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Alina [70]
3 years ago
8

__________ is/are formed between two air masses that have different temperatures. (4 points) A: Wind B:The jet stream C:Clouds D

:Rain
Physics
2 answers:
lidiya [134]3 years ago
8 0
The answer is B: the jet stream.


Westkost [7]3 years ago
5 0
Is/formed between two air masses that ave different temperatures wind
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A dog sits 2.6 m from the center of a merry- go-round. a) If the dog undergoes a 1.7 m/s^2 centripetal acceleration, what is the
alexgriva [62]

Answer:

a) v= 2.1 m/s

b) ω = 0.807 rad/s

Explanation

Conceptual analysis :

The dog and the merry-go- round describes a circular motion, then, the following formulas apply :

a_{c} =\frac{v^{2} }{r} Formula (1)

v = ω *r   Formula (2)

Where:

a_{c} : Centripetal acceleration(m/s²)

v: linear speed or tangential (m/s)

r :  radius of the circle (m)

ω : angular speed ( rad/s)

Data

r= 2.6 m

a_{c} =  1.7 m/s²

Problem develpment

a) We replace data in the formula 1 to calculate the dog's linear speed(v):

a_{c} =\frac{v^{2} }{r}

1.7 =\frac{v^{2} }{2.6}

v^{2} =1.7*2.6 = 4.42

v=(\sqrt{4.42})\frac{m}{s}

v= 2.1 m/s

b)We replace data in the formula 2 to calculate the angular speed of the merry-go- round (ω).

v = ω *r

2.1 = ω *2.6

ω = 2.1/2.6

ω = 0.807 rad/s

6 0
3 years ago
During a total lunar eclipse, the moon
Anit [1.1K]

A,  C,  and  D  all happen at different stages
of a total lunar eclipse.

I'll describe the stages of the eclipse, but before I do, I just need
to clarify:  The Earth doesn't have an umbra or a penumbra, but
its shadow does.
 

-- the eclipse begins when the first edge of the moon
   moves into the penumbra of Earth's shadow; ( C )
   this part of the moon grows steadily.

-- After a while, the first edge of the moon begins to move
   into the umbra of Earth's shadow ( A ), and gets very dark.

-- The total phase of the eclipse begins when the ENTIRE
    moon is in the umbra of Earth's shadow.

Then everything happens in reverse.

--  Eventually, the leading edge of the moon moves out
     of the shadow's umbra, into the penumbra.  This part
     steadily grows.  

-- After a while, none of the moon is in the umbra, and
   the whole thing is in the penumbra.  The moon is
   fully illuminated, but not quite as bright as it should be.

--  Soon, the leading edge of the moon leaves the penumbra
    of Earth's shadow, and gets brighter.  This portion of the moon
    steadily grows, until ...

--  the moon completely leaves the penumbra, all of it is as bright
    as it's supposed to be.  The eclipse is completely over.  ( B )


==>  The whole process lasts several hours.

==>  Everybody on the night side of the Earth sees the same thing
         at the same time.  It doesn't matter WHERE you are on the night
         side ... if you can see the moon in the sky, you see the present
         phase of the eclipse.

==>  The lunar eclipse can only happen at the Full Moon.  In fact, the
         mid-point of the total phase is the exact moment of Full Moon.

8 0
3 years ago
Read 2 more answers
Which of the following statements is true? A. Whenever an object moves, work is being done. B. A larger force always means more
melamori03 [73]
The Correct answer would be D. Work equals force times distance.
8 0
3 years ago
delivery ladies is Shivam noted at the thundering sound of 6 second after the lightning was seen by him​
vitfil [10]

Answer:

this is due to difference in speeds of sound and light

Explanation:

light has a speed of 3×10^8 m/s and it is seen at once because it takes negligible time due to very high speed and short distance  that is why as soon as lightning occurs we can see it . since thundering sound travels with speed of sound which is about 330  to 340 m/s in air hence it takes some time as described in question as 6 seconds

8 0
3 years ago
Read 2 more answers
A spinning disc rotating at 130 rev/min slows and stops 31 s later. how many revolutions did the disc make during this time?
gayaneshka [121]
F = 130 revs/min = 130/60 revs/s = 13/6 revs/s
t = 31s
wi = 2πf = 2π × 13/6 = 13π/3 rads/s
wf = 0 rads/s = wi + at
a = -wi/t = -13π/3 × 1/31 = -13π/93 rads/s²
wf² - wi² = 2a∅
-169π²/9 rads²/s² = 2 × -13π/93 rads/s² × ∅
∅ = 1209π/18 rads
n = ∅/2π = (1209π/18)/(2π) = 1209/36 ≈ 33.5833 revolutions.
3 0
3 years ago
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