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3 years ago

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Occurs when an air mass and its clouds encounter a mountain. This forces the air mass to move from a low elevation to a high ele

vation as it crosses over the mountain.
Frontal wedging
Orographic lifting
Localized convective lifting
Convergence
Jet streams

1 answer:

RUDIKE [14]3 years ago

4 0

Answer:

Localized convective lifting

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An explanation can be consistent but may not be the best explanation for a phenomenon

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Answer:

true

Explanation:

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A dog pulls on a pillow with a force of 5 N at an angle of 37o to the horizontal. Find the x and y components of this force.

Paha777 [63]

Answer:

$x=3.99\ N\approx4\ N$

$y=3\ N$

Explanation:

Given :

Force with which dog pulls the pillow = 5 N

Angle of force to the horizontal = 37°

To find the $x$ and $y$ components of force.

Solution:

For a force $F$ making an angle $\theta$ with the horizontal, the

1) $x$ component or the horizontal component is given by:

$x=F\cos\theta$

2) $y$ component or the vertical component is given by:

$y=F\sin\theta$

Plugging the know values to find $x$ and $y$ components of force applied by the dog on pillow

1) $x=5\cos37\°$

∴ $x=3.99\ N\approx4\ N$ (Answer)

2) $y=5\sin37\°$

∴ $y=3\ N$ (Answer)

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A rock hanging from a string has a net force of zero acting on it ? true or false ?

sattari [20]

Answer:

true cuase it is true its not false

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3 years ago

A person on a road trip drives a car at different constant speeds over several legs of the trip. She drives for 10.0 min at 50.0

irakobra [83]

<h2>The average speed for the entire trip is 47.5 m/s .</h2>

It is given that for different time span car have different speed and also the person spend $40\ min=\dfrac{40}{60}\ hrs =0.67\ hrs\ .$ in eating lunch and buying gas.

We know , average speed is total distance covered by total time taken .

Therefore , average speed , $v=\dfrac{total\ distance }{total\ times}$

$v=\dfrac{\dfrac{10}{60}\times 50+\dfrac{19}{60}\times 100+\dfrac{60}{60}\times 55}{\dfrac{10}{60}+\dfrac{10}{60}+\dfrac{60}{60}+ \dfrac{40}{60}}\\\\\\v=47.5\ m/s$

Hence, this is the required solution.

Learn More :

Average speed

https://brainly.in/question/12701198

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3 years ago

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler

ki77a [65]

Answer:

a) $t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

b) $\theta_1=\frac{w_1^2}{2\alpha}rad$

c) $t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

Explanation:

1) Basic concepts

Angular displacement is defined as the angle changed by an object. The units are rad/s.

Angular velocity is defined as the rate of change of angular displacement respect to the change of time, given by this formula:

$w=\frac{\Delat \theta}{\Delta t}$

Angular acceleration is the rate of change of the angular velocity respect to the time

$\alpha=\frac{dw}{dt}$

2) Part a

We can define some notation

$w_o=0\frac{rad}{s}$ ,represent the initial angular velocity of the wheel

$w_1=?\frac{rad}{s}$ , represent the final angular velocity of the wheel

$\alpha$ , represent the angular acceleration of the flywheel

$t_1$ time taken in order to reach the final angular velocity

So we can apply this formula from kinematics:

$w_1=w_o +\alpha t_1$

And solving for t1 we got:

$t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

3) Part b

We can use other formula from kinematics in order to find the angular displacement, on this case the following:

$\Delta \theta=wt+\frac{1}{2}\alpha t^2$

Replacing the values for our case we got:

$\Delta \theta=w_o t+\frac{1}{2}\alpha t_1^2$

And we can replace $t_1$ from the result for part a, like this:

$\theta_1-\theta_o=w_o t+\frac{1}{2}\alpha (\frac{w_1}{\alpha})^2$

Since $\theta_o=0$ and $w_o=0$ then we have:

$\theta_1=\frac{1}{2}\alpha \frac{w_1^2}{\alpha^2}$

And simplifying:

$\theta_1=\frac{w_1^2}{2\alpha}rad$

4) Part c

For this case we can assume that the angular acceleration in order to stop applied on the wheel is $\alpha_1 =-5\alpha \frac{rad}{s}$

We have an initial angular velocity $w_1$ , and since at the end stops we have that $w_2 =0$

Assuming that $t_2$ represent the time in order to stop the wheel, we cna use the following formula

$w_2 =w_1 +\alpha_1 t_2$

Since $w_2=0$ if we solve for $t_2$ we got

$t_2=\frac{0-w_1}{\alpha_1}=\frac{-w_1}{-5\alpha}$

And from part a) we can see that $w_1=\alpha t_1$ , and replacing into the last equation we got:

$t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

5 0

3 years ago