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3 years ago

What type of subsurface material is able to store groundwater?

Physics

1 answer:

notka56 [123]3 years ago

4 0

<span>Porous material has many spaces that can hold(store) groundwater.</span>

You might be interested in

The approximate mass of a nickel in kilograms is

mart [117]

<span>The nickel has a mass of approximately 0.005kg</span>

5 0

3 years ago

Find the ratio of the diameter of aluminium to copper wire, if they have the same

kicyunya [14]

Answer:

1.24

Explanation:

The resistivity of copper $\rho_1=2.65\times 10^{-8}\ \Omega-m$

The resistivity of Aluminum, $\rho_2=1.72\times 10^{-8}\ \Omega-m$

The wires have same resistance per unit length.

The resistance of a wire is given by :

$R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi (\dfrac{d}{2})^2}\\\\\dfrac{R}{l}=\rho \dfrac{1}{\pi (\dfrac{d}{2})^2}$

According to given condition,

$\rho_1 \dfrac{1}{\pi (\dfrac{d_1}{2})^2}=\rho_2 \dfrac{1}{\pi (\dfrac{d_2}{2})^2}\\\\\rho_1 \dfrac{1}{{d_1}^2}=\rho_2 \dfrac{1}{{d_2}^2}\\\\(\dfrac{d_2}{d_1})^2=\dfrac{\rho_1}{\rho_2}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{\rho_1}{\rho_2}}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{2.65\times 10^{-8}}{1.72\times 10^{-8}}}\\\\=1.24$

So, the required ratio of the diameter of Aluminum to Copper wire is 1.24.

3 0

3 years ago

The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate

ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

$\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}$ ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

$\theta=\frac{3.7m}{0.162m}=22.83rad$

$22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}$

to calculate the angular speed w we can use $\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}$

Thus, wf=12.68rad/s

C) We can use our result in B)

$\alpha=3.52\frac{rad}{s^2}$

I hope this is useful for you

regards

3 0

3 years ago

Read 2 more answers

The athlet at point A runs 150m east, then 70m west and then 100 east

denpristay [2]

Answer:

180m to the east

Explanation:

Displacement is the distance traveled in a specific direction. It is a vector quantity with both magnitude and direction. Therefore, the start and finish position is very paramount.

point A runs 150m east,

70m west

100m east

150m

--------------------------------------------------------→

70m

←---------------------

100m

-----------------------------------→

The displacement of the athlete = 150 - 80 + 100 = 180m to the east

6 0

3 years ago