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3 years ago

The Hurricane's "eye" is surrounded by a "wall" of fast moving air? True or False

2 answers:

7 0

True because The Eye Wall of a hurricane's most devastating region. Located just outside of the eye is the eye wall. This is the location within a hurricane where the most damaging winds and intense rainfall is found. At the surface, the winds are rushing towards the center of a hurricane forcing air upwards at the center.

natulia [17]3 years ago

3 0

Answer:TRUE

Explanation:

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Use the information below for the next five questions:

sergey [27]

Answer: Please see answer in explanation column.

Explanation:

Given that

v≈(331 + 0.60T)m/s

where Temperature, T = 14°C

v≈(331 + 0.60 x 14)m/s

v =331+ 8.4 = 339.4m/s

In our solvings, note that

f= frequency

λ=wavelength

L = length

v= speed of sound

a) Length of the pipe is calculated using the fundamental frequency formulae that

f=v/2L

Length = v/ 2f

= 339.4m/s/ 2 x 494Hz ( s^-1)= 0.3435m

b) wavelength of the fundamental standing wave in the pipe

L = nλ/2,

λ = 2L/ n

λ( wavelength )= 2 x 0.3435/ 1

= 0.687m

c) frequency of the fundamental standing wave in the pipe

F = v/ λ

= 339.4m/s/0.687m=

494.03s^-1 = 494 Hz

d) the frequency in the traveling sound wave produced in the outside air.

This is the same as the frequency in the open organ pipe = 494Hz

e)The wavelength of the travelling sound wave produced in the outside air is the same as the wavelength calculated in b above = 0.687m

f) To play D above middle c . the distance is given by

L =v/ 2 f

= 343/ 2 x 294

=0.583m

7 0

3 years ago

A weight lifter lifts a dumbbell a certain height in 2.0 s, while a competitor does the same workin 1.0 s. Compared to the power

Aloiza [94]

Answer:

a. one-half as great

Explanation:

The power developed by the first lifter is one-half as great as that of the second person.

Power is defined as the rate at which work is done;

Power = $\frac{workdone}{time}$

Since the two lifters do the same work at different time, let us estimate their power;

P₁ = $\frac{workdone}{2}$ P₂ = $\frac{workdone }{1}$

We see that for P₁, power is half of the work done whereas in P₂ power is the same as the work done.

Therefore,

The power of the first weight lifter is one-half the second lifter.

4 0

2 years ago

A force F=0.12N is aplied on spring and spring elongates by 3cm . specific constant of spring

PilotLPTM [1.2K]

The spring constant is 4 N/m

Explanation:

When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:

$F=kx$