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3 years ago

What happens to water when it changes to ice?

Physics

2 answers:

adelina 88 [10]3 years ago

7 0

Answer:

Density decreases, Volume increases

Explanation:

As the temperature decreases, the water contracts. Water has maximum density at 4° C. Thereafter, on further decrease in temperature, the water expands on cooling. Actually the water molecules crystallizes into open hexagonal structure. Thus, volume increases and density decreases. This is the reason that ice floats on water. The mass remains the same.

Vinvika [58]3 years ago

3 0

Answer:

Its density decreases, and its volume increases

Explanation:

Density is defined as Mass/Volume. What happens when water transforms into ice is that its density decreases because the volume the molecules now occupy increases.

The denominator of the rational expression $\frac{Mass}{Volume}$ is now larger (while the numerator stays the same), so the quotient gives a smaller number.

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Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinalandtransverse moduli

Masja [62]

Answer:

Not possible

Explanation:

$E_{cl}$ = longitudinal modulus of elasticity = 35 Gpa

$E_{ct}$ = transverse modulus of elasticity = 5.17 Gpa

$E_m$ = Epoxy modulus of elasticity = 3.4 Gpa

$V_{\rho l}$ = Volume fraction of fibre (longitudinal)

$V_{\rho t}$ = Volume fraction of fibre (transvers)

$E_f$ = Modulus of elasticity of aramid fibers = 131 Gpa

Longitudinal modulus of elasticity is given by

$E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764$

Transverse modulus of elasticity is given by

$E_{ct}=\frac{E_mE_f}{(1-V_{\rho t})E_f+V_{\rho t}E_m}\\\Rightarrow 5.17=\frac{3.4\times 131}{(1-V_{\rho t})131+V_{\rho t}3.4}\\\Rightarrow \frac{3.4\times 131}{5.17}-131=-127.6V_{\rho t}\\\Rightarrow V_{\rho t}=\frac{\frac{3.4\times 131}{5.17}-131}{-127.6}\\\Rightarrow V_{\rho t}=0.35148$

$V_{\rho l}\neq V_{\rho t}$

Hence, it is not possible to produce a continuous and oriented aramid fiber.

5 0

3 years ago

What happened to the maximum height of consecutive swings

GarryVolchara [31]

Answer:

we need more info

Explanation:

3 0

3 years ago

When the mass of the bottle is 0.125 kg, the average maximum height of the beanbag is m. When the mass of the bottle is 0.250 kg

Jet001 [13]

Answer:

when the mass of the bottle is 0.125 kg, the average height of the beanbag is 0.35 m.

when the mass of the bottle is 0.250 kg, the average maximum height of the beanbag is 0.91m.

when the mass of the bottle is 0.375 kg, the average maximum height of the beanbag is 1.26m.

when the mass of the bottle is 0.500 kg, the average maximum height of the beanbag is 1.57m.

Explanation:

5 0

3 years ago

A new landowner has a triangular piece of flat land she wishes to fence. Starting at the first corner, she measures the first si

Lera25 [3.4K]

Answer:

Length of third side = 3.97m

Explanation:

First of all, let we draw the triangle from the given information assuming our first corner is A. second corner is B and third corner is C.

From A-B we have distance = 5.5m = Say it c

From B-C we have distance = 4.3m = Say it a

From A-C we have distance = ? = Say it b which we have to find out.

Using the Law of Cosines: The square of the unknown side equals to the sum of squares of other 2 sides and subtracting 2*(Product of other sides)*(Cos(Angle opposite to the unknown side)

For our case it is:

b² = a² + c² - 2acCos(B) - Say it equation 1

From the attached triangle you may see that, a & c are our known sides and B is the angle opposite to the side b.

There values are:

a = 4.3m; c = 5.5m ; Angle B = 0.8 rad = 0.8 * 57.3 = 45.84 degrees where 1 rad = 57.3 degrees

Now by putting the respectve values in equation 1 we have:

b² = (4.3)² + (5.5)² - 2*4.3*5.5*Cos(45.84)

b² = 18.49 + 30.25 - 32.95

b² = 15.79

b = √15.79

b = 3.97m

Thus the length of third side is 3.97m.

PS: The picture of triangle is being attached for yours understanding.

6 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

Yuliya22 [10]

Answer:

Radio waves have a wavelength between $10^{-9}m$ and $10^{-12}m$

While,

X rays have a wavelength between 1m and 10km.

=> It is one of the condition of diffraction that the obstacle (coming in the way) must be comparable with the size of the wavelength.

=> This shows, that radio waves have a wavelength which is comparable with the size of buildings and can really easily diffract through it

=> While, X-rays are big enough to diffract through the wall.

So, if an X-ray technician stands behind a wall during the use of her machine, she will remain safe.

6 0

3 years ago