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3 years ago

What occurs in an updraft

2 answers:

5 0

Updrafts and downdrafts alsooccur as part of the turbulence that is created when air passes over topographic barriers such as mountains. ... Updraftscharacterize a storm's early development, during which warm air rises to the level where condensation begins and precipitation starts to develop.

mezya [45]3 years ago

4 0

Correct answer: Warm air from the south rises.

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Describe the motion of the box.

Andrej [43]

I think its D but I'm not sure :/

6 0

3 years ago

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In a real system of levers, wheels, or pulleys, the AMA is less than the IMA because _____.

grandymaker [24]

In a real system of levers, wheel or pulleys, the AMA (actual mechanical advantage) is less than the IMA (ideal mechanical advantage) because of the presence of friction.

In fact, the IMA and the AMA of a machine are defined as the ratio between the output force (the load) and the input force (the effort):
$IMA= \frac{F_{out}}{F_{in}}$
however, the difference is that the IMA does not take into account the presence of frictions, while the AMA does. As a result, the output force in the AMA is less than the output force in the IMA (because some energy is dissipated due to friction), and the AMA is less than the IMA.

5 0

3 years ago

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Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

4 years ago

Savvas realize hot on the inside

Liula [17]

Answer:

Thats sounds like a club

Explanation:

8 0

3 years ago

Read 2 more answers

A catcher catches a 150g baseball traveling horizontally at 25 m/s. If the ball takes 20.ms to stop once it is in contact with t

rosijanka [135]

Answer:

The magnitude of the force exerted by the ball on the catcher is 1.9 × 10² N

Explanation:

Hi there!

Let´s find the acceleration of the ball that makes it stop when caught by the catcher. The acceleration can be calculated from the equation of velocity considering that it is constant:

v = v0 + a · t

We know that initially the ball was traveling at 25 m/s, so, if we consider the position of the catcher as the origin of the frame of reference, then, v0 = -25 m/s. We also know that it takes the ball 20 ms (0.02 s) to stop (i.e. to reach a velocity of 0). Then using the equation of velocity:

v = v0 + a · t

0 m/s = -25 m/s + a · 0.020 s

25 m/s/ 0.020 s = a

Now, using the second law of Newton, we can calculate the force exerted by the catcher on the ball:

F = m · a

Where:

F = force.

m = mass of the ball.

a = acceleration.

F = 0.150 kg · (25 m/s/ 0.020 s) = 1.9 × 10² N

According to Newton´s third law, the force exerted by the ball on the catcher will be of equal magnitude but opposite direction. Then, the force exerted by the ball on the catcher will have a magnitude of 1.9 × 10² N.

3 0

4 years ago