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inysia [295]
4 years ago
13

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:
Rudiy274 years ago
3 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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Answer:

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3 years ago
A tiny object carrying a charge of +44 μC and a second tiny charged object are initially very far apart. If it takes 21 J of wor
STatiana [176]

Answer:

The magnitude of the second charge is \rm 1.062\times 10^{-7}\ C or \rm 0.1062\ \mu C.

Explanation:

The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.

The electric potential energy of some charge q_o at a point in the electric field of another charge q is given by the product of the amount of charge q_o and electric potential at that point due to the charge q.

U = q_o\ V.

The electric potential at that point is given by

V = \dfrac{kq}{r}.

where k is the Coulomb's constant.

Therefore,

U=q_o\ \dfrac{kq}{r}.

Now, We have given two charges q_1 = +44\ \mu C = +44\times 10^{-6}\ C and q_2, whose value is to be found.

When the two charges are infinitely dar apart, the electric potential energy of the system is given by

U_i = \dfrac{kq_1q_2}{\infty}=0.

When the coordinates of position of the two charges are

(x_1,\ y_1) = (1.00\ mm,\ 1.00\ mm).\\(x_2,\ y_2) = (1.00\ mm,\ 3.00\ mm).

The distance between the two charges is given by

r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(1.00-1.00)^2+(3.00-1.00)^2}=2.00\ mm = 2.00\times 10^{-3}\ m.

The electric potential energy of the charges in this configuration is given by

U_f = \dfrac{kq_1q_2}{r}\\=\dfrac{(8.99\times 10^9)\times (+44\times 10^{-6})\times q_2}{2.00\times 10^{-3}}\\=1.9778\times 10^8\times q_2.

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.

Therefore,

W = U_f-U_i\\21=(1.9778\times 10^8\times q_2)-0\\\Rightarrow q_2 = \dfrac{21}{1.9778\times 10^8}\\=1.062\times 10^{-7}\ C\\=0.1062\times 10^{-6}\ C\\=0.1062\ \mu C.

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3 years ago
The image below shows two opposite forces acting on a rope, what can we say is true about the affect of the forces on the rope?
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The answer to the question is true
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During a goal-line stand, a 112-kg fullback moving eastward with a speed of 6 m/s
11111nata11111 [884]

Answer:

-1.24 m/s

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Total momentum before collision = (mass of full back * velocity of fullback) + (mass of lineman * velocity of line man).

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The total momentum after collision = (mass of full back + mass of line back) * velocity after collision.

Let velocity after collision be v, hence:

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Total momentum before collision = total momentum after collision

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