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zalisa [80]
3 years ago
6

A dog runs down his driveway with an initial speed of 5 m/s for 8 s, and then uniformly increases his speed to 10 m/s in 5 s.

Physics
1 answer:
MakcuM [25]3 years ago
8 0
Acceleration is given by change in velocity divided by change in time, so his acceleration should just be (10-5)/5 which is [tex] \frac{5}{2} \frac{m}{s^{2}} [tex]

The driveway is 40 meters plus 225/4. You can do the math.
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A bicyclist starts at point P and travels around a triangular path that takes her through points Q and R before returning to the
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Answer:

Displacement by cyclist is zero.

Explanation:

In the given question bicyclist is travelling in a rectangular track having P , Q and R edges.

The bicyclist starts from P and travel through Q and R and returned to P again.

We need to find its displacement.

We know displacement  of a body is its difference between its initial position to final position.

Here in the given question the bicyclist returns to P again.

Therefore, total displacement by bicyclist is zero.

Hence, this is the required solution.

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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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