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3 years ago

Which process produces rising air when mountains push the air upward?

Physics

2 answers:

Schach [20]3 years ago

7 0

Answer:

The answer is A) orographic lifting

Explanation:

kicyunya [14]3 years ago

5 0

Answer;

A. orographic lifting

Explanation;

Orographic lifting is a process that takes place when an air mass is forced from a low elevation to a higher elevation as it moves over rising terrain.

This can be explained by; When air is blocked by mountains, it cannot go through these mountains, As it ascends or moves up the mountain, the air then cools as it rises and when it cools to its saturation point, the water vapor condenses and cloud forms.

These clouds formed are known as orographic clouds, that develop in response to the lifting forced by the topography the earth.

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In Europe, gasoline efficiency is measured in km/L. If your car's gas mileage is 36.0mi/gal , how many liters of gasoline would

Zigmanuir [339]

$36 \frac{mi}{gal} * \frac{1 km}{0.6214 mi}* \frac{1 gal}{3.78 L} = 15.3 \frac{km}{L}

142/15.3=9.27L$

3 0

3 years ago

A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

4 0

3 years ago

A student releases a marble from the top of a ramp. The marble increases

Fed [463]

Answer:

Vf = 69.56 cm/s

Explanation:

In order to find the final speed of the ramp, we will use the equations of motion. First we use second equation of motion to find out the acceleration of marble:

s = Vi t + (1/2)at²

where,

s = distance traveled = 160 cm

Vi = Initial Speed = 0 cm/s (since, marble starts from rest)

t = time interval = 4.6 s

a = acceleration = ?

Therefore,

160 cm = (0 cm/s)(4.6 s) + (1/2)(a)(4.6 s)²

a = (320 cm)/(4.6 s)²

a = 15.12 cm/s²

Now, we use first equation of motion:

Vf = Vi + at