1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
OlgaM077 [116]
3 years ago
13

A. If we increase the wind velocity, the maximum vertical dispersal height and rate of diffusion will decrease____.

Physics
2 answers:
vovikov84 [41]3 years ago
8 0

Answer:

a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase

b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.

c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase

Explanation:

a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase

b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.

c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase

Archy [21]3 years ago
7 0

Answer:

a decrease

b increase

c increase

You might be interested in
What is true about solar energy hitting earth?
polet [3.4K]
B. It’s the same roughly at all latitudes
6 0
2 years ago
A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at
Vaselesa [24]

Answer:24.70 ^{\circ}C

Explanation:

Given

mass of lead piece m_l=234 gm\approx 0.234 kg

mass of water in calorimeter m_w=611 gm\approx 0.611 kg

Initial temperature of water T_w=24^{\circ}C

Initial temperature of lead piece T_l=24^{\circ}C

we know heat capacity of lead and water are 125.604 J/kg-k and 4.184 kJ/kg-k respectively

Let us take T ^{\circ}C be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

m_lc_l(T_l-T)=m_wc_w(T-T_w)

0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)

86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)

86-T=86.97T-2087.49

T=\frac{2173.491}{87.97}=24.70^{\circ}C

3 0
3 years ago
If you are an astronaut in the middle of the near side of the moon during a full moon,how would the ground around you look?How w
Radda [10]
The moon would be bright and the earth would be darker because the sun is on the opposite side of the earth at that time and the light from it is reflecting off the moon to produce light upon the nigh also.......

You wouldn’t see the sun a night...

Unless you lived in the north/south pole
3 0
3 years ago
Read 2 more answers
I need an answer to question number 4 ASAP
Nat2105 [25]

Answer:

Screwing

Explanation:

3 0
3 years ago
A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioact
ArbitrLikvidat [17]

Answer:

  • 0.09 % of the original radioactive nucllde its left after 10 half-lives
  • It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

Explanation:

The equation for radioactive decay its:

N ( t) \ = \ N_0 \ e^{ \ -  \frac{t}{\tau}},

where N(t) its quantity of material at time t, N_0 its the initial quantity of material and \tau its the mean lifetime of the radioactive element.

The half-life t_{\frac{1}{2}} its the time at which the quantity of material its the half of the initial value, so, we can find:

N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}

so:

\ N_0 \ e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}

e^{ \ -  \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}

-  \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )

t_{\frac{1}{2}}\ = \tau ln( 2 )

So, after 10 half-lives, we got:

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  t_{\frac{1}{2}}}{\tau}}

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  \frac{10 \  \tau \ ln( 2 ) }{\tau}}

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ -  10 \  \ ln( 2 ) }

N ( 10 \  t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

10 \ * \ 24,110 \ years \ = \ 241,100 \ years

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

7 0
3 years ago
Read 2 more answers
Other questions:
  • In your textbook reading Chapter 26, the author suggests that an electric vehicle (EV) fleet can be used as a kind of distribute
    10·1 answer
  • How did Archimedes measure the mass of the block of gold?
    12·1 answer
  • Pilots can be tested for the stresses of flying high-speed jets in a whirling "human centrifuge," which takes 1.2min to turn thr
    11·2 answers
  • Physics help please ;-; . An astronaut has a mass of 82.0 kg. What is the astronaut's stationary weight at a position 4230 kilom
    5·1 answer
  • A Canadian college student who has taken an astronomy class goes home for the holidays and persuades his parents to let him borr
    6·1 answer
  • Substances which naturally attract each other called what
    15·1 answer
  • A wire 1.0 m long experiences a magnetic force of 0.50 N due to a
    8·1 answer
  • If it is brittle, dissolves easily in water, has a high melting point, and conducts electric current it is a(n) __________.
    6·1 answer
  • 1. Which wave phenomenon is illustrated by this image?
    5·2 answers
  • how long would it take for a resultant upward force of 100N to increase the speed of 50kg object from 100m/s to 150m/s​
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!