The speed with which force is developed by skeletal muscles is determined by the

The tidal lung volume of human breathing, representing the amount of air inhaled and exhaled in a normal breath, is 500 cm3. (As

Darina [25.2K]

Answer:

Explanation:

Temperature of air = 18°C = 273 + 18 = 291 K .

volume = 500 cc = 0 .5 litre .

pressure = one atmosphere ( atm) .

From gas equation , we can calculate this volume at NTP as follows.

volume = .5 x ( 273 / 291 ) litre

= 0.469 litre .

In any gas at NTP , 22.4 litre contains 6.02 x 10²³ molecules

.469 litre will contain 6.02 x 10²³ x .469 / 22.4 molecules

= 126 x 10²⁰ molecules .

b )

one mole = 6.02 x 10²³ molecules

6.02 x 10²³ molecules has weight of 28.96 grams

126 x 10²⁰ molecules has weight of 28.96 x 126 x 10²⁰ / 6.02 x 10²³ grams

= .606 gram .

c )

volume of all the air in the atmosphere = volume of sphere

= 4 / 3 x π x R³

= ( 4 / 3) x 3.14 x (999.5 x 10³ )³ m³

= 4.18 x 10¹⁸ m³

density of air = 1.225 kg / m³

mass of air = 1.225 x 4.18 x 10¹⁸ kg

= 5.12 x 10¹⁸ kg

d )

volume of air inhaled by 7 billion people

= . 5 x 7 x 10⁹ litre

= 3.5 x 10⁶ m³ .

Total volume of air in atmosphere = 4.18 x 10¹⁸ m³

required percentage

= 3.5 x 10⁶ x 100 / 4.18 x 10¹⁸

= .8373 x 10⁻¹⁰ % .

6 0

3 years ago

If the weight of a submerged object is equal to the buoyant force, what net force acts upon the object?

Y_Kistochka [10]

Answer:

A

Explanation:

The weight is acting downwards where as the buoyant force acting upwards (opposite) direction with equal amount of force. so the opposite forces cancel out each other (because of the force amount being equal) and no net force is acting on the object.

Hope i have helped you

Thanks.

7 0

3 years ago

The aqueduct passes under Johnson Road in Lancaster through a siphon. The maximum capacity of the aqueduct is 350 m3/s. The heig

Mariulka [41]

Answer:

D ≈ 8.45 m

L ≈ 100.02 m

Explanation:

Given

Q = 350 m³/s (volumetric water flow rate passing through the stretch of channel, maximum capacity of the aqueduct)

y₁ - y₂ = h = 2.00 m (the height difference from the upper to the lower channels)

x = 100.00 m (distance between the upper and the lower channels)

We assume that:

the upper and the lower channels are at the same pressure (the atmospheric pressure).
the velocity of water in the upper channel is zero (v₁ = 0 m/s).
y₁ = 2.00 m (height of the upper channel)
y₂ = 0.00 m (height of the lower channel)
g = 9.81 m/s²
ρ = 1000 Kg/m³ (density of water)

We apply Bernoulli's equation as follows between the point 1 (the upper channel) and the point 2 (the lower channel):

P₁ + (ρ*v₁²/2) + ρ*g*y₁ = P₂ + (ρ*v₂²/2) + ρ*g*y₂

Plugging the known values into the equation and simplifying we get

Patm + (1000 Kg/m³*(0 m/s)²/2) + (1000 Kg/m³)*(9.81 m/s²)*(2 m) = Patm + (1000 Kg/m³*v₂²/2) + (1000 Kg/m³)*(9.81 m/s²)*(0 m)

⇒ v₂ = 6.264 m/s

then we apply the formula

Q = v*A ⇒ A = Q/v ⇒ A = Q/v₂

⇒ A = (350 m³/s)/(6.264 m/s)

⇒ A = 55.873 m²

then, we get the diameter of the pipe as follows

A = π*D²/4 ⇒ D = 2*√(A/π)

⇒ D = 2*√(55.873 m²/π)

⇒ D = 8.434 m ≈ 8.45 m

Now, the length of the pipe can be obtained as follows

L² = x² + h²

⇒ L² = (100.00 m)² + (2.00 m)²

⇒ L ≈ 100.02 m

6 0

3 years ago

Read 2 more answers

What is the magnitude of electrical force of attraction between an copper nucleus (29 protons) and its innermost electron if the

Agata [3.3K]

The charge of the copper nucleus is 29 times the charge of one proton:
$Q=29 q= 29 \cdot 1.6 \cdot 10^{-19}C=4.64 \cdot 10^{-18}C$
the charge of the electron is
$e=-1.6 \cdot 10^{-19}C$
and their separation is
$r=1.0 \cdot 10^{-12} m$

The magnitude of the electrostatic force between them is given by:
$F=k_e \frac{Qe}{r^2}$
where $k_e$ is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)
$F=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(4.64 \cdot 10^{-18}C)(1.6 \cdot 10^{-19}C)}{(1.0 \cdot 10^{-12} m)^2}=6.68 \cdot 10^{-3} N$

3 0

3 years ago

An object in a certain direction with an acceleration in the perpendicular direction

geniusboy [140]

Answer:

An object moving in certain direction with an acceleration in the perpendicular direction. The above condition is possible . Example of such situation in life would be when stone tied to a string whirling in a circular path

Hope this helps and pls mark as BRAINLIEST :)

3 0

3 years ago