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2 years ago

The average rate of disappearance of ozone in the reaction is

Physics

1 answer:

olasank [31]2 years ago

7 0

Answer:

1.3 x 10^(-2) atm/s

Explanation:

It follows the stoichiometry. For every mole of O3 that disappears, 1.5 moles (that is, 3/2) of O2 appears:

1.5 * 0.009 atm/s = 0.0135 atm/sec; the answer is 1.3 x 10^(-2) atm/s

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The drawing shows a plot of the output emf of a generator as a function of time t. The coil of this device has a cross-sectional

adell [148]

This question is incomplete, the missing image uploaded along this answer below;

Answer:

a) frequency is 2.381 Hz

b) the angular frequency/speed is 14.96 rad/s

c) the magnitude of the magnetic field is 0.5199 T

Explanation:

Given that;

cross sectional Area A = 0.018 m²

Number of turns N = 200

from the diagram. maximum time T = 0.42 sec

a) the frequency f of the generator in hertz

frequency = 1 / T

we substitute

frequency = 1 / 0.42 = 2.381 Hz

Therefore, frequency is 2.381 Hz

b) the angular frequency in rad/s

angular speed ω = 2πf

we substitute

ω = 2π × 2.381

ω = 14.96 rad/s

Therefore, the angular frequency/speed is 14.96 rad/s

c) the magnitude of the magnetic field.

to determine the magnitude of the magnetic field, we use the following expression;

e = NBAω

from the diagram, e = 28.0 V

so we substitute

28.0 V = 200 × B × 0.018 × 14.96

28 = 53.856B

B = 28 / 53.856

B = 0.5199 T

Therefore, the magnitude of the magnetic field is 0.5199 T

6 0

3 years ago

When dribbling you should?

Hoochie [10]

D all of the above (:

6 0

2 years ago

A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.

Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

Under the assumption that the can is a closed system, the conservation law applied to the system would be: $E_{in}-E_{out}=E_{change}$ , where $E_{in}$ is all energy entering the system, $E_{out}$ is the total energy leaving the system and, $E_{change}$ is the change of energy of the system.
As the purpose is to kept the beverage can at constant temperature, the change of energy ( $E_{change}$ ) would be 0.
The energy that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)$ where $\varepsilon$ is the emissivity of the surface, $\sigma=5.67*10^{-8}\frac{W}{m^2K}$ known as the Stefan–Boltzmann constant, $A_S$ is the total area of the exposed surface, $T_S$ is the temperature of the surface in Kelvin, $T_{\infty}$ is the environment temperature in Kelvin.
For the can the surface area would be ta sum of the top and the sides. The area of the top would be $A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2$ , the area of the sides would be $A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2$ . Then the total area would be $A_{total}=A_{top}+A_{sides}=0.01624m^2$
Then the radiation heat transferred to the can would be $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W$ .
The can would lost heat evaporating water, in this case would be $Q_{out}=\frac{dm}{dt}*h_{fg}$ , where $\frac{dm}{dt}$ is the rate of mass of water evaporated and, $h_{fg}$ is the heat of vaporization of the water ( $2257\frac{J}{g}$ ).
Then in the conservation balance: $Q_{in}-Q_{out}=Q_{change}$ , it would be $1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0$ .
Recall that $1W=1\frac{J}{s}$ , then solving for $\frac{dm}{dt}$ : $\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}$