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fredd [130]
3 years ago
8

As the result of a thermal inversion the prevailing air temperature profile increases 1°C/100m above the ground level. To what m

aximum height will polluted air rise if it is released at ground level 15°C warmer than the atmosphere? The adiabatic lapse rate is −0.65°C/100m.
Physics
1 answer:
astra-53 [7]3 years ago
5 0

Answer:

909.1 m

Explanation:

Rate of temperature increase with 100 m elevation = 1°C

h = Maximum Height

Adiabatic lapse rate = -0.65°C/100 m

We have the relation

\dfrac{1^{\circ}C}{100}=\dfrac{15^{\circ}C}{h}+\left(-\frac{0.65}{100}\right)\\\Rightarrow \dfrac{1^{\circ}C}{100}+\left(-\frac{0.65}{100}\right)=\dfrac{15^{\circ}C}{h}\\\Rightarrow h=\dfrac{15^{\circ}C}{1.65^{\circ}C}\times 100\\\Rightarrow h=909.1\ m

The maximum height is 909.1 m

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Two light bulbs are 2.0 m apart. From what distance can these light bulbs be marginally resolved by a small telescope with a 4.5
andrezito [222]

Answer:

R = 1.2295 10⁵  m

Explanation:

After reading your problem they give us the diameter of the lens d = 4.50 cm = 0.0450 m, therefore if we use the Rayleigh criterion for the resolution in the diffraction phenomenon, we have that the minimum separation occurs in the first minimum of diffraction of one of the bodies m = 1 coincides with the central maximum of the other body

            θ = 1.22 λ / D

where the constant 1.22 leaves the resolution in polar coordinates and D is the lens aperture

             

how angles are measured in radians

          θ = y / R

where y is the separation of the two bodies (bulbs) y = 2 m and R the distance from the bulbs to the lens

            \frac{y}{R} = 1.22 \frac{ \lambda}{D}

            R = \frac{ y \ D}{1.22 \lambda}

let's calculate

            R = \frac{ 2 \ 0.045}{ 1.22 \ 600 \ 10^{-9}}

            R = 1.2295 10⁵  m

3 0
3 years ago
The tires on your truck have 0.35 m radius. In a straight line, you drive 2600 m. What is the angular displacement of the tire,
Travka [436]
1) In a circular motion, the angular displacement \theta is given by
\theta =  \frac{S}{r}
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire: S=2600 m. Using the information about the radius, r=0.35 m, we find the total angular displacement:
\theta =  \frac{2600 m}{0.35 m} =7428 rad

2) If we put larger tires, with radius r=0.60 m, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
\theta =  \frac{2600 m}{0.60 m}=4333 rad
8 0
4 years ago
In a weak acid solution, _____.
erastovalidia [21]

Answer:

A)

Explanation:

8 0
3 years ago
A 0.45kg baseball is pitched towards home plate at 20 m/s. The ball is hit back towards the pitcher with a speed of 30 m/s. What
Inessa05 [86]

Answer:

4.5kgm/s

Explanation:

Change in momentum is expressed as

Change in momentum = m(v-u)

M is the mass

V is the final velocity

u is the initial velocity

Given

m=0.45kg

v = 30m/s

u = 20m/s

Substitute

Change in momentum = 0.45(30-20)

Change in momentum = 0.45×10

Change in momentum = 4.5kgm/s

3 0
3 years ago
A person wants to make a metronome for music practice. He uses a 35-g object attached to a spring to serve as the time standard.
alukav5142 [94]

To develop this problem it will be necessary to apply the concepts related to the frequency of a spring mass system, for which it is necessary that its mathematical function is described as

f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

Here,

k = Spring constant

m = Mass

Our values are given as,

m = 35g = 35*10^{-3}kg

f = 1 Hz

Rearranging to find the spring constant we have that,

k = (2\pi f \sqrt{m})^2

k = 4\pi^2 f^2 m

k = (4) (\pi)^2 (1) (35*10^{-3})

k = 1.38N/m

Therefore the spring constant is 1.38N/m

7 0
4 years ago
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