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Xelga [282]
3 years ago
10

What are the 5 steps in the process of air circulation in order

Physics
1 answer:
creativ13 [48]3 years ago
4 0
Yes,Physics 5pts just now<span>are the 5 steps in the process of air circulation in order</span>
You might be interested in
A spaceship starting from a resting position accelerates at a constant rate of 9.8 meters per second per second. How long and ho
Verizon [17]

Accelerating at 9.8 m/s² means that every second, the speed is 9.8 m/s faster than it was a second earlier.  It's not important to the problem, but this number (9.8) happens to be the acceleration of gravity on Earth.

1% of the speed of light = (300,000,000 m/s) / 100 = 3,000,000 m/s .

Starting from zero speed, moving (9.8 m/s) faster every second,
how long does it take to reach  3,000,000 m/s ?

           (3,000,000 m/s) / (9.8 m/s²)  =  306,122 seconds .
                                                   (That's  5,102 minutes.)
                                                        (That's  85 hours.)
                                                     (That's  3.54 days.)

Speed at the beginning . . . zero .
Speed at the end . . . 3,000,000 m/s
Average speed . . . . . 1,500,000 m/s

Distance = (average speed) x (time)

               = (1,500,000 m/s) x (306,122 sec) = 4.592 x 10¹¹ meters

                                                                     =  459 million kilometers

                         That's like from Earth
                                                  to       Sun
                                                             to    Earth
                                                                    to        Sun. 

4 0
2 years ago
A robin in flight has 20.8 J of PE when it is 27.6 m high. What is the mass of the robin? (Unit = kg)
schepotkina [342]

Answer:

<h2>\boxed{  \bold{ \purple{0.0769 \: kg \: }}}</h2>

Explanation:

\sf{Potential   Energy ( P.E ) \:  =  \: 20.8 \: joule}

\sf{distance \:  = 27.6 \: metre}

\sf{mass = } ?

\sf{acceleration \: due \: to \: gravity = 9.68 \:  {metre \: per \: second}^{2} }

Now, let's find the mass:

\sf{PE  \:  =  mass \times gravity \:  \times  \: distance}

plug the values

⇒\sf{20.8 = m \times 9.8 \times 27.6}

Multiply the numbers

⇒\sf{20.8 = 270.48 \: m}

Swap the sides of the equation

⇒\sf{270.48m = 20.8}

Divide both sides of the equation by 270.48

⇒\sf{ \frac{270.48m}{270.48}  =  \frac{20.8}{270.48} }

Calculate

⇒\sf{0.0769} kg

Hope I helped!

Best regards!!

7 0
3 years ago
Read 2 more answers
Two bar magnets are labeled A and B. The ends of each magnet are numbered 1 or 2, but the poles are not labeled. When A1 is brou
vichka [17]
It is definitely letter D. <span>A1 and B1 are like poles, but there is not enough information to tell whether they are north poles or south poles.

A1 and B1 is either both north poles or both south poles. Repulsion of both magnets says it all--like poles always repel while opposite poles always attract. Thus, the best conclusion to this would be choice D.</span>
3 0
3 years ago
Read 2 more answers
How many foot-pounds of work does it take to throw a baseball 90 mph? a baseball weighs 5 oz, or 0.3125 lb?
NARA [144]

Kinetic energy of the ball is (mv²) / 2, where m is the mass and v is the velocity

So plugging in the mass and the velocity into the kinetic energy expression, you get:

Kinetic energy of the ball = (mv²) / 2

(0.3125/32) times (132)² divided by 2 = 85 ft-lbs


Kinetic energy of the ball = 85 ft-lbs


5 0
2 years ago
A 1.25-kg ball begins rolling from rest with constant angular acceleration down a hill. If it takes 3.60 s for it to make the fi
miv72 [106K]

Answer:

The time taken is  \Delta t  = 1.5 \ s

Explanation:

From the question we are told that

   The mass of the ball is  m =  1.25 \ kg

    The time taken to make the first complete revolution is  t= 3.60 s

    The displacement of the first complete revolution is  \theta  =  1 rev  =  2 \pi \  radian

Generally the displacement for one  complete revolution is mathematically represented as

       \theta =  w_i t  +  \frac{1}{2} *  \alpha  * t^2

Now given that the stone started from rest w_i  = 0 \ rad / s

     2 \pi =0   +  0.5*  \alpha  *(3.60)^2

     \alpha   =  0.9698 \  s

Now the displacement for two  complete revolution is

         \theta_2  =  2 *  2\pi

         \theta_2  = 4\pi

Generally the displacement for two complete revolution is mathematically represented as  

     4 \pi =   0  +  0.5 * 0.9698 * t^2

=>   t^2  =  25.9187

=>   t=  5.1 \ s

So

 The  time taken to complete the next oscillation is mathematically evaluated as

     \Delta t  =  t_2  - t

substituting values

      \Delta t  = 5.1 -  3.60

     \Delta t  = 1.5 \ s

           

 

7 0
2 years ago
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