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defon
4 years ago
8

How does adding more of a substance affect it's density?

Physics
2 answers:
fomenos4 years ago
5 0
I’m going to use molasses as an example of a substance.

The mass and volume both change when changing the amount of molasses.
However, the density does not change. This is because the mass and volume increase at the same rate/proportion!

Even though there is more molasses (mass) in test tube A, the molasses also takes up more space (volume). Therefore, the spacing between those tiny particles that make up the molasses is constant (does not change).

The size or amount of a material/substance does not affect its density.
Tanya [424]4 years ago
5 0

Adding more of a substance has no effect on its density.  The formula for density is:

Density = (mass) divided by (volume).

If you doubled the volume of your sample of the substance, then you also doubled the mass of the sample.  So when you divide (mass) by (volume), you still get the same number.  The density doesn't change.


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Assume that the upward direction is positive and the downward direction is negative. What is the ball's velocity (in m/s) when i
LenaWriter [7]

The given question is incomplete. The complete question is as follows.

You throw a ball vertically upward, and as it leaves your hand, its speed is 26.0 m/s.

(a) How high (in m) does it rise above the level where it leaves your hand?

(b) How long (in s) does it take to reach its highest point?

(c) How long (in s) does the ball take to return to the level where it left your hand after it reaches its highest point?

(d) Assume that the upward direction is positive and the downward direction is negative. What is the ball's velocity (in m/s) when it returns to the level where it left your hand? (Indicate the direction with the sign of your answer.)

Explanation:

(a) For maximum height, the formula will be as follows.

           v^{2} = u^{2} + 2as

                 a = v^{2} - 2gh

or,                 h = \frac{v^{2}}{2g}

                        = \frac{(26)^{2}}{2 \times 9.8}

                        = \frac{676}{19.6}

                        = 34.5 m/s

Hence, it rises 34.5 m/s above the level where it leaves your hand.

(b) Time to reach maximum height is as follows.

            v = u + at

or,           v - gt = 0

                 t = \frac{v}{g}

                   = \frac{26}{10}

                   = 2.6 sec

Therefore, it will take 2.6 sec to reach its highest point.

(c)  Time taken by the ball to ascent is equal to the time it has taken to descent.

Therefore, time taken by the ball to return to the level where it left your hand after it reaches its highest point? is also 2.6 sec.

(d)  Speed of the ball will be 26 m/s in the downward direction. Hence, the velocity will be -26 m/s.

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Answer:

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3 years ago
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horsena [70]

I  = pressure amplitude given = 0.2 W/m²

dB = decibel reading

decibel reading from the pressure amplitude is given as

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inserting the values in the above equation

dB = 10 log₁₀ (0.2/10⁻¹²)

dB = 10 log₁₀ (2 x 10⁻¹/10⁻¹²)

dB = 10 log₁₀ (2 x 10⁻¹.10¹²)

dB = 10 log₁₀ (2 x 10¹²⁻¹)

dB = 10 log₁₀ (2 x 10¹¹)

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3. Which statement best describes the research projects that are funded by private
SOVA2 [1]

Answer:

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Explanation:

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