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Aleonysh [2.5K]
2 years ago
12

An object is moving with constant non-zero velocity. Which of thw following statements about it must be true

Physics
1 answer:
hram777 [196]2 years ago
5 0

Answer:

The net force on the object is zero.

Explanation:

An object is moving with constant non-zero velocity. If velocity is constant, it means that the change in velocity is equal to 0. As a result, acceleration of the object is equal to 0. Net force is the product of mass and acceleration. Hence, the correct option is (d) "The net force on the object is zero".

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g In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period
NeTakaya

The question is incomplete. The complete question is :

In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?

Solution :

The underdamped RLC circuit

$v_{t} = ve^{-\frac{R}{2L}t} \cos \omega t$

$\omega = \sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}= \frac{2 \pi}{T}$

We know in one time period, v = 2v, at t = T, $v_t = 3.8 v$

so, $9.8 = 5 e^{-\frac{R}{2L}T} \cos \frac{2 \pi}{T}T$

   $e^{-\frac{R}{2L}T} = \frac{3.8}{5} \times 1$

   $\frac{R}{2L}T= \ln \frac{5}{3.8}$

  $\frac{R}{L}= \frac{2}{1.2 \times 10^{-6}} \ln \frac{5}{3.8}$

 $\frac{R}{L} = 457.3 \times 10^3$

Now, Q value $= \frac{1}{R}\sqrt{\frac{L}{C}}$

                     $=\sqrt{\frac{L}{R^2C}\times \frac{L}{L}}$

                     $=\sqrt{(\frac{L}{R})^2 \times \frac{1}{LC}}$

              $\frac{1}{LC}=27.43 \times 10^{12}$

∴ $Q=\sqrt{\left(\frac{1}{457.3 \times 10^3}\right)^2 \times 27.43 \times 10^{12}}$

  $Q=\sqrt{131.166}$

      = 11.45

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