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2 years ago

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Which statement describes the relationship between bonding and surface

2 answers:

Leni [432]2 years ago

7 0

Answer:

The more hydrogen bonds a molecule can make, the higher the surface tension.

Explanation:

Hydrogen bonds provide higher surface tension to a liquid

More hydrogen molecules - stronger cohesive forces

photoshop1234 [79]2 years ago

3 0

Answer:

The statement describes the relationship between bonding and surface

Explanation:

C. The more hydrogen bonds a molecule can make, the higher the surface tension.

Hope this helps

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The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

7 0

3 years ago

Ph11_UnitPacket2019

frozen [14]

Let's see

Use snells law

$\\ \rm\Rrightarrow \dfrac{n_1}{n_2}=\dfrac{sini}{sinr}$

$\\ \rm\Rrightarrow \mu=\dfrac{sin30}{sin19.9}$

$\\ \rm\Rrightarrow \mu=0.5/0.34$

$\\ \rm\Rrightarrow \mu=1.47$

It may be glass

3 0

2 years ago

Rex makes himself a bowl of cereal. He puts 28 g of cereal in a bowl and then adds 245 g of milk. Before eating the cereal, he r

Galina-37 [17]

Answer:

A.)

Explanation:

Because you didn't add anything or take anything away.

5 0

3 years ago

A Canadian driving from Quebec to Montreal finds he has traveled 271 km. How many miles is this? There are 1.61 km in 1 mi

kotegsom [21]

Answer:

168.32 mile

Explanation:

1 mile = 1.61 km

1.61 km = 1 mile

1 km = 1 / 1.61 mile

So, 271 km = 271 / 1.61 = 168.32 mile

3 0

3 years ago

LIst properties common to many organic compounds

stich3 [128]

Some properties I know are melting point<span>, </span>boiling point<span>, and index of </span>refraction<span>.</span>

7 0

3 years ago