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professor190 [17]
2 years ago
5

An airplane has a mass of 12,000kg. If it is accelerating at 16m/sec2, what is its force?(round to the nearest whole number) NEE

D THIS ASAP
Chemistry
1 answer:
icang [17]2 years ago
8 0

Answer:

F=1.92x10^5N

Explanation:

Hello!

In this case, since the force exerted by an object with a certain mass an acceleration is:

F=m*a

We can plug in the given mass of 12,000 kg and acceleration of 16 m/s^2 to obtain:

F=12,000kg*16m/s^2\\\\F=192,000kg*m/s^2\\\\F=1.92x10^5N

Best regards!

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1) The densities of air at −85°C, 0°C, and 100°C are 1.877 g dm−3, 1.294 g dm−3, and 0.946 g dm−3, respectively. From these data
earnstyle [38]

Answer:

1) The absolute zero temperature is -272.74 °C

2) The absolute zero temperature is -269.91°C

Explanation:

1) The specific volume of air at the given temperatures are;

At -85°C, v =  1/(1.877) = 0.533 dm³/g, the pressure =

At 0°C, v = 1/1.294 ≈ 0.773 dm³/g

At 100°C, v = 1/0.946 ≈ 1.057 dm³/g

We therefore have;

v₁ = 0.533 T₁ = 188.15  K

v₂ = 0.773 T₂ = 273.15  K

v₃ = 1.057 T₃ = 373.15  K

The slope of the graph formed by the above data is therefore given as follows;

m = (1.057 - 0.533)/(373.15 - 188.5) = 0.00284 dm³/K

The equation is therefor;

v - 0.533 = 0.00284×(T - 188.15)

v = 0.00284×T - 0.534 + 0.533  = 0.00284×T - 0.001167

v =  0.00284×T - 0.001167

Therefore, when the temperature, at absolute 0, we have;

v = 0

Which gives;

0 =  0.00284×T - 0.001167

0.00284×T = 0.001167

T = 0.001167/0.00284 ≈ 0.411 K

Which is 0.411  -273.15 ≈ -272.74 °C

The absolute zero temperature is -272.74 °C

2) The given volume of the gas = 20.00 dm³ at 0°C and 1.000 atm

The slope of the volume temperature graph at constant pressure = 0.0741 dm³/(°C)

The temperature is converted to Kelvin temperature, in order to apply Charles law as follows;

0°C = 0 + 273.15 K = 273.15 K

Therefore, the equation of the graph can be presented as follows;

v - 20.00 = 0.0741 × (T - 273.15)

Which gives;

v = 0.0741·T - 0.0741 ×(273.15) + 20

v = 0.0741·T - 20.2404125 + 20

v = 0.0741·T - 0.240415

Therefore at absolute 0, v = 0, we have;

0 = 0.0741·T - 0.240415

0.0741·T = 0.240415

T = 0.240415/0.0741 = 3.2445 K

The temperature in degrees is therefore;

3.2445 K - 273.15 ≈ -269.91°C

The absolute zero temperature is therefore, -269.91°C

3 0
3 years ago
Pls help me in this:(​
antiseptic1488 [7]

Answer:

a) 26.98

b) 3

c) 3* 1.01 = 3.03

d) 3 *16 = 48

e) 26.98 + 3.03 + 48 = 78.01

f) 6.023 * 10²³

3 0
3 years ago
7. A rectangular block of lead (Pb) measures 20.0 mm X 30.0 mm X 45.0 mm. If the density of Pb is 11.34 g/cm3, calculate the mas
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Mass of Lead is .00042 g
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3 years ago
If the container is closed and the ethanol is allowed to reach equilibrium with its vapor, how many grams of liquid ethanol rema
SVETLANKA909090 [29]

Explanation:

Let us assume that the given data is as follows.

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       P = 40 torr    (1 atm = 760 torr)

So,     P = \frac{40 torr}{760 torr} \times 1 atm

             = 0.053 atm

          n = ?

According to the ideal gas equation, PV = nRT.

Putting the given values into the above equation to calculate the value of n as follows.

                 PV = nRT

   0.053 atm \times 3.10 L = n \times 0.0821 L atm/mol K \times 292 K

                 0.1643 = n \times 23.97

                    n = 6.85 \times 10^{-3}

It is known that molar mass of ethanol is 46 g/mol. Hence, calculate its mass as follows.

               No. of moles = \frac{mass}{\text{molar mass}}

                 6.85 \times 10^{-3} = \frac{mass}{46 g/mol}  

                    mass = 315.1 \times 10^{-3} g

                              = 0.315 g

Thus, we can conclude that the mass of liquid ethanol is 0.315 g.

4 0
3 years ago
1s2 2s2 2p6 3s2 3p6 is the electron configuration for
kvasek [131]

Answer: The electron configuration is for the element Argon

Explanation: 3p6 is all the way on the right side of the periodic table in the third row or period

3 0
3 years ago
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