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soldier1979 [14.2K]

3 years ago

12

What is demonstrated by water moving up a straw?

2 answers:

Alja [10]3 years ago

8 0

<span>The answer is C: water is drawn up a straw by cohesion and adhesion. Water molecules stick to one another and the walls of the straw, just like in a capillary.
Cohesion is the attractive force between like materials (between water
molecules).
Adhesion is the attractive force between twounlike materials (such as between
water and a solid container).
Capillary action is the tendency of a liquid to rise innarrow tubes or small openings as a result of adhesion and cohesion.
The liquid water molecules bind to the straw—a process known as adhesion. In the narrow space of the straw, the interaction of cohesion and adhesion causes theliquid to be drawn upward in the straw.</span>

Novay_Z [31]3 years ago

7 0

It's C. Cohesion and Adhesion

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4 years ago

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Tju [1.3M]

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3 years ago

Three capacitors with individual capacitances of 2 μF, 5 μF, and 10 μF

rodikova [14]

Answer:

1.25C

Explanation:

When capacitance is in series we add them like this: 1/Ctotal = 1/C1 +1/C2 + 1/C3.....

1/C = 1/2 + 1/5 + 1/10 = 5 + 2 + 1/10 = 8/10

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5 0

3 years ago

the acceleration on the moon is 1.6 m/s^2 what is the maximum height of a ball you throw at 20 m/s straight up?

Artist 52 [7]

Answer:

125 m

Explanation:

Given:

v₀ = 20 m/s

v = 0 m/s

a = -1.6 m/s²

Find: Δy

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(0 m/s)² = (20 m/s)² + 2 (-1.6 m/s²) Δy

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3 0

4 years ago

A 15-ft3 tank contains oxygen initially at 14.7 psia and 80°F. A paddle wheel within the tank is rotated until the pressure insi

Nikolay [14]

Explanation:

Equation for energy balance will be as follows.

$\Delta E_{system} = E_{in} - E_{out}$

$\Delta U = W_{in} - Q_{out}$

Hence, $W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

Therefore, we will calculate the final temperature as follows.

$\frac{P_{1}V}{T_{1}} = \frac{P_{2}V}{T_{2}}$

$T_{2} = \frac{20 psia}{14.7}(638 R)$

= 868.03 R

Now, we will calculate the mass as follows.

m = $\frac{P_{1}V}{RT_{1}}$

= $\frac{14.7 psia \times 15 ft^{3}}{0.3353 psi ft^{3}/lbm R \times 638 R}$

= 1.031 lbm

Hence,

$W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

Putting the values into the above equation as follows.

$W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

$W_{in} = 20 Btu + 1.031 lbm (\frac{0.160 Btu}{lbm R})(735 - 540)R$

$W_{in}$ = 655.2 Btu

Thus, we can conclude that work done by paddle wheel is 655.2 Btu.

6 0

3 years ago