During which process do cells use oxygen to release stored energy?

A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 - L cans per min

jekas [21]

Answer:

a) 1.301 kg/s

b) 0.001301 m³/s

c) V₁ = 6.505 m/s, V₂ = 1.626 m/s

d) 118.93 kPa

Explanation:

Given:

The number of cans = 220

The volume of can, V = 0.355 L = 0.355 × 10⁻³ m³

time = 1 minute = 60 seconds

gauge pressure at point 2, P₂ = 152 kPa

b) Thus, the volume flow rate, Q = Volume/ time

Q = (220 × 0.355 × 10⁻³)/60 = 0.001301 m³/s

a) mass flow rate = Volume flow rate × density

since it is mostly water, thus density of the drink = 1000 kg/m³

thus,

mass flow rate = 0.001301 m³/s × 1000 kg/m³ = 1.301 kg/s

c) Given:

Cross section at point 1 = 2.0 cm² = 2 × 10 ⁻⁴ m²

Cross section at point 2 = 8.0 cm² = 8 × 10 ⁻⁴ m²

also,

Q = Area × Velocity

thus, for point 1

0.001301 m³/s = 2 × 10 ⁻⁴ m² × velocity at point 1 (V₁)

or

V₁ = 6.505 m/s

for point 2

0.001301 m³/s = 8 × 10 ⁻⁴ m² × velocity at point 1 (V₂)

or

V₂ = 1.626 m/s

d) Applying the Bernoulli's theorem between the points 1 and 2 we have

$P_1+\rho gV_1 + \frac{\rho V_1^2}{2}=P_2+\rho gV_2 + \frac{\rho V_2^2}{2}$

or

$P_1=P_2+\rho\timesg(y_2-y_1)+\frac{\rho}{2}(V_2^2-V_1^2))$

on substituting the values in the above equation, we get

$P_1=152+1000\times 9.8(1.35)+\frac{1000}{2}(1.626^2-6.505^2))$

it is given that point 1 is above point 2 thus, y₂ -y₁ is negative

or

$P_1=118.93\ kPa$

thus, gauge pressure at point 1 is 118.93 kPa

8 0

3 years ago

If the temperature of water warming on a hot plate rises 10 degrees Celsius [°C], what is the change in temperature in units of

Luba_88 [7]

Answer:

Delta_temp = 18[°F]

Explanation:

°F = 9/5*(10)

°F = 18

Note: It is important to clarify that it is only a temperature increase, that it is only a temperature increase. The question is not related to converting from 10°C to fahrenheit degrees

4 0

3 years ago

Find the acceleration produced.

krok68 [10]

Answer:

0.67m/s²

Explanation:

Given parameters:

Mass of toy = 1.2kg

Force applied = 0.8N

Unknown:

Acceleration = ?

Solution:

According to newton's second law of motion;

Force = mass x acceleration

Now,

Acceleration = $\frac{Force}{mass}$

Acceleration = $\frac{0.8}{1.2}$ = 0.67m/s²

4 0

3 years ago

Read 2 more answers

An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec

lana66690 [7]

Answer:

91.87 m/s

Explanation:

<u>Given:</u>

x = initial distance of the electron from the proton = 6 cm = 0.06 m

y = initial distance of the electron from the proton = 3 cm = 0.03 m
u = initial velocity of the electron = 0 m/s

<u>Assume:</u>

m = mass of an electron = $9.1\times 10^{-31}\ kg$
v = final velocity of the electron
e = magnitude of charge on an electron = $1.6\times 10^{-19}\ C$

p = magnitude of charge on a proton = $1.6\times 10^{-19}\ C$

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

$\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\$

$\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\$

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

8 0

3 years ago

Rock formed from harden magma that rises to the surface is called

salantis [7]

Answer:

Igneous rocks

Explanation:

3 0

3 years ago