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lisabon 2012 [21]

4 years ago

6

PLEASE ANSWER QUICK!! What happens at the condensation point?A. Molecules of a liquid heat up and begin to convert to a gaseous

state.
B. Molecules of a solid speed up and begin to turn to a liquid state.
C. Molecules of a liquid slow down and start to convert to a solid state.
D. Molecules of a gas slow down and change to a liquid state.

1 answer:

Nastasia [14]4 years ago

4 0

Answer:

D. Molecules of a gas slow down and change to a liquid state.

Explanation:

Condensation refers to a process by which a gas changes from gaseous state to liquid state. For example, water vapor changes to from the state of being a gas to liquid state water.
Condensation is the opposite of evaporation and occurs when gaseous particles slow down and change into liquid state.
Heat energy is lost during condensation and gaseous molecules lose kinetic energy making them to slow down and thus changing to liquid state,

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Answer:

3100 m/s

Explanation:

The relationship between frequency and wavelength of a wave is given by the wave equation:

$v=f\lambda$

where

v is the speed of the wave

f is its frequency

$\lambda$ is the wavelength

For the wave in this problem,

f = 15,500 Hz

$\lambda=0.20 m$

Therefore, the wave speed is

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3 years ago

a toy dart gun generates a dart with .140kg.m/s momentum and a velocity of 4m/s. what is the mass of the dart in grams?

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Answer: m = 0.035kg = 35g

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Formula-

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The equation for momentum is

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2 years ago

A plane flies from base camp to lake a, 200 km away in the direction 20.0° north of east. after dropping off supplies it flies t

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Distance of lake a is 200 km at 20 degree north of east

distance between lake a and b is 230 km at 30 degree west of north

now the distance between base and lake b is given as

$d = d_1 + d_2$

given that

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$d_1 = 187.94 i + 68.4 j$

$d_2 = -230 sin30 i + 230 cos30 j$

$d_2 = -115 i + 199.2 j$

now the total distance is

$d = (187.94 - 115)i + (199.2 + 68.4)j$

$d = 72.94 i + 267.6 j$

now the magnitude of the distance is given as

$d = \sqrt{72.94^2 + 267.6^2}$

$d = 277.4$

also the direction is given as

$\theta = tan^{-1}\frac{267.6}{72.94}$

$\theta = 74.7 degree$

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3 years ago

The vacuum pressure of a condenser is given to be 80 kpa. if the atmospheric pressure is 98 kpa, what is the gage pressure and a

dimaraw [331]

The absolute pressure is given by the equation,

$P_{abs}=P_{atm}-P_{vac}$

Here, $P_{abs}$ is absolute pressure, $P_{atm}$ is atmospheric pressure and $P_{vac}$ is vacuum pressure.

Therefore,

$P_{abs}=98 kPa-80 kPa=18kPa$

The gage pressure is given by the equation,

$P_{gage}=P_{abs}-P_{atm}$ .

Thus,

$P_{gage}=18kPa-98 kPa=-80 kPa$ .

In kn/m^2,

The absolute pressure,

$P_{abs}=18kPa(\frac{1kN/m^2}{kPa}) =18\ kN/m^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1kN/m^2}{kPa}) =-80\ kN/m^2$ .

In lbf/in2

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=2.6\ lbf/in^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=-11.6\ lbf/in^2$

In psi,

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa})=2.610\ psi$ .

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa} )=-11.6030\ psi$

In mm Hg

The absolute pressure,

$P_{abs}=18kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})= 135\ mm\ of\ Hg$

The gage pressure,

$P_{gage}=-80kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})=-600\ mm\ of\ Hg$

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3 years ago

If you measure the amount of work accomplished in a particular time interval, u have measured-

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d = distance [meters] [m]

W = work [Joules]

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P = W/t

where:

P = power [watts] [w]

W = work [Joules] [J]

t = time [seconds] [s]

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3 years ago