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3 years ago

Which of these processes in the water cycle occur at a very high rate, to cause a hurricane?

1 answer:

vladimir1956 [14]3 years ago

7 0

The correct answer is evaporation and condensation

When these occur rapidly hurricanes are made because of a lot of water vapor entering the atmosphere and creating cyclonic behavior. Precipitation happens because of this, but it is not a cause of it.

You might be interested in

Why do raindrops fall with a constant speed during the later stages of their descents?

BartSMP [9]

Firstly they have a acceleration downwards due the force downwards due they gravitational field acting on it's mass.
as it falls it gains speed, and as it gains speed the air Resistance which is a upward force actin on the drop increases, eventually the rain drop's upward and downward forces are balanced and hence there is no RESULTANT force therefore no acceleration, so the drops falls in constant speed (terminal verlocity is a better term)

Are you wondering that why is the raindrop still moving given that the forces are balanced? If so according to Newton's 1st law an object will keep moving or Remain at rest until a RESULTANT force acts on it.

6 0

3 years ago

You look down an old well, cannot see the bottom, and mutter to yourself "Oh well!". In order to estimate the depth of the well,

telo118 [61]

Answer:

The best estimate of the depth of the well is 2.3 sec.

Explanation:

Given that,

Record time,

$t_{1}=2.19\ sec$

$t_{2}=2.30\ sec$

$t_{3}=2.26\ sec$

$t_{4}=2.29\ sec$

$t_{5}=2.27\ sec$

We need to find the best estimate of the depth of the well

According to record time,

We can write of the record time

$t_{1}=2.19\approx 2.2\ sec$

$t_{2}=2.30\approx 2.3\ sec$

$t_{3}=2.26\approx 2.3\ sec$

$t_{4}=2.29\approx 2.3\ sec$

$t_{5}=2.27\approx 2.3\ sec$

Here, all time is nearest 2.3 sec.

So, we can say that the best estimate of the depth of the well is 2.3 sec.

Hence, The best estimate of the depth of the well is 2.3 sec.

6 0

3 years ago

B. Jerome plays middle linebacker for South's varsity football team. In a game against

weqwewe [10]

Answer:

Option D

670 Kg.m/s

Explanation:

Initial momentum is given by mv=82*5.6=459.2 Kg.m/s (taking eastward as positive)

Final momentum is also mv but v being westward direction, we take it negative

Final momentum=82*-2.5= -205 Kg.m/s

Change in momentum=Final momentum-Initial momentum=-205-459.2=-664.2 Kg.m/s

Impulse=change in momentum=664.2 Kg.m/s rounded off as 670 Kg.m/s

4 0

3 years ago

The half-life of caffeine is 5 hours. If you ingested a 30 oz Big Gulp, how many oz of caffeine is left after one half life? * Y

xeze [42]

Answer:

The amount of caffeine left after one half life of 5 hours is 15 oz.

Explanation:

Half life is the time taken for a radioactive substance to degenerate or decay to half of its original size.

The half life of caffeine is 5 hours. So ingesting a 30 oz, this would be reduced to half of its size after the first 5 hours.

So that:

After one half life of 5 hours, the value of caffeine that would be left is;

$\frac{30}{2}$ = 15 oz

The amount of caffeine left after one half life of 5 hours is 15 oz.

8 0

4 years ago

Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur

BARSIC [14]

Answer:

$v_A=7667m/s\\\\v_B=7487m/s$

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

$F_g=\frac{GMm}{R^{2} }$

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

$F_c=m\frac{v^{2}}{R}$

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

$\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}$

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So $R_A=6774km=6.774*10^6m$ and $R_B=7103km=7.103*10^6m$ (Since $R_{earth}=6371km$ ). Then, we get:

$v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s$

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0

3 years ago