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igomit [66]
3 years ago
9

15) On a cold day, you take in 4.2 L (i.e., 4.2 x 10-3 m3) of air into your lungs at a temperature of 0°C. If you hold your brea

th until the temperature of the air in your lungs reaches 37°C, what is the volume of the air in your lungs at that point, assuming the pressure does not change?
Physics
1 answer:
Naily [24]3 years ago
8 0

Answer:

4.8L ( i.e 4.8 x 10^-3 m3)

Explanation:

Step 1:

Data obtained from the question.

Initial volume (V1) = 4.2L

Initial temperature (T1) = 0°C

Final temperature (T2) = 37°C

Final volume (V2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below

K = °C + 273

T1 = 0°C = 0°C + 273 = 273K

T2 = 37°C = 37°C + 273 = 310K

Step 3:

Determination of the final volume.

Since the pressure is constant,

Charles' Law equation will be applied as shown below:

V1 /T1 = V2/T2

4.2/273 = V2 /310

Cross multiply to express in linear form

273 x V2 = 4.2 x 310

Divide both side by 273

V2 = (4.2 x 310)/273

V2 = 4.8L ( i.e 4.8 x 10^-3 m3)

Therefore, the volume of the air in the lungs at that point is 4.8L ( i.e 4.8 x 10^-3 m3)

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Skylar travels 50 meters N and then goes 30 meters W before coming straight back south 20 meters. What distance did she travel?
lesya [120]

Answer:

100\ \text{meters}.

Explanation:

Distance Skylar traveled North is 50\ \text{meters}

Then she traveled 30\ \text{meters} Westward.

After which she traveled 20\ \text{meters} towards the South.

The total distance traveled would be the sum of the distances.

50+30+20=100\ \text{meters}

The distance traveled by Skylar was is 100\ \text{meters}.

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The area of the large syringe in an experiment is 18cm2 and that of the smaller one is 3.0cm2. A force of 2N is applied on the s
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Explanation:

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Convection currents in air and water occur because _
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warm fluids are less dense than cold fluids

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Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

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3 years ago
Which of the following is an example of speed?
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