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3 years ago

Where are a hurricanes fastest winds and heaviest rain found?

Physics

1 answer:

PSYCHO15rus [73]3 years ago

8 0

The fastest winds and heaviest rain is found in the eye of the hurricane.

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An electric bulb marked 40W-200V is used in a circuit of supply voltage 100V. Now its power is:

svp [43]

Answer:

Power of the bulb is 10 Watts.

Explanation:

We have,

Power and voltage of electric bulb is 40 W and 200 V. Using this data we can find the resistance of the bulb.

Power, $P=\dfrac{V^2}{R}$

R is resistance

$R=\dfrac{V^2}{P}\\\\R=\dfrac{(200)^2}{40}\\\\R=1000\ \Omega$

The supplied voltage is 100 V.

Now power is given by :

$P=\dfrac{V^2}{R}\\\\P=\dfrac{100^2}{1000}\\\\P=10\ W$

So, now the power of the bulb is 10 Watts.

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3 years ago

A ball is rolling across the floor. Why does the ball come to a stop?

OLEGan [10]

Explanation:

the notmal focrce stopped it.

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3 years ago

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A 180 cm length of string has a mass of 5.0 g. it is stretched with a tension of 8.6 n between fixed supports. (a) what is the w

Svetlanka [38]

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4 years ago

The positions of the consellations appear to change throughout the year because

slava [35]

The earth is revolving around the sun and when the seasons change the earth tilts at a different angle and it changes your perspective on the constellations<span />

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3 years ago

A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.90 m/s2 . At 20.0s after blastoff, t

Brilliant_brown [7]

Answer:

A) 580m

B) 0 m/s

C) 9.8m/s^2

D) downward

E) 10.87s

F) 106.62 m/s

Explanation:

A) The distance traveled by the rocket is calculated by using the following expression:

$y=\frac{1}{2}at^2$

a: acceleration of the rocket = 2.90 m/s^2

t: time of the flight = 20.0 s

$y=\frac{1}{2}(2.90\frac{m}{s^2})(20.0s)^2=580m$

B) In the highest point the rocket has a velocity with magnitude zero v = 0m/s because there the rocket stops.

C) The engines of the rocket suddenly fails in the highest point. There, the acceleration of the rocket is due to the gravitational force, that is 9.8 m/s^2

D) The acceleration points downward

E) The time the rocket takes to return to the ground is given by:

$t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2(580m)}{9.8m/s^2}}=10.87s$

10.87 seconds

F) The velocity just before the rocket arrives to the ground is:

$v=\sqrt{2gy}=\sqrt{2(9.8m/s ^2)(580m)}=106.62\frac{m}{s}$

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3 years ago