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o-na [289]
3 years ago
7

How do you calculate the total pressure?

Physics
1 answer:
neonofarm [45]3 years ago
4 0

Answer:

You could put a pressure stick against the pressure and see the pressure or estimate it from the power its coming out.

Explanation:

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The rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3 × 10-11 e-250/T and 2
Vlada [557]

Answer:

Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x10^{-11} e^{-255/T}  and 2x10^{-12} e^{-940/T}  

T = -50 °C = 223 K

The reaction rate will be given by [Cl] [O3] 3x10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3]  

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x10^{-12} e^{-940/223} = 2.95^{-14} [OH] [O3]

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 330 * [Cl] / [OH]

Than, the concentration of OH is approximately 100 times of Cl, and the result will be that the reaction with Cl is 3.3 times faster than the  reaction with OH

Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80 °C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 × 105 molecules cm-3

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x10^{-11} e^{-255/T}  and 2x10^{-12} e^{-940/T}  

T = -80 °C = 193 K

The reaction rate will be given by [Cl] [O3] 3x10^{-11} e^{-255/193} = 8.21^{-12} [Cl] [O3]  

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x10^{-12} e^{-940/193} = 1.53^{-14} [OH] [O3]

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 535 * [Cl] / [OH]

Than, considering the concentration of Cl increases by a factor of 100 to about 4 × 10^{5} molecules cm^{-3}, the result will be that the reaction with OH will be 535 + (100 to about 4 × 10^{5} molecules cm^{-3}) times faster than the  reaction with Cl

Explanation:

4 0
3 years ago
Which of the following can be contracted from contact with bloodborne pathogens?
sergij07 [2.7K]

HIV can be contracted from contact with bloodborne pathogens.

Other bloodborne diseases are HBV, malaria, syphilis and brucellosis

<h3>What are bloodborne pathogens?</h3>

Bloodborne pathogens can be defined as those microorganisms or pathogenic organisms that cause disease and are present in human blood.

Blood borne pathogens can also be contacted through the following means

  • Se- xual contact
  • Needle contact

In conclusion; HIV can be contracted from contact with bloodborne pathogens.

Learn more about bloodborne pathogens:

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6 0
2 years ago
A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The bo
WITCHER [35]

Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

acceleration=\frac{30}{5}=6m/s^{2}

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.

5 0
4 years ago
Velocity is defined as:
gogolik [260]

Velocity is defined as a change in position.

6 0
3 years ago
Question 2 (10 points)
kenny6666 [7]

The force required to slow the truck was -5020 N

Explanation:

First of all, we find the acceleration of the truck, which is given by

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time

For the truck in this problem,

v = 11.5 m/s

u = 21.9 m/s

t = 2.88 s

So the acceleration is

a=\frac{11.5-21.9}{2.88}=-3.6 m/s^2

where the negative sign means that this is a deceleration.

Now we can find the force exerted on the truck, which is given by Newton's second law:

F=ma

where

m = 1390 kg is the mass of the truck

a=-3.6 m/s^2 is the acceleration

And substituting,

F=(1390)(-3.6)=-5004 N

So the closest answer among the option is -5020 N.

Learn more about acceleration and forces:

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8 0
3 years ago
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