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2 years ago

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

Physics

2 answers:

vovikov84 [41]2 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]2 years ago

7 0

Answer:

a decrease

b increase

c increase

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Answer:

Explanation:

We know the force is (16N-12), which is 4N, and we know the acceleration is 2 m/s^2. Meaning, we can solve the formula m = F / a (mass equals force divided by acceleration), and we get 2kg.

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2 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

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2 years ago

How rolling friction is less than sliding friction?

vovangra [49]

Rolling friction is considerably less than sliding friction as there is no work done against the body that is rolling by the force of friction. For a body to start rolling a small amount of friction is required at the point where it rests on the other surface, else it would slide instead of roll.

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2 years ago

A 50-ohm resistor in a 0.5 Amp circuit produces how much voltage drop?

ahrayia [7]

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Use Ohm's law to find the potential drop:

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2 years ago

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