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2 years ago

A pendulum completes 2 oscillation in 5s.

Physics

1 answer:

Otrada [13]2 years ago

6 0

Answer:

Explanation:

<h3>Given :-</h3><h2>Completes 2 oscillation in 5s.</h2><h2>A) Time period = ?</h2><h2 /><h2> $t \: = \frac{total \: time \: }{total \: no \: of \: oscillation}$ </h2><h2 /><h2> $\frac{5}{2} = 2.5s$ </h2>

<em>Thus 1 oscillation is completed in 2.5 </em><em>s.</em>

<h2>B) If g = 9.8 ms−², find its length.</h2><h2> $t \: = 2\pi \frac{ \sqrt{l} }{g }$ </h2><h3>= 2.5s </h3><h3> $2\pi \frac{ \sqrt{l} }{g}$ </h3><h3>Length = ? </h3><h2> $\frac{ {t}^{2} }{4\pi {}^{2} } \times g$ </h2><h3 /><h3> $\frac{2.5}{ {4}^{2} } \times 9.8 = 1.55m$ </h3>

Thus the length of g = 9.8 ms −² = 1.55m

<h2>Hope it helps u.......</h2><h2>STAY SAFE, STAY HEALTHY AND BLESSED .</h2><h2> HAVE A MARVELOUS DAY AHEAD !</h2><h2>THANK YOU !</h2>

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A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040

Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force ( $F_{B}$ ), as follows:

$F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB$ (1)

<u>Where:</u>

<em>m: is the proton's mass = 1.67*10⁻²⁷ kg</em>

<em>v: is the proton's velocity</em>

<em>r: is the radius of the proton's orbit</em>

<em>q: is the proton charge = 1.6*10⁻¹⁹ C</em>

<em>B: is the magnetic field = 0.040 T </em>

Solving equation (1) for r, we have:

$r = \frac{mv}{qB}$ (2)

By conservation of energy, we can find the velocity of the proton:

$K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V$ (3)

<u>Where:</u>

<em>K: is kinetic energy</em>

<em>U: is electrostatic potential energy</em>

<em>ΔV: is the potential difference = 1.0 kV </em>

Solving equation (3) for v, we have:

$v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s$

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

$r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m$

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!

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The vehicles are fitted with shockers why

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[I copied this from the internet, but hope it helps!]

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