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3 years ago

Which weather condition commonly occurs along a cold front

2 answers:

7 0

Precipitation is your answer. according to my science textbook, in a cold front, warm air and cold air interact and cause precipitation.

lbvjy [14]3 years ago

7 0

Cold fronts form when a cooler air mass moves into an area of warmer air in the wake of a developing extratropical cyclone. The warmer air interacts with the cooler air mass along the boundary, and usually produces precipitation. Cold fronts often follow a warm front or squall line.

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A 1.11 kg piece of aluminum at 78.3 c is put into a glass with 0.210 kg of water at 15.0

ValentinkaMS [17]

M = mass of aluminium = 1.11 kg

$c_{a}$ = specific heat of aluminium = 900

$T_{ai}$ = initial temperature of aluminium = 78.3 c

m = mass of water = 0.210 kg

$c_{w}$ = specific heat of water = 4186

$T_{wi}$ = initial temperature of water = 15 c

T = final equilibrium temperature = ?

using conservation of heat

Heat lost by aluminium = heat gained by water

M $c_{a}$ ( $T_{ai}$ - T) = m $c_{w}$ (T - $T_{wi}$ )

(1.11) (900) (78.3 - T) = (0.210) (4186) (T - 15)

T = 48.7 c

7 0

3 years ago

Energy that flows from an objective with a higher temperature to an object with a lower temperature

dalvyx [7]

That's THERMAL energy, often referred to as "heat".

7 0

4 years ago

The rate at which a candle burns in millimeters per minute is:

anzhelika [568]

since both components, length and time, are measurable
<span>since Rate = length ÷ time </span>
<span>∴ rate is also measurable and ∴ quantitative.

</span>

3 0

3 years ago

Calculate the number of free electrons per cubic meter for some hypothetical metal, assuming that there are 1.3 free electrons p

boyakko [2]

Answer:

The number of free electrons per cubic meter is $7.61\times 10^{28}\ m^{-3}$

Explanation:

It is given that,

The number of free electrons per cubic meter is, 1.3

Electrical conductivity of metal, $\sigma=6.8\times 10^7\ \Omega^{-1}m^{-1}$

Density of metal, $\rho=10.5\ g/cm^3$

Atomic weight, A = 107.87 g/mol

Let n is the number of free electrons per cubic meter such that,

$n=1.3\ N$

$n=1.3(\dfrac{\rho N_A}{A})$

Where

$\rho$ is the density of silver atom

$N_A$ is the Avogadro number

A is the atomic weight of silver

$n=1.3\times (\dfrac{10.5\ g/cm^3\times 6.02\times 10^{23}\ atoms/mol}{107.87\ g/mol})$

$n=7.61\times 10^{22}\ cm^{-3}$

$n=7.61\times 10^{28}\ m^{-3}$

Hence, this is the required solution.

6 0

3 years ago

You walk forward at 1.5 m/s for 8s. Your friend decides to walk faster and starts out at 2.0 m/s for the first 4 s. Then she slo

Troyanec [42]

The first thing you should know is that the distance is equal to the speed per time.
Therefore if You walk forward at 1.5 m / s for 8s
d = 1.5t [0,8] s
Your friend decides to walk faster and starts at 2.0 m / s for the first 4 s.
d = 2t [0,4] s
Then she slows down and walks forward at 1.0 m / s for the next 4s
d = t + 4 [4,8] s
Who walked farther?
They both walked the same distance
12 meters

6 0

3 years ago