_p=\frac{m_pv}{qB}

And substitute the quotient between velocity and charge:

r_p=\frac{m_p}{B}(\frac{Br_d}{m_d})

Now, we cancel out the magnetic field:

r_p=\frac{m_pr_d}{m_d}

Now, we substitute the values:

r_p=\frac{(1.67\times10^{-27}kg)(0.385m)}{(3.34\times10^{-27}kg)}

Solving the operations:

r_p=0.193m=19.3cm

Therefore, the radius is 19.3 cm.

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2 years ago
How are dunes and deltas alike?
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It might be a or d not too sure though
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