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3 years ago

Explain the process of anomalous expansion of water

1 answer:

8 0

The anomalous expansion of water refers to the fact that water, unlike most substances, expands when it freezes. The density of water increases as temperature decreases but reaches a maximum at 4° C then begins to expand.

When liquid water is cooled, it contracts like one would expect until a temperature of approximately 4 degrees Celsius is reached. After that, it expands slightly until it reaches the freezing point, and then when it freezes it expands by approximately 9%.

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PLS HELP FOR THIS PHYSICS TEST <br><br> (pls only answer if u rlly know, this is an important test)

lesya692 [45]

Answer:

C. 8.01 m/s²

Explanation:

vf²= vi² + 2 • a • d

2ad = vf² - vi²

a = (vf²- vi²)/2d

d=25.00 -5.00=20.00 m

vi =0

vf=17.90 m/s

a =(17.90² -0²)/(2*20) = 8.01 m/s²

4 0

3 years ago

What is the role of neutrons in the nucleus.

Serjik [45]

Answer:

<em><u>To be neutral</u></em>

Explanation:

Neutrons are electrically neutral but contribute to the mass of a nucleus to nearly the same extent as the protons. Neutrons can explain the phenomenon of isotopes (same atomic number with different atomic mass). The main role of neutrons is to reduce electrostatic repulsion inside the nucleus.

5 0

3 years ago

Read 2 more answers

A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does

Irina-Kira [14]

Answer:

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

$y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)$ (1)

Where:

$y(t)$ - Position of the mass as a function of time, measured in meters.

$A$ - Amplitude, measured in meters.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the block, measured in kilograms.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

$k = \frac{m\cdot g}{\Delta y}$ (2)

Where:

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

$\Delta y$ - Deformation of the spring due to gravity, measured in meters.

If we know that $m=1.65\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta y = 0.260\,m$ , then the spring constant is:

$k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}$

$k = 62.237\,\frac{N}{m}$

If we know that $A = 0.130\,m$ , $k = 62.237\,\frac{N}{m}$ , $m=1.65\,kg$ , $x(t) = 0\,m$ and $\phi = 0\,rad$ , then (1) is reduced into this form:

$0.130\cdot \cos (6.142\cdot t)=0$ (1)

And now we solve for $t$ . Given that cosine is a periodic function, we are only interested in the least value of $t$ such that mass reaches equilibrium position. Then:

$\cos (6.142\cdot t) = 0$