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andre [41]
3 years ago
11

I need help on this

Physics
1 answer:
ZanzabumX [31]3 years ago
6 0
What is the question asking?
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The mining-resource cycle determines the use of ore materials. Which step is not part of the cycle?
ExtremeBDS [4]

Answer:

c) marketing and selling products that use ore materials

8 0
3 years ago
During a flood, a stream overflows its banks, and water covers the adjacent ____.
nata0808 [166]

Answer:

Flood Plain

Explanation:

The amount of water that circulates through a river, the flow, varies in time and space. These variations define the hydrological regime of a river. Temporary variations occur during or just after episodes of rains or thaws. Much of the water that falls in the catchment basin circulates underground, or feeds underground aquifers and takes much longer to feed the river flow and can reach it days, weeks or months after the rain generated by the runoff. The runoff that goes to the river is what increases its flow. In extreme cases, flooding can occur when the water supply is greater than the river's ability to evacuate it, overflowing and covering nearby flat areas or floodplain. In this distribution between the runoff water (or stream) that goes directly to the channel and water that infiltrates, feeds the aquifers and maintains the flow in the river in times without precipitation depends largely on the geomorphological integrity of the entire river system .

In natural dynamics, the river systems have their own space that has been modeled by the floodwaters and is made up of the channel, the banks and the plain or flood plain. Its dimensions have been defined by the main flood events that this river has attended. Floodplains are wide and flat areas built by the river in its floodwaters. They are flooded frequently and are covered by sediments and nutrients that fertilize the soil act as natural reservoirs, reducing the speed of the downstream current. They store floodwater and rainfall in aquifers (underground area).

6 0
3 years ago
A cube of side 6.50 cm is placed in a uniform field E = 7.50 × 10^3 N/C with edges parallel to the field lines (field enters the
Svetllana [295]

Answer:

a) \Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}, b) \Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}, c) \Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}

Explanation:

a) The net flux through the cube is:

\Phi_{net}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}+(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}

\Phi_{net} = 0\,\frac{N\cdot m^{2}}{C}

b) The flux through the right face is:

\Phi_{right}=-(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}

\Phi_{right} = -31.688\,\frac{N\cdot m^{2}}{C}

c) The flux through the left face is:

\Phi_{left}=(7.50\cdot 10^{3}\,\frac{N}{C} )\cdot (0.065\,m)^{2}

\Phi_{left} = 31.688\,\frac{N\cdot m^{2}}{C}

5 0
3 years ago
Read 2 more answers
In the design of an optical fibre, what
MakcuM [25]

Answer:

silica glass

Explanation:

7 0
2 years ago
Read 2 more answers
What is the
Len [333]

Answer:

d=5\ g/cm^3

Explanation:

Given that,

Mass of the object, m = 100 grams

Volume of the object, V = 20 cm³

We need to find the density of the object. We know that, density is equal to mass per unit volume. So,

d=\dfrac{m}{V}\\\\d=\dfrac{100\ g}{20\ cm^3}\\\\d=5\ g/cm^3

So, the density of the object is equal to 5\ g/cm^3.

6 0
3 years ago
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