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3 years ago

What are cyclones? How are they formed? Explain the structure of a cyclone.

Physics

1 answer:

mixer [17]3 years ago

7 0

Answer:

To form a cyclone, warm, moist air over the ocean rises upward from near the surface. As this air moves up and away from the ocean surface, it leaves is less air near the surface. So basically as the warm air rises, it causes an area of lower air pressure below.

The main parts of a tropical cyclone are the rainbands, the eye, and the eyewall. Air spirals in toward the center in a counter-clockwise pattern in the northern hemisphere (clockwise in the southern hemisphere), and out the top in the opposite direction.

Hope this helps, have a great day/night, and stay safe!

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The rate of spin of a disk is decreasing. A point on the disk that is near the rim of the disk will have...

NNADVOKAT [17]

At a point near the rim of the disk, it will have a non-zero radial acceleration and a zero tangential acceleration. Also known as centripetal acceleration, radial acceleration takes place along the radius of the disk. On the other hand, the tangential acceleration is along the path of disk's motion.

8 0

3 years ago

A science class puts a balloon containing 1.25 l of air at 101 kpa into a bell jar. using an air pump, the class removes some of

storchak [24]

The first thing you should know in this case is the following definition:
PV = nRT
Then, as the temperature is constant, then:
PV = k
Then, we have two states:
P1V1 = k
P2V2 = k
We can then equalize both equations:
P1V1 = P2V2
Substituting the values:
(1.25) * (101) = (2.25) * (P2)
Clearing P2:
P2 = ((1.25) * (101)) /(2.25)=56.11Kpa
answer:
the new pressure inside the jar is 56.11Kpa

8 0

4 years ago

A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found

kati45 [8]

Answer:

1) $\rm q_2$ is positive.

2) $\rm q_2=4.56\times 10^{-10}\ C.$

Explanation:

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., $\rm q_2$ is positive.

Given:

Mass of the balloon, m = 0.00275 kg.
Charge on the balloon, $\rm q_1 = -3.50\times 10^{-8}\ C.$
Distance between the rod and the balloon, d = 0.0640 m.
Acceleration due to gravity, $\rm g = 9.81\ m/s^2.$

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, $\rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.$

The magnitude of the electrostatic force on the balloon due to the rod is given by

$\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.$

$\rm \dfrac{1}{4\pi \epsilon_o}$ is the Coulomb's constant.

For the elecric force and the weight to be balanced,

$\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.$

3 0

3 years ago

(m/s) 20 10. 100 -10 total distance

Sergio039 [100]

Answer:

sum of them 4

Explanation:

8 0

3 years ago

A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.

JulsSmile [24]

Answer:

$\mu =0.75$

Explanation:

Frictional Force

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

$F_c=m.a_c$

The centripetal acceleration a_c is computed as

$\displaystyle a_c=\frac{v^2}{r}$

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

$F_c=m.\frac{v^2}{r}$

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

$F_r=\mu N$