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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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A cat is chasing a mouse across a 1.3 m tall dining table. The mouse darts to the side and the cat accidentally slides off, land

erastovalidia [21]

Answer:

1) 0.51 seconds.

2) 1.45 m/s.

Explanation:

given, height from which cat falls = 1.3 m

we know that, s = ut + $\frac{1}{2}$ at².

here if we consider cat moment only in downward direction,

intial velocity of cat in downward direction , u = 0.

so, time, t = $\sqrt{\frac{2h}{g} }$ .

⇒ t = $\sqrt{\frac{2(1.3)}{9.81} }$ = 0.51 seconds.

t = 0.51 seconds.

now, consider cat moment only in forward direction

s = ut , since acceleration is zero in forward direction

⇒ u = $\frac{s}{t}$ .

so, u = $\frac{0.75}{0.51}$ = 1.45 m/s .

6 0

3 years ago

Racing cars driven by chris and kelly are side by side at the start of a race. the table shows the velocities of each car (in mi

Mamont248 [21]

Solution

distance travelled by Chris

\Delta t=\frac{1}{3600}hr.

X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}

=\frac{579.5}{3600}=0.161miles

Kelly,

\Delta t=\frac{1}{3600}hr.

X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}

=\frac{657.5}{3600}

\Delta X=X_{k}-X_{C}=0.021miles

4 0

3 years ago

What sphere on earth includes mountains

Lostsunrise [7]

I believe it is lithosphere

3 0

3 years ago

Please help me with this physics prooblem

zaharov [31]

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

$x=v_0\cos20.0^\circ t+\dfrac12a_xt^2$

$y=v_0\sin20.0^\circ t+\dfrac12a_yt^2$

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

$x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ$

$y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ$

So we have enough information to solve for the components of the acceleration vector, $a_x$ and $a_y$ :

$x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}$

$y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}$