How much energy is required to move an electron through a potential difference of

Choose ALL that apply

Keith_Richards [23]

The answers a a,b, and d

6 0

3 years ago

Where does a phase change occur

il63 [147K]

A phase change is occuring; the liquid water is changing to gaseous water, or steam. On a molecular level, the intermolecular forces between the water molecules are decreasing. The heat is providing enough energy for the water molecules to overcome these attractive forces.

3 0

3 years ago

Read 2 more answers

Basketball player Darrell Griffith is on record as

Gre4nikov [31]

Explanation:

1.

We use the equation

h = $\frac{gt^2}{2}$ , where

h is the height traveled,

g is the acceleration due to gravity and

t is the time taken to reach height h.

We can now calculate t to be

$\sqrt{\frac{2*1.2 m}{9.81 m/s^2} }$

= 0.495 s

Let v be the initial velocity of the player.

The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.

v = 9.81 m/s^2 x 0.495 s = 4.85 m/s

2.

The player takes 0.3 s to increase his velocity from 0 m/s to 4.85 m/s. So his average accelaration is

4.85 m/s / 0.3 s = 16.2 m/s^2

5 0

3 years ago

Saturn moves in an orbit around the Sun with radius 10 AU. How many degrees does it move on the Celestial in one year? (Hint: Ca

Lana71 [14]

Answer:

B. About 12 degrees

Explanation:

The orbital period is calculated using the following expression:

T = 2π*( $\sqrt{\frac{r^3}{Gm}}$ )

Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.

Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.

Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:

$\frac{Tsaturn}{Tearth}$ = $\sqrt{\frac{rSaturn^3}{rEarth^3} } = \sqrt{10^3}}$