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3 years ago

Which takes place as water cycles from the bottom of the pot toward the top?

2 answers:

8 0

The option that takes place as water cycles from the bottom of the pot toward the top is that A. thermal energy is transferred.
As the pot gets warmer and warmer, the heat flows everywhere inside the pot, ultimately reaching the top, and heating the water at the top as well. There is no chemical energy here, and molecules don't gain thermal energy, it is just transferred to the top of the pot.

aleksklad [387]3 years ago

6 0

Answer:

Explanation:

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James threw a ball vertically upward with a velocity of 41.67ms-1 and after 2 second David threw a ball vertically upward with a

Reptile [31]

Answer:

When have passed 3.9[s], since James threw the ball.

Explanation:

First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.

We'll use the kinematics equations to find these two unknowns.

$y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can find the velocity after 2 seconds.

$v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

$y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\$

As we can see the equation is based on Time (t).

Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

$y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]$

Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.

Time = 2 + 1.9 = 3.9[s]

4 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

UkoKoshka [18]

Answer:

Explanation:

1) Hypermetropia (better known as Farsighted- this is why nearby objects seem blurry for him)

2) In such instances, image are typically formed farther from the near point

3) Such defects are quite common so there are common procedures such as using convex lens which can restore the sight to normal.

7 0

3 years ago

All but two gaps within a set of Venetian blinds have been blocked off to create a double-slit system. These gaps are separated

Alecsey [184]

Answer:

Explanation:

separation between two gaps, d = 5 cm

angle between central and second order maxima, θ = 0.52°

use

d Sinθ = n λ

n = 2

0.05 x Sin 0.52° = 2 x λ

λ = 2.27 x 10^-4 m

λ = 226.9 micro metre

6 0

3 years ago

Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average o

Musya8 [376]

Explanation:

It is given that,

A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

Radius of Mercury's orbit, $r=5.79\times 10^{10}\ m$

Radius of discovered planet, $R=\dfrac{2}{3}r$

$R=\dfrac{2}{3}\times 5.79\times 10^{10}\ m=3.86\times 10^{10}\ m$

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :

$T^2\propto R^3$

$T^2=\dfrac{4\pi^2R^3}{GM}$

$T^2=\dfrac{4\pi^2\times (3.86\times 10^{10})^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}}$

$T=\sqrt{1.71\times 10^{13}}$

T = 4135214.625 s

T = 47.86 days

So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.

6 0

3 years ago

The city council is considering discussing whether or not to put fluoride in the city's water supply. Many other towns add it al

umka2103 [35]

They should look for <span>a report from an independent scientific research firm,
even if they have to pay for it.

In preparing its report, the firm would have already surveyed many of the </span>
<span>citizens from several other towns that currently add fluoride to their water,
plus a lot of other relevant medical research on the subject.</span>

8 0

2 years ago

Read 2 more answers