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3 years ago

Why can water dissolve some ionic compounds, like NH4Cl, as well as some nonionic compounds, like methanol?

Physics

2 answers:

nirvana33 [79]3 years ago

4 0

I think that’s the answer u r looking for

Inga [223]3 years ago

3 0

Answer:

Water is polar molecule. ... The ionic compounds are dissolved as these ions attract the hydrogen( negative charge attraacts them) and oxygen (positive charge attracts them) whereas methanol is a polar molecule and forms hydrogen bond with water molecules and hence dissolves.

Explanation:

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A 65 kg box is lifted by a person pulling a rope a distance of 15 meters straight up at a constant speed. How much Power is requ

Y_Kistochka [10]

Answer:

Power is 1061.67W

Explanation:

Power=force×distance/time

Power=65×9.8×15/9 assuming gravity=9.8m/s²

Power=3185/3=1061.67W

8 0

2 years ago

Abdou was explaining to a classmate that graphite is a good lubricant

balu736 [363]

The fact that the layers of graphite are held together by only weak Van der Walls forces implies that they can slide over each other.

<h3>Why is graphite a solid lubricant?</h3>

We know that graphite is composed of layers. These hexagonal layers are held together by weak Van Der Walls forces and as such are able to slide over each other. The carbon atom in each layer are held together by strong covalent bonds.

The fact that the layers of graphite are held together by only weak Van der Walls forces implies that they can slide over each other and as such make the graphite fluid.

Thus, the image that shows these layers of graphite is attached to this an answer

Learn more about graphite:brainly.com/question/11095487

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3 0

2 years ago

PLEASE HELP ME 45 POINTS

sergij07 [2.7K]

Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is 3.551 meters per second.

Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:

Mass A

$\Sigma F = T - m_{A}\cdot g = -m_{A}\cdot a$ (1)

Mass B

$\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a$ (2)

Where:

$T$ - Tension force in the cord, measured in newtons.

$m_{A}$ , $m_{B}$ - Masses of blocks A and B, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

$m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g$

$(m_{B}+m_{A})\cdot a = (m_{A}-m_{B})\cdot g$

$a = \frac{m_{A}-m_{B}}{m_{B}+m_{A}} \cdot g$

If we know that $m_{A} = 5\,kg$ , $m_{B} = 2\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the acceleration of the masses is:

$a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)$

$a = 4.203\,\frac{m}{s^{2}}$

The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

$T = m_{B}\cdot (a+g)$

If we know that $m_{B} = 2\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $a = 4.203\,\frac{m}{s^{2}}$ , then the tension force in the cord:

$T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)$

$T = 28.02\,N$

The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

$\Delta y = \frac{1}{2}\cdot a\cdot t^{2}$ (3)