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3 years ago

If ice didn't float, but sank instead, this change wouldn't affect aquatic life. true of false

Physics

2 answers:

ruslelena [56]3 years ago

6 0

Answer:

No it will not (false)

Explanation:

I say this because if anything the ice will sink and melt from a solid to a liquid and that will lead to i guess more water for the aquatic life. (THIS IS A THEORY SHAWTY)

Volgvan3 years ago

5 0

False

the aquatic lif would have to make significant changes to how they hunt and find shelter. Especially in Antartica

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Which of the five themes of geography would a person most likely use to study the importation of Chinese goods into the United S

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B. Movement would be the correct answer.

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2 years ago

3. When a person is outside of the system and they add energy to the

NISA [10]

Answer:

We show added energy to a system as +Q or -W

Explanation:

The first law of thermodynamics states that, in an isolated system, energy can neither be created nor be destroyed;

Energy is added to the internal energy of a system as either work energy or heat energy as follows;

ΔU = Q - W

Therefore, when energy is added as heat energy to a system, we show the energy as positive Q (+Q), when energy is added to the system in the form of work, we show the energy as minus W (-W).

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3 years ago

The MAC is 58 inches, The CG limits are from 26% to 43% MAC. If the CG is found to be

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By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.

<h3>What are the limits?</h3>

First, we need to find the limits.

We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.

So if 58 in is the 100%, the 26% and 43% of that are:

26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.

But we know that the CG is found to be 45.5% MAC, then it measures:

(45.5%/100%)*58in = 0.455*58in = 26.39 in

We need to compare it with the largest limit, so we get:

26.39 in - 24.94 in = 1.45 in

This means that the CG is 1.45 inches out of limits.

If you want to learn more about percentages, you can read:

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2 years ago

A rigid, nonconducting tank with a volume of 4 m3 is divided into two unequal parts by a thin membrane. One side of the membrane

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The final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

To find the answer, we need to know about the thermodynamic processes.

<h3>How to find the final temperature of the gas?</h3>

Any processes which produce change in the thermodynamic coordinates of a system is called thermodynamic processes.
In the question, it is given that, the tank is rigid and non-conducting, thus, dQ=0.
The membrane is raptured without applying any external force, thus, dW=0.
We have the first law of thermodynamic expression as,

$dU=dQ-dW$

Here it is zero.

$dU=0$ ,

As we know that,

$dU=C_pdT=0\\\\thus, dT=0\\\\or , T=constant\\\\i.e, T_1=T_2$

Thus, the final temperature of the system will be equal to the initial temperature,

$T_1=T_2=100^0C=373K$

We found that the process is isothermal,
Thus, the work done will be,

$W=RT*ln(\frac{V_2}{V_1} )=373R*ln(\frac{4}{\frac{4}{3} })\\ \\W=409.8R J$

Where, R is the universal gas constant.

<h3>What is a reversible process?</h3>

Any process which can be made to proceed in the reverse direction is called reversible process.
During which, the system passes through exactly the same states as in the direct process.

Thus, we can conclude that, the final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

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2 years ago

Read 2 more answers

You are investigating how objects move when they are dropped from different heights. To collect your data, you drop a 1 kg weigh

ASHA 777 [7]

The time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

The given parameters;

<em>Mass of the first object, m1 = 1 kg</em>

<em>Mass of the second object, m2 = 5 kg</em>

The final velocity of the objects during the downward motion is calculated as follows;

$v_f = v_0 + gt\\\\v_f = 0 + gt\\\\\v_f = gt$

The time of motion of the object from the given height is calculated as;

$h = v_0 t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} }$

The time of motion of each object is independent of mass of the object.

Thus, the time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

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2 years ago