1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kari74 [83]
3 years ago
8

Which describes the eye of a hurricane?

Physics
2 answers:
777dan777 [17]3 years ago
7 0
<h2><em><u>B</u></em></h2><h2><em><u>THE EYE OF THE STORM IS CALM</u></em></h2>
Lemur [1.5K]3 years ago
4 0

Answer:

<h2>b. has warm, calm air and light winds</h2>

Explanation:

The eye of a hurricane is calm and "peaceful", it's the region with calm weather. But the eyewall is really dangerous, due to its pressure and speed difference. Actually, the eyewall has the most severe winds and weather. Another important characteristic is that the eye has the lower pressure.

So, among the options, b. is the right answer.

b. has warm, calm air and light winds

You might be interested in
Consider the same 70kg/686N student on the surface on another planet from the table above: Jupiter. Tompared to the gravitationa
Yuliya22 [10]

Answer:

Dont mind me

Explanation:

6 0
3 years ago
Write a short paragraph that explains the central idea of the article. Use at least two details from the article to support your
zhannawk [14.2K]

Answer:

you should write about a book you read

Explanation:

because maybe you got really good things in it

or here is an example

3 0
2 years ago
If you set the cruise control of your car to a certain speed and take a turn, the speed of the car will remain the same. Is the
Bumek [7]

Answer:

Yes, the car has acceleration.

Explanation:

Acceleration is defined as the rate of change of velocity. The velocity is a vector quantity. If a car is moving with constant speed but taking a turn, it means the velocity is changing, so the car have some acceleration.

5 0
3 years ago
A professional baseball player can throw the ball around 45 m/s if the distance between the pitcher and the batter is 18.39 m. H
svetlana [45]

Answer:

The time taken for the ball to get to the batter is 0.41 s.

Explanation:

Given;

initial velocity of the baseball, u = 45 m/s

horizontal distance between the pitcher and the batter, X = 18.39 m

The horizontal distance or range of a projectile is given as;

X = ut

where;

t is the time of flight

u is the initial velocity

t = X / u

t = 18.39 / 45

t = 0.41 s

Therefore, the time taken for the ball to get to the batter is 0.41 s.

6 0
2 years ago
A ball is tossed from an upper-story window of a building. the ball is given an initial velocity of 8.00 m/s at an angle of 20.0
otez555 [7]
A) The motion of the ball consists of two indipendent motions on the horizontal (x) and vertical (y) axis. The laws of motion in the two directions are:
x(t)=v_0 \cos \alpha t
y(t)=h-v_0 \sin \alpha t - \frac{1}{2}gt^2
where
- the horizontal motion is a uniform motion, with constant speed v_0 \cos \alpha, where v_0 = 8.00 m/s and \alpha=20.0^{\circ}
- the vertical motion is an uniformly accelerated motion, with constant acceleration g=9.81 m/s^2, initial position h (the height of the building) and initial vertical velocity v_0 \sin \alpha (with a negative sign, since it points downwards)

The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
x(3.00 s)=v_0 \cos \alpha t=(8 m/s)(\cos 20^{\circ})(3.0 s)=22.6 m

b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
0=h-v_0 \sin \alpha t -  \frac{1}{2}gt^2
which becomes
h=(8 m/s)(\sin 20^{\circ})(3.0 s)+ \frac{1}{2}(9.81 m/s^2)(3.0 s)^2=52.3 m

c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that 
y(t)=h-10
If we substitute this into the equation of y(t), we have
h-10 = h-v_0 \sin \alpha t-  \frac{1}{2}gt^2
\frac{1}{2}gt^2+v_0 \sin \alpha t -10 =0
4.9 t^2 +2.74 t-10 =0
whose solution is t=1.18 s (the other solution is negative, so it has no physical meaning). Therefore, the ball reaches a point 10 meters below the level of launching after 1.18 s.

4 0
3 years ago
Other questions:
  • What is the name of the upward force that enables a bird to fly?
    5·1 answer
  • Unpolarized light with intensity S is incident on a series of polarizing sheets. The first sheet has its transmission axis orien
    10·1 answer
  • Calculate the electric field at the center of a square 46.4 cm on a side, if one corner is occupied by a +42.0 µc charge and the
    11·1 answer
  • A block of mass 0.240 kg is placed on top of a light, vertical spring of force constant 5 200 N/m and pushed downward so that th
    7·1 answer
  • How does the image distance and magnification give information about the type of image is produced bya lens or multiple lenses/
    7·1 answer
  • What is the magnitude of vector K ?
    14·1 answer
  • If a toy car has a centripetal acceleration of 50 m/s2 and was making the turn at 10 m/s. what was his radius
    9·1 answer
  • Lab report of experimrnt to find the series resonance of acceptor circuit and parallel resonance of rejector circuit in RLC?
    13·1 answer
  • Which two types of air masses would likely form a subtropical jet stream?
    8·2 answers
  • How Do I get A Picture For My Profile, It keeps saying my pistures won't work
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!