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3 years ago

Which mass would have the greatest acceleration if the same unbalanced force was applied to each?

1 answer:

3 0

I don't know which masses you're looking at, so I can't describe it to you.
But whatever they look like, it would be the one with the smallest mass.
That's not necessarily the smallest one, just the one that weighs the least.

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A sample of oxygen gas at 25.0Â°c has its pressure tripled while its volume is halved. What is the final temperature of the gas?

Marat540 [252]

Answer:

447 K

Explanation:

25 C = 25 + 273 = 298 K

Assuming ideal gas, we can apply the ideal gas law

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

$T_2 = T_1\frac{P_2}{P_1}\frac{V_2}{V_1}$

Since pressure is tripled, then $P_2 / P_1 = 3$ . Volume is halved, then $V_2 / V_1 = 0.5$

$T_2 = 298*3*0.5 = 447 K$

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3 years ago

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A box having a mass of 0.2 kg is dragged across a horizontal floor by

Lyrx [107]

you can check attachment for answer.

kind regards

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3 years ago

The phase change in which a substance changes from a gas directly to a solid is

vlabodo [156]

<span>Condensation is the change of the substance from liquid to solid phase. Example of this is the formation of ice. Vaporization is the change of substance from liquid to gas phase. Example of this is the boiling of water. Deposition is the change of a substance from gas to solid phase. Example of this is the formation of ice on a winter day. Sublimation is the change of a substance from solid to gas phase. Example of this is dry ice. The answer is letter C.</span>

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3 years ago

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A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the

zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

$V = \frac{\pi d^2}{4}h$

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

$V_0 = \frac{\pi d^2}{4}H$

$V_f = \frac{\pi d^2}{4}h$

We know as well by definition that

$1gal = 3.785*10^{-3}m^3$

Then we have for the statement that

$V_f = V_0 -1gal$

$V_f = V_0 - 3.785*10^{-3}$

Replacing the previous data

$\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}$

$\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}$

Solving to get h,

$h = 1.99963m$

Therefore the change is

$\Delta h = H-h$

$\Delta h = 2- 1.99963$

$\Delta h = 3.7*10^{-4}m=0.37mm$

Therefore te change in the height of the water in the tank is 0.37mm

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3 years ago

An object is released from height of 17m. <br> The object will hit the ground approximately in

zzz [600]

$\text{Given that,}\\\\\text{Height, h = 17 m}\\\\\\\text{We know that,}\\\\h = v_0t + \dfrac 12 gt^2\\\\\implies h = \dfrac 12 gt^2\\\\\implies t^2 = \dfrac{2h}g\\\\\implies t =\sqrt{\dfrac{2h}g} = \sqrt{\dfrac{2(17)}{9.81}} = 1.87 ~ \text{sec}$

5 0

2 years ago