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andre [41]
4 years ago
11

Answer these two questions please

Physics
1 answer:
Grace [21]4 years ago
8 0
1.The answer is True
2.The answer is False
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A deuteron particle consists of one proton and one neutron and has a mass of 3.34x10^-27 kg. A deuteron particle moving horizont
nikklg [1K]

_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.

3 0
2 years ago
A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA. What amount of vars must be added to bring the pf to 0.85
kvasek [131]

Answer:

\mathtt{Q_{sh} = 600.75 \ vars}

Explanation:

Given that:

A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA

Here:

the initial power factor  i.e cos θ₁ = 0.7 lag

θ₁ = cos⁻¹ (0.7)

θ₁ = 45.573°

Active power P = 1500 watts

Apparent power S = 2100 VA

What amount of vars must be added to bring the pf to 0.85

i.e the required power factor here is cos θ₂ = 0.85 lag

θ₂ =  cos⁻¹   (0.85)

θ₂ = 31.788°

However; the initial reactive power Q_1 = P×tanθ₁

the initial reactive power Q_1 = 1500 × tan(45.573)

the initial reactive power Q_1 = 1500 × 1.0202

the initial reactive power Q_1 =  1530.3 vars

The amount of vars that must therefore be added to bring the pf to 0.85

can be calculated as:

Q_{sh} = P( tan \theta_1 - tan \theta_2)

Q_{sh} = 1500( tan \ 45.573 - tan \ 31.788)

Q_{sh} = 1500( 1.0202 - 0.6197)

Q_{sh} = 1500( 0.4005)

\mathtt{Q_{sh} = 600.75 \ vars}

3 0
3 years ago
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