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statuscvo [17]
3 years ago
7

How does water get up to the atmosphere, and how does it get back down to earth surface

Physics
1 answer:
IrinaK [193]3 years ago
6 0

Answer:

Water gets up to the Earth's atmosphere by evaporating from a body of water, which is then they become water vapor. It returns back to the surface by returning back to its water state and falling back down (as rain). The water vapor turns into clouds (clouds are really just water droplets), and when it cannot hold anymore waters, it disperses all the water (by raining).

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A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro
tekilochka [14]

Answer:

distance between object and image =  18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

distance between object and image

solution

we know lens formula that is

1/f = 1/v + 1/u     ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm   ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm     ......................3

so from equation 2 and 3

distance between object and image =  5.4 + 13.5

distance between object and image =  18.9 cm

6 0
3 years ago
Which body is in equilibrium?
sergey [27]
"(1) a satellite moving around Earth in a circular <span>orbit" is the only option from the list that describes an object in equilibrium, since velocity and gravity are working together to keep the orbit constant. </span>
6 0
3 years ago
Read 2 more answers
The speed at which a light aircraft can take off is 120 km/h. (A) What is the minimum constant acceleration required for the pla
kondor19780726 [428]

Answer:

A) a = 2.31[m/s^2]; B) t = 14.4 [s]

Explanation:

We can solve this problem using the kinematic equations, but firts we must identify the data:

Vf= final velocity = take off velocity = 120[km/h]

Vi= initial velocity = 0, because the plane starts to move from the rest.

dx= distance to run = 240 [m]

v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\  a=2.31[m/s^2]\\

To find the time we must use another kinematic equation.

v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]

7 0
3 years ago
What real-world examples show no work being done? Can you think of examples other than resisting the force of gravity?
GalinKa [24]

-- pushing on a brick wall

-- standing on your little brother's back so that he can't get up

-- taking a nap while on the job

-- squeezing anything that doesn't yield to your squeeze, such as a glass bottle or your girl friend

-- watching TV

-- solving math problems in your head

-- making pictures out of clouds in the sky

8 0
3 years ago
An electric model train is accelerated at a rate of 8 m/s^2 by a 12 N force? What is the
jok3333 [9.3K]

Answer:

m = 1.5 kg

Explanation:

Data:

  • Aceleration (a) = 8 m/s²
  • Force (F) = 12 N
  • Mass (m) = ?

Use formula:

  • \boxed{m = \frac{F}{a}}

Replace in the formula:

  • \boxed{m = \frac{12N}{8\frac{m}{s^{2}}}}

Equate the newtons:

  • \boxed{m = \frac{12kg*\frac{m}{s^{2}}}{8\frac{m}{s^{2}}}}

Simplify m/s²:

  • \boxed{m = \frac{12kg}{8}}

It divides:

  • \boxed{m =1.5kg}

What is the mass of the train?

The mass of the train is <u>1.5 kilograms.</u>

4 0
3 years ago
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