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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

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dem82 [27]

Answer:

Frequency, f=2.30\times 10^{18}\ Hz

Explanation:

Given that,

The wavelength of the x-rays, \lambda=0.13\ nm=0.13\times 10^{-9}\ m

We need to find the frequency of an x-ray. All electromagnetic wave travel with a speed of light. It is given by the formula as :

c=f\lambda

f is the frequency

f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{0.13\times 10^{-9}}\\\\f=2.30\times 10^{18}\ Hz

So, the frequency of an x-ray is 2.30\times 10^{18}\ Hz. Hence, this is the required solution.

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3 years ago
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Answer:

egocentric

Explanation:

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3) A tolley of mass 4kg, moving with a velocity
Leokris [45]

Answer:

3.57 m/s

Explanation:

The sum of the 2 momentums Is equal the finale momentums. so if momentums Is q, v Is velocity and m Is Mass, q3=m1*v1+m2**v2=16+9=25 m*kg/s

q3=m3*v3

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3 years ago
A light bulb dissipates 100 Watts of power when it is supplied a voltage of 220 volts.
ycow [4]

Given Information:

Power = P = 100 Watts

Voltage = V = 220 Volts

Required Information:

a) Current = I = ?

b) Resistance = R = ?

Answer:

a) Current = I = 0.4545 A

b) Resistance = R = 484 Ω

Explanation:

According to the Ohm’s law, the power dissipated in the light bulb is given by

P = VI

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.

Re-arranging the above equation for current I yields,

I = \frac{P}{V}  \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\

Therefore, 0.4545 A current is flowing through the light bulb.

According to the Ohm’s law, the voltage across the light bulb is given by

V = IR

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.

Re-arranging the above equation for resistance R yields,

R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega

Therefore, the resistance of the bulb is 484 Ω

3 0
3 years ago
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Write a collision scenario here. If you choose your own collision, you can have neither, one, or both of the objects break. Be s
OleMash [197]

Answer:

My scenario would be A Car vs. a guard rail on a road.  You have a car that is coming down a Highway at a speed of 43 Mph Miles per hour (69.2018 Kmh)

And it hits a steel guardrail and the car smashes in at the front and the guardrail is only bent while the car has the bumper and the hood along with the headlights and windshield along with the passenger side window break.

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