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3 years ago

Beats are the difference in (a) Frequency (b) Amplitude (c) Intensity (d) None

Physics

1 answer:

sasho [114]3 years ago

7 0

Answer:

Beats are the difference in frequency.

(a) is correct option.

Explanation:

Beat :

Beat is the difference of the frequency of two waves.

The difference in frequency is equal to the number of beat per second.

Amplitude :

Amplitude of the wave is the maximum displacement.

Frequency :

Frequency is the number oscillations of wave in per second.

Intensity :

Intensity is the power per unit area.

Hence, Beats are the difference in frequency.

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A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points

mr Goodwill [35]

Answer:

Approximately $11.0\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81 \; \rm N \cdot kg^{-1}$ , and that the tabletop is level.)

Explanation:

Weight of the book:

$W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N$ .

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, $F(\text{normal force}) \approx 10.202\; \rm N$ .

As a side note, the $F_N$ and $W$ on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction $F(\text{kinetic friction})$ on it will be equal to

$\mu_{\rm k}$ , the coefficient of kinetic friction, times
$F(\text{normal force})$ , the normal force that's acting on it.

That is:

$\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}$ .

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that $15.0\; \rm N$ applied force. The net force on the book shall be:

$\begin{aligned}& F(\text{net force}) \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}$ .

Apply Newton's Second Law to find the acceleration of this book:

$\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}$ .

6 0

2 years ago

Outside our solar system the closest star to earth is Proxima century life from the start takes 2200000 minutes to reach earth.

s2008m [1.1K]

Answer:

2200000 = 2.2E6 min for light from Proxima to reach earth

8.3 min from light sun to reach earth

2.2E6/8.3 = 2.56E5 times for light from Proxima

Proxima is about 256,000 times farther away than the sun

Since the sun is about 93,000,000 = 9.3E7 miles from earth

Proxima is then 9.3E7 * 2.56E5 = 2.4E13 miles away

Note - the speed of light is

3.00E8 m/s * 60 s/min / 1000 m/km = 1.8E7 km/min as given

5 0

2 years ago

If a specimen was being viewed using a 20x objective lens and 10x ocular lens, what would be the total magnification

Paraphin [41]

Answer:

As Per Given Information

20x objective lens was used by specimen

10x ocular lens was also used by him.

we have to find the total magnification.

For calculating the total magnification we 'll simply do multiplication

Total Magnification = 20x × 10x

Total Magnification = 200x

So , the total magnification will be 200x .

6 0

2 years ago

You skip north for 12 minutes to your best friend's house that is 1.5 kilometers away. What is your average velocity?

bagirrra123 [75]

Answer:

The average velocity is 7.5 km/h

Explanation:

Let's convert minutes to hours so our answer can be given in a common units of km/hour:

12 minutes = 12/60 hours = 0.2 hours

Now we estimate the average velocity calculating the distance travelled over the time it took:

1.5 / 0.2 km/h = 7.5 km/h

3 0

3 years ago

A man jogs at a speed of 1.6 m/s. His dog

FromTheMoon [43]

I believe it is
1.6x=2.7(x-1.8)
1.1x=2.7*1.8
x~4.4
4.4*1.6
~7.1m

5 0

3 years ago