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3 years ago

A thermometer is placed directly sunlight will it read the temperature of the air or of the sun or of something else??

1 answer:

7 0

It will read the temperature of the atmosphere.It is unable to read the temperature of the sun because some of the heat energy from the sun is absorbed by the layer of air molecules on Earth.

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The upward force on an airplane's wing is thrust.

kompoz [17]

No. 'Thrust' is what most people in aviation call the force
that pushes the aircraft forward.

The same people generally call the upward force on the wing "lift".

3 0

4 years ago

What force is required to accelerate to 10 kg object to 5.9 m/s/s?

g100num [7]

Force required to accelerate 10 kg object to 5.9 m/s/s ?

Mass = 10 kg

Acceleration = 5.9 m/s^2

Force = Mass * Acceleration

Force = 10 kg * 5.9 m/s^2

Force = 59 kg m /s^2 = 59 N

3 0

3 years ago

A piston–cylinder device initially contains 2 L of air at 100 kPa and 25°C. Air is now compressed to a final state of 600 kPa an

bagirrra123 [75]

Answer:

a. The energy of the air at the initial and the final states is 0kJ and 0.171kJ respectively

b. 0.171kJ

c. 0.143

Explanation:

Because there are same conditions of the state of air at the surroundings and at the Initial stage, the energy of air at the Initial stage is 0kJ.

Calculating energy at the final state;

We start by calculating the specific volume of air in the environment and at the final state.

U2 = At the final state, it is given by

RT2/P2

U1= At the Initial state, it is given by

RT1/P1

Where R = The gas constant of air is 0.287 kPa.m3/kg

T2 = 150 + 273 = 423K

T1 = 25 + 273 = 298K

P2 = 600KPa

P1 = 100KPa

U2 = 0.287 * 423/600

U2 = 0.202335m³/kg

U1 = 0.287 * 298/100

U1 = 0.85526m³/kg

Then we Calculate the mass of air using ideal gas relation

PV = mRT

m = P1V/RT1 where V = 2*10^-3kg

m = 100 * 2 * 10^-3/(0.287 * 298)

m = 0.00234kg

Then we calculate the entropy difference, ∆s. Which is given by

cp2 * ln(T2/T1) - R * ln(P2/P1)

Where cp2 = cycle constant pressure = 1.005

∆s = 1.005 * ln (423/298) - 0.287 * ln(600/100)

∆s = -0.1622kJ/kg

Energy at the final state =

m(E2 - E1 + Po(U2 - U1) -T0 * ∆s)

E2 and E1 are gotten from energy table as 302.88 and 212.64 respectively

Energy at the final state

= 0.00234(302.88 - 212.64 + 100(0.202335 - 0.85526) - 298 * -0.1622)

Energy at the final state = 0.171kJ

Minimum Work = ∆Energy

Minimum Work = Energy at the final state - Energy at the initial state

Minimum Work = 0.171 - 0

Minimum Work done = 0.171kJ

c. The second-law efficiency of this process is calculated by ratio of meaningful and useful work

= 0.171/1.2

= 0.143

3 0

3 years ago

A student combined equal amounts of two solutions one solution had a ph of 2 and the other had a ph 12 which would most likely b

motikmotik

The answer would be 6.

8 0

3 years ago

Read 2 more answers

A sample of nitrogen occupies 5.50 liters under a pressure of 900 torr at 25oC. At what temperature will it occupy 10.0 liters a

vitfil [10]

Answer:

<u>At 268.82°C</u> volume occupied by nitrogen is 10 liters at pressure of 900 torr.

Explanation:

Given:

Volume of a sample of nitrogen = 5.50 liters

Pressure = 900 torr

Temperature = 25°C

To find the temperature at which the nitrogen will occupy 10 liters volume at same pressure.

Solution:

Since the pressure is kept constant, so we can apply the temperature-volume law also called the Charles Law.

Charles Law states that the volume of a gas held at constant pressure is directly proportional to the temperature of the gas in Kelvin.

Thus, we have :

$V$ ∝ $T$

$\frac{V}{T}=k$

where $k$ is a constant.

For two samples of gases, the law can be given as:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

From the data given:

$V_1=5.5\ l$

$T_1=25\ \°C =(273+25)K= 298 K$

$V_2=10\ l$

We need to find $T_2$ .

Plugging in values in the formula.

$\frac{5.5}{298}=\frac{10}{T_2}$

Multiplying both sides by $T_2$ .

$T_2\times\frac{5.5}{298}=\frac{10}{T_2}\times T_2$

$\frac{5.5}{298}T_2={10}$

Multiplying both sides by $\frac{298}{5.5}$

$\frac{298}{5.5}\times\frac{5.5}{298}T_2=\frac{10\times 298}{5.5}$

$T_2=541.82\ K$

$T_2=541.82\ K-273\ K = 268.82\°C$

Thus, at 268.82°C volume occupied by nitrogen is 10 liters at pressure of 900 torr.

7 0

3 years ago