Pressing two objects together with more force _________

First, water evaporates from the ocean to the atmosphere, then ____________. put each step in the correct order.

oksian1 [2.3K]

<span>First, water evaporates from the ocean to the atmosphere, then... put each step in the correct order.
</span>Second- atmospheric circulation advects the water.
Third- water condensates to form clouds.
Fourth- precipitation falls from clouds to the land.
Fifth- some water stored as snow or ice.
Sixth- water either ±ows along the surface in rivers and lakes or infiltrates the ground.
Seventh- water returns to the ocean.

3 0

3 years ago

draw a velocity graph with Vi = 4 m/s and decreasing uniformly so that velocity at 2 seconds is 2 m/s and remaining constant fro

Shkiper50 [21]

For the velocity graph: start at 0s and 4m/s and draw a straight line to 2s and 2 m/s. Then draw a straight horizontal line to 4s and 2m/s

For the acceleration graph: start with a horizontal line from 0s and 2m/s/s to 2s and 2m/s/s. The draw another line from 2 s and 0m/s/s to 4 s and 0m/s/s

6 0

3 years ago

In the 2016 Olympics in Rio, after the 50 m freestyle competition, a problem with the pool was found. In lane 1 there was a gent

Dmitry_Shevchenko [17]

Answer:

Explanation:

A )

speed of swimming in still water is given by the expression

distance / time

= 50 / 25

= 2 m /s

In lane 1 , 1.2 cm/s current is flowing in the direction that the swimmers are going so swimmers will cover distance at the rate of 2 + 1.2 = 3.2 m /s.

time to cover distance of 50 m in lane 1

= distance / speed

= 50 / 3.2 = 15.625 s

In lane 8 , 1.2 cm/s current is flowing against the direction that the swimmers are going so swimmers will cover distance at the rate of 2 - 1.2 = .8 m /s.

time to cover distance of 50 m in lane 1

= distance / speed

= 50 / .8 = 62.5 s

8 0

3 years ago

A capacitor of capacitance C has charge Q and stored energy is U if the charge is increased to 2Q then energy will be A)4U B)2U

Lubov Fominskaja [6]

Answer:

2u

Explanation:

2u

W = Vq

q = CV

W = V.CV

W = CV²

W/C = V²

V = √(W/C)

√(W1/C1) = √(W2/C2)

√(u/c) = √(x/2c)

x = 2u

8 0

3 years ago

Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T

Zina [86]

a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

a)

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

b)

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

$f_0 = 60 MHz$

so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

c)

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$