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3 years ago

1. refers to all liquid fresh water located above ground.

Physics

1 answer:

AleksandrR [38]3 years ago

7 0

Answer:

1- Surface water

2- Groundwater

3- Lake

4- River

5- Watershed/Catchment basin

6- Wetland

Explanation:

Surface water refers to all the water bodies that are present on the surface of the earth. This includes rivers, streams, ponds, lakes.

Groundwater refers to the subsurface flow of water. When precipitation occurs, the water percolates downward through the rocks in the region where the rocks are porous and permeable, and this water then becomes groundwater.

A lake is usually defined as a large water body that is comprised of nutrient minerals and has extensive areas of open water. The depth of the lakes from one place to another and some are so deep that sunlight cannot penetrate into the deeper zone and the plants cannot carry out the process of photosynthesis.

A river is defined as a water body that flows from a region of higher elevation to the region of lower elevation and it flows across the land surface and configures the landscape.

Catchment basin is usually defined as a large area in which the precipitation occurs and the waters are accumulated and flow downward and mix up with the river channel and streams.

Wetlands are usually defined as an ecosystem that is comprised of shallow standing water, and a sufficient amount of vegetation.

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MA_775_DIABLO [31]

Answer:

Resolving horizontally. :

$\sum d_{x} = - (17.0 \cos 20.0) - (11.0 \cos 35.0) + (30.0 \cos 50.0) + 0 \\ { \underline{d _{x} = - 5.702 \: m}} \\ \\ \sum d _{y} = (17.0 \sin 20.0) + 12.0 - (11.0 \sin 35.0) - (30.0 \sin 50.0) \\ { \underline{d _{y} = - 11.476 \: m}}$

therefore, for resultant:

$d = \sqrt{ {d _{y} }^{2} + d _{x} {}^{2} }$

substitute:

$d = \sqrt{ {( - 5.702)}^{2} + {( - 11.476)}^{2} } \\ \\ d = \sqrt{164.211} \\ \\ { \boxed{ \boxed{ \bf{d = 12.8 \: m}}}}$

6 0

2 years ago

A certain resistor dissipates 0.5 W when connected to a 3 V potential difference. When connected to a 1 V potential difference,

Stels [109]

Answer:

Explanation:

$Power = IV$

From ohms law we know that

$V= IR\\\\I= \frac{V}{R} \\\\Power= \frac{V}{R}*V\\\\Power= \frac{V^2}{R}$

Given data

P1 = 0.5 Watt

P2 = ?

V1= 3 Volts

V2= 1 Volt

Thus we can solve for the power dissipated as follows

$P1= \frac{V1^2}{R1}\\\\P2= \frac{V2^2}{R2}$

$\frac{P1}{P2} = \frac{V1^2}{V2^2}\\\\ P2=\frac{ V2^2}{ V1^2} *P1\\\\ P2=\frac{ 1^2}{ 3^2} *0.5= 0.055= 0.056 W$

The resistor will dissipate 0.056 Watt

7 0

3 years ago

A circular loop of wire has radius of 9.50 cmcm. A sinusoidal electromagnetic plane wave traveling in air passes through the loo

m_a_m_a [10]

Answer:

The induced emf is $\epsilon = 0.1041 \ V$

Explanation:

From the question we are told that

The radius of the circular loop is $r = 9.50 \ cm = 0.095 \ m$

The intensity of the wave is $I = 0.0215 \ W/m^2$

The wavelength is $\lambda = 6.90\ m$

Generally the intensity is mathematically represented as

$I = \frac{ c * B^2 }{ 2 * \mu_o }$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4 \pi *10^{-7} N/A^2$

B is the magnetic field which can be mathematically represented from the equation as

$B = \sqrt{ \frac{ 2 * \mu_o * I }{ c} }$

substituting values

$B = \sqrt{ \frac{ 2 * 4\pi *10^{-7} * 0.0215 }{ 3.0*10^{8}} }$

$B = 1.342 *10^{-8} \ T$

The area is mathematically represented as

$A = \pi r^2$

substituting values

$A = 3.142 * (0.095)^2$

$A = 0.0284$

The angular velocity is mathematically represented as

$w = 2 * \pi * \frac{c}{\lambda }$

substituting values

$w = 2 * 3.142 * \frac{3.0*10^{8}}{ 6.90 }$

$w = 2.732 *10^{8} rad \ s^{-1}$

Generally the induced emf is mathematically represented as

$\epsilon = N * B * A * w * sin (wt )$

At maximum induced emf $sin (wt) = 1$

$\epsilon = N * B * A * w$

substituting values

$\epsilon = 1 * 1.342 *10^{-8} * 0.0284 *2.732 *10^{8}$

$\epsilon = 0.1041 \ V$

7 0

3 years ago

Helpppppppp please !!!!

agasfer [191]

Answer:

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nikdorinn [45]

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3 years ago