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3 years ago

you ride a bike for 2 hours at 25 km/hr then 3 hours at 34 km/hr. what is your average velocity (in km/hr)?

Physics

1 answer:

mamaluj [8]3 years ago

4 0

<h3>Answer:</h3>

30.4 km/hr

<h3>Explanation:</h3>

<u>We are given</u>;

Speed in the first 2 hours as 25 km/hr
Speed in the next 3 hours as 34 km/hr

We are required to determine the average velocity in km/hr

To get the average velocity we divide total distance by total time.
Thus, we need to determine the total distance

Distance = Speed × time

Distance covered in the first 2 hours;

= 25 km/hr × 2 hours

= 50 km

Distance in the next 3 hours

= 34 km/hr × 3 hours

= 102 km

Therefore, total distance = 50 km + 102 km

= 152 km

Total time = 2 hrs + 3 hrs

= 5 hours

Therefore;

Average speed = 152 km ÷ 5 hours

= 30.4 km/hr

Thus, the average speed is 30.4 km/hr

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A source emits sound uniformly in all directions. There are no reflections of the sound. At a distance of 12 m from the source,

yaroslaw [1]

Answer:

1.58 W

Explanation:

Since the sound spreads uniformly in all directions, it must be in a form of a circle with radius of 12 m. So the area of the circle is

$A = \pi r^2 = \pi 12^2 = 452.389 m^2$

From the intensity of the sound we can calculate the power at 12 m

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3 years ago

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

konstantin123 [22]

Answer:

$I=2.71\times 10^{-5}\ A$

Explanation:

A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

Let given is,

The diameter of a parallel plate capacitor is 6 cm or 0.06 m

Separation between plates, d = 0.046 mm

The potential difference across the capacitor is increasing at 500,000 V/s

We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :

$C=\dfrac{A\epsilon_o}{d}\\\\C=\dfrac{\pi r^2\epsilon_o}{d}$ , r is radius

Let I is the displacement current. It is given by :

$I=C\dfrac{dV}{dt}$

Here, $\dfrac{dV}{dt}$ is rate of increasing potential difference

$I=\dfrac{\pi r^2\epsilon_o}{d}\times \dfrac{dV}{dt}\\\\I=\dfrac{\pi (0.03)^2\times 8.85\times 10^{-12}}{0.46\times 10^{-3}}\times 500000\\\\I=2.71\times 10^{-5}\ A$

So, the value of displacement current is $2.71\times 10^{-5}\ A$ .

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3 years ago

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fredd [130]

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3 years ago