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3 years ago

The area of land where water is drained downhill into a body of water is known as a drainage basin, or _______.

Physics

2 answers:

o-na [289]3 years ago

8 0

Answer: Watershed

A watershed or drainage basin is a down the slope land where the water is drained off from sources of water flowing comparatively at some elevation. It includes water from rivers, river streams that progressively drain into larger areas. The initial source of watershed is called as headwater. The endpoint of watershed is open up into another water body is called as mouth.

GenaCL600 [577]3 years ago

6 0

Water sheds i hope this helps give me a brainiest answer

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The combination of all forces that act on an object are (2 points)

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Answer:

A) The net force

Explanation:

If two forces of equal strength act on an object in opposite directions, the forces will cancel, resulting in a net force of zero and no movement.

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2 years ago

2. A student drives 7.8-km trip to school and averages a speed of

Alekssandra [29.7K]

Answer:

The total time is: t=451.22 sec

The average speed is: V=34.57 m/s

Explanation:

Average speed

The average speed is calculated by dividing the total distance traveled by an object (x) by the total time it took it to travel that distance (t).

$\displaystyle V=\frac{x}{t}$

Since the student makes the trip in two parts, we have to calculate the total distance and the total time.

We know the distance to school is 7.8 Km = 7,800 m. The student makes his way home over the same distance, thus the total distance is

x=2*7,800 m=15,600 m

The first trip to school was done at an average speed of v1=32.6 m/s. Knowing the distance and speed, we can calculate the time:

$\displaystyle t1=\frac{x1}{v1}=\frac{7,800}{32.6}=239.26\ sec$

The second trip back home was done at an average speed of v2=36.8 m/s. Let's calculate the second time:

$\displaystyle t2=\frac{x2}{v2}=\frac{7,800}{36.8}=211.96\ sec$

The total time is:

$t=239.26\ sec+211.96\ sec=451.22\ sec$

$\boxed{t=451.22\ sec}$

The average speed is:

$\displaystyle V=\frac{15,600}{451.22}=34.57\ m/s$

$\boxed{\displaystyle V=34.57\ m/s}$

6 0

2 years ago

You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.

Neporo4naja [7]

Answer:

The distance is $r_2 = 0.24 \ m$

Explanation:

From the question we are told that

The distance from the conversation is $r_1 = 24.0 \ m$

The intensity of the sound at your position is $\beta _1 = 40 dB$

The intensity at the sound at the new position is $\beta_2 = 80.0dB$

Generally the intensity in decibel is is mathematically represented as

$\beta = 10dB log_{10}[\frac{d}{d_o} ]$

The intensity is also mathematically represented as

$d = \frac{P}{A}$

$\beta = 10dB * log_{10}[\frac{P}{A* d_o} ]$

=> $\frac{\beta}{10} = log_{10} [\frac{P}{A (l_o)} ]$

From the logarithm definition

=> $\frac{P}{A * d_o} = 10^{\frac{\beta}{10} }$

=> $P = A (d_o ) [10^{\frac{\beta }{ 10} } ]$

Here P is the power of the sound wave

and A is the cross-sectional area of the sound wave which is generally in spherical form

Now the power of the sound wave at the first position is mathematically represented as

$P_1 = A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]$

Now the power of the sound wave at the second position is mathematically represented as

$P_2 = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

Generally power of the wave is constant at both positions so

$A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ] = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

$4 \pi r_1 ^2 [10^{\frac{\beta_1 }{ 10} } ] = 4 \pi r_2 ^2 [10^{\frac{\beta_2 }{ 10} } ]$

$r_2 = \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}$

substituting value

$r_2 = \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}$

$r_2 = 0.24 \ m$

7 0

3 years ago