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3 years ago

How permeable and porous would an aquifer be?

1 answer:

7 0

Answer: An aquifer is a body of saturated rock through which water can easily move. Aquifers must be both permeable and porous and include such rock types as a sandstone.

Explanation: However, if these rocks are highly fractured, they make good aquifers.

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The energy that generates wind comes from what source?

Rudiy27

Answer:

we can say that wind energy is due to

D) Severe thunderstorms

Explanation:

As we know that wind energy is converted into kinetic energy of wind mills

This kinetic energy of wind mill is then converted into electrical energy using turbine

now we can consider here energy conservation theory that energy is only converted from one form to other form

it neither be destroyed nor be created but it can transfer from one form to other form

So here we can say that wind energy is due to

D) Severe thunderstorms

3 0

3 years ago

Why would physics be used to study the movement of ocean waves?

Lunna [17]

I think it’s c because the other ones are just options not facts

5 0

3 years ago

A cubical Gaussian surface surrounds two positive charges, each has a charge q 1 1 = + 3.90 × 10 − 12 3.90×10−12 C, and three ne

Masteriza [31]

Answer:

The electric flux is zero because charge is zero.

Explanation:

Given that,

Positive charge $q_{1}=3.90\times10^{-12}\ C$

Negative charge $q_{2}=-2.60\times10^{-12}\ C$

We need to calculate the total charged

Using formula of charge

$Q_{enc}=2q_{1}+3q_{2}$

Put the value into the formula

$Q_{enc}=2\times3.90\times10^{-12}+3\times(-2.60\times10^{-12})$

$Q_{enc}=0$

We need to calculate the electric flux

Using formula of electric flux

$\phi=\dfrac{Q_{enc}}{\epsilon_{0}}$

Put the value into the formula

$\phi=\dfrac{0}{8.85\times10^{-12}}$

Hence, The electric flux is zero because charge is zero.

7 0

3 years ago

Which of the following best explains how earth is heated through radiation

lawyer [7]

Answer:

The sun touches earth during daytime and the suns rays heat our earth giving us heat. The sun heating the earth is also considered radiation.

Explanation:

6 0

2 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago