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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.

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How does the resistance change as you add bulbs to a series circuit?

Ronch [10]

The resistance is increased when more and more bulbs are added to the circuit.

7 0

3 years ago

Read 2 more answers

A positive charge of 18nC is evenly distributed along a straight rod of length 4.0 m that is bent into a circular arc with a rad

sergey [27]

Explanation:

Formula for angle subtended at the center of the circular arc is as follows.

$\theta = \frac{S}{r}$

where, S = length of the rod

r = radius

Putting the given values into the above formula as follows.

$\theta = \frac{S}{r}$

= $\frac{4}{2}$

= $2 radians (\frac{180^{o}}{\pi})$

= $114.64^{o}$

Now, we will calculate the charge density as follows.

$\lambda = \frac{Q}{L}$

= $\frac{18 \times 10^{-9} C}{4 m}$

= $4.5 \times 10^{-9} C/m$

Now, at the center of arc we will calculate the electric field as follows.

E = $\frac{2k \lambda Sin (\frac{\theta}{2})}{r}$

= $\frac{2(9 \times 10^{9} Nm^{2}/C^{2})(4.5 \times 10^{-9}) Sin (\frac{114.64^{o}}{2})}{2 m}$

= 34.08 N/C

Thus, we can conclude that the magnitude of the electric eld at the center of curvature of the arc is 34.08 N/C.

5 0

3 years ago

saad has mass 80kg when resting on the ground at the equator what will be the centripetal acceleration on saad if the radius of

Sidana [21]

Thank you for posting your question here at brainly. Below is the answer:

m = 80 kg
R = 6.4 x 10^6 m
Earth completes one rotation in 24 hours
Thus, T = 24 x 3600 s
Centripetral acceleration is given by
ac = w^2R
ac = (2 pie/T)^2R
ac = (2 pie/T)^2R = 4 pie^2/(24 x 3600)^2 x 6.4 x 10^6
ac = 0.034 m/s^2

8 0

4 years ago

The fastest pitched baseball was clocked at 46 m/s. assume that the pitcher exerted his force (assumed to be horizontal and cons

Zielflug [23.3K]

Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance

We know that vi = 0 because the ball was at rest initially.

Therefore,

Solving Eq.1 for acceleration,

 v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·s
Substituting the values,
a = 46²/2×1
a = 1058 m/s²

Now applying Newton's 2nd law of motion,

F = ma
= 0.145×1058

F = 153.4 N

8 0

3 years ago

Consider a cylindrical segment of a blood vessel 2.70 cm long and 3.10 mm in diameter. What additional outward force would such

Lyrx [107]

This question is incomplete, the complete question is;

- Calculate the difference in blood pressure between the feet and top of the head of a person who is 1.80m Tall

- Consider a cylindrical segment of a blood vessel 2.70 cm long and 3.10 mm in diameter. What additional outward force would such a vessel need to withstand in the person's feet compared to a similar vessel in her head

Answer:

- the difference in blood pressure is 18698.4 Pa

- the additional outward force F is 4.86 N

Explanation:

Given the data in the question;

we know that the expression for difference in blood pressure is;

ΔP = pgh

where p is density = 1060 kg/m³

g is acceleration due to gravity = 9.8 m/s²

and h is height = 1.80 m

now we substitute

ΔP = 1060 × 9.8 × 1.80

ΔP = 18698.4 Pa

therefore the difference in blood pressure is 18698.4 Pa

Also given that;

diameter of blood vessel d = 3.10 mm

radius r = 3.10 mm / 2 = 1.55 mm = 0.00155 m

length l = 2.70 cm = 0.027 m

Surface area of the cylindrical segment of a blood vessel is

A = 2πrl

we substitute

A = 2 × π × 0.00155 × 0.027

A = 2.6 × 10⁻⁴ m²

the required for will be;

F = PA

we substitute

F = 18698.4 Pa × 2.6 × 10⁻⁴ m²

F = 4.86 N

Therefore, the additional outward force F is 4.86 N

5 0

3 years ago