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3 years ago

A rock layer that stores and transmits water

1 answer:

5 0

The ability of the ground water to pass through the pore spaces in the rock is described as the rock's permeability. Permeable layers of rock that store and transport water are called aquifers.

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Help solve these two problems im having trouble trying to start these problems?

belka [17]

Answer:

25. Approximately 8.1 meters

26. North 1.31 km, and East 2.81 km

Explanation:

25.

Notice that the displacements: 6 meters east and 5.4 south create the legs of a right angle triangle. The hypotenuse of that triangle will be the distance (d) needed to cover in order to get the ball in the hole in one putt. That is:

$d=\sqrt{6^2+5.4^2} =\sqrt{65.16} \approx 8.072\,\,\,m$

which can be rounded to 8.1 m.

26.

Notice that the 3.1 km at an angle of 25 degrees north of east, is the hypotenuse of a right angle triangle that has for legs the east and north components of that distance.

We can find the leg corresponding to the east displacement using the cosine function (that relates adjacent side with hypotenuse):

$east\,\, comp=3.1 * cos(25^o)\approx 2.809\,\,km$

and we can calculate the north component using the sine function that relates the opposite side to the angle with the hypotenuse.

$north\,\,component = 3.1 * sin(25^o) \approx 1.31 \,\,km$

6 0

3 years ago

A 1100 kg car pushes a 1800 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the

12345 [234]

Answer:The answer is 3000 N.

Force (F) is the multiplication of mass (m) and acceleration (a).

F = m · a

It is given:

mc = 1000 kg

mt = 2000 kg

total force: F = 4500 N

total mass: m = mc + mt

Let's calculate acceleration which is common:

a = F/m = F/(mc + mt) = 4500/(1000 + 2000) = 4500/3000 = 1.5 m/s²

Now, when we know acceleration, let's calculate force on the truck:

Ft = mt · a = 2000 · 1.5 = 3000 N

Explanation:

6 0

3 years ago

Cart A, with a mass of 0.20 kg, travels on a horizontal air trackat 3.0m/s and hits cart B, which has a mass of 0.40 kg and is i

inysia [295]

Answer:

$\large \boxed{\text{C. 2.3 m/s}}$

Explanation:

Data:

$m_{\text{A} } = \text{0.20 kg};\,v_{\text{Ai}} = \text{3.0 m/s}\\m_{\text{B} } = \text{0.40 kg};\,v_{\text{Bi}} = \text{2.0 m/s}\\$

Calculation:

This is a perfectly inelastic collision. The two carts stick together after the collision and move with a common final velocity.

The conservation of momentum equation is

$\begin{array}{rcl}m_{\text{A}}v_{\text{Ai}} +m_{\text{B}} v_{\text{Bi}}&=&(m_{\text{A}} + m_{\text{B}})v_{\text{f}}\\0.20\times 3.0 + 0.40\times 2.0 & = & (0.20 + 0.40)v_{\text{f}}\\0.60 + 0.80 & = & 0.60v_{\text{f}}\\1.40 & = & 0.60v_{\text{f}}\\v_{\text{f}}&=& \dfrac{1.40}{0.60}\\\\& = & \textbf{2.3 m/s}\\\end{array}\\\text{The centre of mass has a velocity of $\large \boxed{\textbf{2.3 m/s}}$}$

3 0

3 years ago

What is the name for all the electromagnetic waves that exist

zheka24 [161]

The Electromagnetic spectrum.

8 0

3 years ago

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ is the pressure difference between the two ends

$\mu$ is viscosity of the fluid

L is the length of the pipe

$Q=Av$ is the volumetric flow rate, with $A=\pi r^2$ being the section of the tube and $v$ the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

$\mu=0.001 Pa/s$ is the dynamic water viscosity at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Using these data in the formula, we get:

$\Delta P = 3200 Pa$

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0

3 years ago

A rock layer that stores and transmits water​

A rock layer that stores and transmits water