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tatiyna
2 years ago
5

8. When wind dies down or stops blowing happens.

Physics
2 answers:
timama [110]2 years ago
5 0

Answer:

Deposition happens, if the wind stops blowing.

Deposition is the dropping of sediment by wind, water, ice, or gravity.

yan [13]2 years ago
4 0

Answer:

Deposition

Explanation:

  • Deposition occurs when water slows or ceases moving, the wind dies or stops blowing, or glaciers melt. The deposited material can also be used to construct new landforms. Waves, for example, can dump sediment in places offshore, where it might accumulate to form sand dunes.

When the wind calms down or vegetation stops or slows the breeze, the sediment particles begin to fall. Water is another factor that may erode, move, or deposit sediment. Flowing water is a key erosive agent. Water transports dirt and rock fragments as it moves. Warm, wet air will not travel if wind systems are not present. Water will still evaporate, but it will not move, therefore everywhere else than a major body of water will dry up. Lakes may be fine since evaporating water will flow back into them, and the sea will be fine, but everywhere else will become extremely dry very rapidly. Wind is constantly blowing somewhere on the world at any given time. Winds are usually quiet near the middle of a high pressure system. Wind is the passage of air from a high pressure location to a low pressure area.... So essentially air is always moving. Weathering and erosion are caused by wind. Weathering is caused by wind blowing debris against cliffs and huge rocks. This wears down the rock, reducing it to sand and dust. Sand and dust are also eroded by wind. 2. Rocks are tough and durable, but they don't last forever. Weathering and erosion are processes that occur as a result of forces such as wind and water breaking down rocks. Weathering is the process through which rocks deteriorate. Weathering is caused by a variety of factors, including climate change.

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A client with hypertension who weighs 72.4 kg is receiving an infusion of nitroprusside (Nipride) 50 mg in D5W 250 ml at 75 ml/h
Mkey [24]

To solve this problem it is necessary to simply apply the concepts related to cross-multiply and proportion between units.

Let's start first by relating the amount of dose needed to be supplied per hour, in other words,

The infusion of 250ml should be supplied at a rate of 75ml / hour, so what amount x of mg hour should be supplied with 50Mg.

\frac{x}{75ml/hour} \rightarrow \frac{50mg}{250ml}

x \rightarrow \frac{50mg*75ml/hour}{250ml}

x \rightarrow \frac{3750mg}{250hour}

x \rightarrow 15\frac{mg}{hour}

Converting to mcg units we know that 1mg is equal to 1000mcg and that 1 hour contains 60 min, therefore

x \rightarrow 15\frac{mg}{hour}

x \rightarrow 15\frac{mg}{hour}(\frac{1000mcg}{1mg})(\frac{1hour}{60min})

x \rightarrow 250mcg/min

The dose should be distributed per kilogram of the patient so if the patient weighs 72.4kg,

Dose = \frac{250mcg/min}{72.4kg}

Dose = 3.5 \frac{mcg/min}{kg}

Therefore the client will receive 3.5mcg/kg/min.

8 0
3 years ago
A lamp is on a dimmer switch. As you turn the switch from the lowest setting to the highest setting, which statement would best
avanturin [10]

Answer:

D

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We have three identical metallic spheres A, B, C. Initially sphere A is charged with charge Q, while B and C are neutral. First,
larisa [96]

Answer:

The final charges of each sphere are:   q_A = 3/8 Q , q_B = 3/8 Q ,               q_C = 3/4 Q

Explanation:

This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.

Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point

                q_A = Q / 2

                q_B = Q / 2

Now sphere A touches sphere C, ending with half the charge

                q_A = ½ (Q / 2) = ¼ Q

                q_B = ¼ Q

Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge

                  q = Q / 4 + Q / 2 = ¾ Q

This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q

                  q_A = 3/8 Q

                  q_B = 3/8 Q

The final charges of each sphere are:

                q_A = 3/8 Q

                q_B = 3/8 Q

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7 0
3 years ago
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mafiozo [28]

Answer:

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Sedbober [7]
1950 g This is the answer due to the kilograms of lead being distributed
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