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2 years ago

An artesian system is present when groundwater __________.

Physics

1 answer:

ahrayia [7]2 years ago

4 0

<h2>Answer: under pressure rises above the aquifer level</h2>

Explanation:

An artesian system is one that connects with an accumulation of water (aquifer) whose surface or level is above the surface of the system or well.

In other words, the upper limit of the aquifer is higher than the opening through which the liquid flows in the artesian well. This has the advantage that the water spreads without needing to be pumped.

It should be noted that its name comes from the region of <u>Artois</u>, France, where in 1126 the oldest well in Europe was drilled with these characteristics.

Therefore:

<h2>An artesian system is present when groundwater <u>under pressure rises above the aquifer level.</u></h2>

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The answer is
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2 years ago

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3 years ago

Read 2 more answers

Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the

Over [174]

Answer:

Part a)

$a= 0.32 m/s^2$

Part b)

$F_c = 3.6 N$

Part c)

$F_c = 5.5 N$

Explanation:

Part a)

As we know that the friction force on two boxes is given as

$F_f = \mu m_a g + \mu m_b g$

$F_f = 0.02(10.6 + 7)9.81$

$F_f = 3.45 N$

Now we know by Newton's II law

$F_{net} = ma$

so we have

$F_p - F_f = (m_a + m_b) a$

$9.1 - 3.45 = (10.6 + 7) a$

$a = \frac{5.65}{17.6}$

$a= 0.32 m/s^2$

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

$F_c - F_f = m_b a$

$F_c = \mu m_b g + m_b a$

$F_c = 0.02(7)(9.8) + 7(0.32)$

$F_c = 3.6 N$

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

$F_p - F_f - F_c = m_b a$

$9.1 - 0.02(7)(9.8) - F_c = 7(0.32)$

$F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)$

$F_c = 5.5 N$

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2 years ago

A 645 g block is released from rest at height h0 above a vertical spring with spring constant k = 530 N/m and negligible mass. T

Nina [5.8K]

Answer:

a)5.88J

b)-5.88J

c)0.78m

d)0.24m

Explanation:

a) W by the block on spring is given by

W= $\frac{1}{2}$ kx² = $\frac{1}{2}$ (530)(0.149)² = 5.88 J

b) Workdone by the spring = - Workdone by the block = -5.88J

c) Taking x = 0 at the contact point we have U top = U bottom

So, mg $h_o$ = $\frac{1}{2}$ kx² - mgx

And, $h_o$ = ( $\frac{1}{2}$ kx² - mgx )/(mg) = $[\frac{1}{2} (530)(0.149^2)-(0.645)(9.8)(0.149)$ ]/(0.645x9.8)

$h_o$ = 0.78m

d) Now, if the initial initial height of block is 3 $h_o$

$h_o$ = 3 x 0.78 = 2.34m

then, $\frac{1}{2}$ kx² - mgx - mg $h_o$ =0

$\frac{1}{2}$ (530)x² - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0

265x² - 6.321x - 14.8 = 0

a=265

b=-6.321

c=-14.8

By using quadratic eq. formula, we'll have the roots

x= 0.24 or x=-0.225

Considering only positive root:

x= 0.24m (maximum compression of the spring)

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2 years ago

Polonium, the Period 6 member of Group 6A(16), is a rare radioactive metal that is the only element with a crystal structure bas

Burka [1]

Answer:

Approximate atomic radius for polonium-209 is 167.5 pm .

Explanation:

Number of atom in simple cubic unit cell = Z = 1

Density of platinum = $9.232 g/cm^3$

Edge length of cubic unit cell= a = ?

Atomic mass of Po (M) = 209 g/mol

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

ρ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

$9.232 g/cm3=\frac{1 \times 209 g/mol}{6.022\times 10^{23} mol^{-1}\times (a)^{3}}$

$a = 3.35\times 10^{-8} cm$

Atomic radius of the polonium in unit cell = r

r = 0.5a

$r=0.5\times 3.35\times 10^{-8} cm=1.675\times 10^{-8} cm$

$1 cm = 10^{10} pm$

$1.675\times 10^{-8} cm=1.675\times 10^{-8}\times 10^{10}=167.5 pm$

Approximate atomic radius for polonium-209 is 167.5 pm.

7 0

3 years ago