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Sloan
13 days ago
8

Which activity is a method of habitat restoration?

Physics
You might be interested in
(a) Calculate the absolute pressure at the bottom of a fresh- water lake at a depth of 27.5 m. Assume the density of the water i
kicyunya [3294]

Response:

a) P = 370.993\,kPa, b) F = 25.948\,kN

Clarification:

a) The absolute pressure at a depth of 27.5 meters is:

P = P_{atm} + P_{man}

P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (27.5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)

P = 370.993\,kPa

b) The force applied by the water is:

F = (P - P_{atm})\cdot A

F = (370.993\,kPa-101.3\,kPa)\cdot \left(\frac{\pi}{4} \right)\cdot (0.35\,m)^{2}

F = 25.948\,kN

5 0
3 months ago
Read 2 more answers
For the RC circuit and the RL circuit, assume that the period of the source square wave is much larger than the time constant fo
serg [3582]

Answer with Explanation:

Concepts and reasoning

The principle for addressing this question is that a capacitor in an RC circuit allows current to flow until fully charged. Once charged, it prevents any further current from moving through. Conversely, the situation is different with an inductor in an RL circuit. In accordance with Faraday's law, an inductor generates an electromagnetic force to counteract the applied voltage, but when no change in flux occurs, it behaves akin to a regular wire as if the inductor is absent.

In the accompanying diagram, a resistor is connected in series with a capacitor.

As we observe

the voltage across both the capacitor and the source.

V_{C}Voltage across a resistor in an RC circuit.V_{s}

Voltage across a resistor in an RL circuit.V_{R}=V_S\left ( e^{-\frac{t}{RC}} \right )

V_{R}=V_S\left (1- e^{-\frac{Rt}{L}} \right )

7 0
3 months ago
Consider 500 g of silver (atomic mass 107 g/mol). Assume that the temperature is high enough that equipartition applies. We heat
ValentinkaMS [3465]

Answer:

A. 2.57 K

Explanation:

Using the specific heat capacity,

Q = cmΔT........................ Equation 1

Where Q = Heat energy absorbed by silver, m = mass of silver, c = specific heat capacity for silver, ΔT = temperature change of the silver.

Rearranging gives ΔT as the subject of the equation

ΔT = Q/cm................... Equation 2

Provided: Q = 300 J, m = 500 g = 0.5 kg

Constant: c = 233 J/kg.K

Inserting values into equation 2 yields

ΔT = 300/(0.5×233)

ΔT  = 300/116.5

ΔT = 2.57 K

Therefore, the correct answer is A. 2.57 K

3 0
3 months ago
If you are driving 72 km/h along a straight road and you look to the side for 4.0 s, how far do you travel during this inattenti
ValentinkaMS [3465]
Speed is defined as distance over time. Hence, to determine the distance, we use d = V * t. Plugging in the values yields d = (72 Km / h) * (1h / 3600s) * (4.0 s) = 0.08Km. Thus, during this distracted period, a distance of 0.08Km was covered.
8 0
4 months ago
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