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3 years ago

Why would scientists call solids and liquids a condensed phase of matter?

Physics

1 answer:

lawyer [7]3 years ago

8 0

Explanation:

Liquids and solids are often referred to as condensed phases because the particles are very close together.

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A 270,000 kg rocket in gravity -free space expels 270 kg of fuel by exerting a force of 360 N on it. Find the magnitude of the a

Brut [27]

F = m a
360 = 270 x a
1.3 = a

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3 years ago

What is chemical change ??

IgorC [24]

Answer:

the combination,decomposition or replacement that occurs in the molecules of matter during chemical change called chemical reaction

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2 years ago

Read 2 more answers

You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene

uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is $150\,^{\circ}C$ . According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

$\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T)$ (1)

Where:

$\rho_{k}$ - Density of kerosene, measured in kilograms per cubic meter.

$V_{k}$ - Volume of kerosene, measured in cubic meters.

$c_{k}$ , $c_{t}$ - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

$T$ - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

$V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}$

If we know that $m_{t} = 1.83\,kg$ , $c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{t,o} = 88\,^{\circ}C$ , $T_{k,o} = 24.0\,^{\circ}C$ , $T = 57\,^{\circ}C$ , $c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C}$ and $\rho_{k} = 820\,\frac{kg}{m^{3}}$ , then the volume of the liquid needed to accomplish this task without boiling is:

$V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}$

$V_{k} = 2.273\times 10^{-4}\,m^{3}$

$V_{k} = 0.273\,L$

0.273 liters are needed to accomplish this task without boiling.

3 0

3 years ago

The area of an equilaterla triangle is increasing at a rate of 5 m^2/hr. find the rate at witch the height is chganging when the

Lena [83]

The rate at which the height is changing is ( 5 / x ) m / hr

We know that,

Area of an equilateral triangle A = $\sqrt{3}$ $x^{2}$ / 4

h = $\sqrt{3}$ x / 2

Where,

x = Side

h = Height

Given that,

dA / dt = 5 $m^{2}$ / hr

h = $\sqrt{3}$ x / 2

Differentiate both sides with respect to t

dh / dt = ( $\sqrt{3}$ / 2 ) ( dx / dt )

dx / dt = ( 2 / $\sqrt{3}$ ) ( dh / dt )

A = $\sqrt{3}$ $x^{2}$ / 4

Differentiate both sides with respect to t

dA / dt = ( $\sqrt{3}$ / 4 ) ( 2x ) ( dx / dt )

5 = ( $\sqrt{3}$ / 4 ) ( 2x ) ( 2 / $\sqrt{3}$ ) ( dh / dt )

dh / dt = ( 5 / x ) m / hr

Rate of change of height is defined as the rate at which height of an object changes with respect to time. It is represented as dh / dt

Therefore, the rate at which the height is changing is ( 5 / x ) m / hr

To know more about Rate of change of height

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1 year ago

Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for sl

ikadub [295]

Answer:

Please see below as the answer is self-explanatory.

Explanation:

We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).
Both movements are independent each other, due to they are perpendicular.
In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.
Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:

$v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)$

Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:

$t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s (2)$

In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.
Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:

$\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (3)$

In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:

$v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)$

Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:

$\Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)$

Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.

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3 years ago