Read the following selections from the section "Mickey Guyton's Star-Spangled Banner."

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

vovikov84 [41]

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

8 0

2 years ago

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two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y

svet-max [94.6K]

Answer:

Explanation:

Given

initial velocity component of engines is

$v_0_x=6380 m/s$

$v_0_y=6770 m/s$

time period of engine running=763 s

Displacement in $x=4.50\times 10^6$

$y=7.27\times 10^6$

Using $s=ut+\frac{at^2}{2}$ in x and y direction

$x=v_0_x\times t+\frac{at^2}{2}$

$4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}$

$4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}$

$a=-1.23 m/s^2$

In y direction

$y=v_0_y\times t+\frac{a't^2}{2}$

$7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}$

$7.27\times 10^6-5.16\times 10^6=\frac{a\times 763^2}{2}$

$a=7.24 m/s^2$

x component $=-1.23 m/s^2$

y component $=7.24 m/s^2$

3 0

2 years ago

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A 50g man is walking at a speed of 4m/s. what is the kinetic energy of the walker?

Phantasy [73]

We have: K.E. = 1/2 mv²
Here: m = 50 g = 0.05 Kg
v = 4 m/s

Substitute their values,
K.E. = 1/2 * 0.05 * 4²
K.E. = 1/2 * 0.05 * 16
K.E. = 0.4 J

In short, Your Answer would be 0.4 Joules

Hope this helps!

4 0

3 years ago

How many grams of ice at 0°c will melt if 1.50 kj of heat are added? (the molar heat of fusion of water is 6.01 kj/mol.) 4.49 g

Nata [24]

The computation would be:moles = mass/ Molar Mass, but we are looking for the mass, so rearranging, will give us: mass = moles x MM
Q = moles x Hf
Q = (mass/MM) x Hf
mass = (Q x MM) / Hf
= (1.50-kJ x 18.0-g/mol) / 6.01-kJ/mol
=4.49 g H20 is the answer

7 0

3 years ago

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There is danger of a vehicle skidding on a wet road.

matrenka [14]

Assertion(A):'

There is danger of a vehicle skidding on a wet road.

<h3>Reason (R) : </h3>

The tyres of the vehicle lose their grip on the road due to increase in friction due to presence of water on the road.

Answer:-

Assertion is true
Reason is true
Reason is the perfect explanation of Assertion

6 0

2 years ago