In an electrochemical cell, the anode is ___

A 4.2 kg parachutist is moving straight downward with a speed of 3.85 m/s

statuscvo [17]

Kinetic energy= .5 x m x v^2
KE=.5 x 4.2 x 3.85^2
KE=31.13

4 0

3 years ago

Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is

saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

$\dot{m_{i}} = \rho_{i} v_{i}A_{i}$

Where:

$v_{i}$ : is the initial velocity = 20 m/s

$A_{i}$ : is the inlet area of the nozzle = 60 cm²

$\rho_{i}$ : is the density of entrance = 2.21 kg/m³

$\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s$

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

$\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}$

$0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}$

$A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}$

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!

5 0

3 years ago

Read 2 more answers

It has been argued that power plants should make use of off-peak hours to generate mechanical energy and store it until it is ne

sdas [7]

Answer:

Explanation:

90 rpm = 90 / 60 rps

= 1.5 rps

= 1.5 x 2π rad /s

angular velocity of flywheel

ω = 3π rad /s

Let I be the moment of inertia of flywheel

kinetic energy = (1/2) I ω²

(1/2) I ω² = 10⁷ J

I = 2 x 10⁷ / ω²

=2 x 10⁷ / (3π)²

= 2.2538 x 10⁵ kg m²

Let radius of wheel be R

I = 1/2 M R² , M is mass of flywheel

= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .

1/2 πR⁴ x t x d = 2.2538 x 10⁵

R⁴ = 2 x 2.2538 x 10⁵ / πt d

= 4.5076 x 10⁵ / 3.14 x .1 x 7800

= 184

R= 3.683 m .

diameter = 7.366 m .

b ) centripetal accn required

= ω² R

= 9π² x 3.683

= 326.816 m /s²

3 0

3 years ago

Calculate the Latent Heat of Vaporization. (Please see picture attached)

Hunter-Best [27]

Answer:

20 J/g

Explanation:

In this question, we are required to determine the latent heat of vaporization

To answer the question, we need to ask ourselves the questions:

What is latent heat of vaporization?

It is the amount of heat required to change a substance from its liquid state to gaseous state without change in temperature.

It is the amount of heat absorbed by a substance as it boils.

How do we calculate the latent heat of vaporization?

Latent heat is calculated by dividing the amount of heat absorbed by the mass of the substance.

In this case;

Mass of the substance = 20 g
Heat absorbed as the substance boils is 400 J (1000 J - 600 J)

Thus,

Latent heat of vaporization = Quantity of Heat ÷ Mass

= 400 Joules ÷ 20 g

= 20 J/g

Thus, the latent heat of vaporization is 20 J/g

7 0

3 years ago

What are the milestones of modern phyiscs?

nlexa [21]

Answer:

The articles appearing under "Milestones in Physics" will give an insight into special events or situations that have been decisive for the evolution of Physics

6 0

2 years ago

Read 2 more answers

In an electrochemical cell, the anode is ____.