Ask question

Ask question

All categories

notsponge [240]

4 years ago

9

How must of the roller coaster's kinetic and/or potential energy is in each of the points. For each point, pick one option. (Sor

ry for the report, I found a mistake in my question and could not edit.)
Point A:
a. 1/2 Potential Energy, 1/2 Kinetic Energy
b. 1/2 Potential Energy, 0 Kinetic
c. 100 Potential Energy

Point B:
a. 1/4 Potential, 3/4 Kinetic
b. 100 Kinetic, 0 Potential
b. 1/2 Potential, 1/2 Kinetic

Point C:
a. 1/2 Potential, 1/2 Kinetic
b. 3/4 Potential, 1/4 Kinetic
c. 100 Potential, 0 Kinetic

1 answer:

labwork [276]4 years ago

3 0

Answer

Point A-
Point B-
Point C- 100 Potential, 0 Kinetic

You might be interested in

The standard metric unit of density for a solid is

prisoha [69]

There is no SI "base unit" of density.
(Any unit of mass) divided by (any unit of volume) is
a valid unit of density.

The units of density that are seen most often are

(gram per cm³) and (kgm per meter³) .

5 0

3 years ago

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

<h3>a.</h3>

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

<h3>b.</h3>

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

<h2>b</h2>

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

3 0

3 years ago

Heavier elements require higher temperatures to fuse. After stars run out of hydrogen in their cores, they leave the main sequen

Kruka [31]

Answer:

A. Add mass to the Sun.

Explanation:

A. Add mass to the Sun.

Adding mass will make to take more time for the hydrogen to run out and hence, enough temperature will be developed to fuse helium atom into other heavier elements, and eventually get hot enough to fuse the helium in their cores into carbon.

The only hypothetical solution is that we need to add Mass to the Sun.

5 0

4 years ago

Which subatomic particle has the least mass?

goblinko [34]

Answer:

electron

Explanation:

6 0

3 years ago

Suppose a logs mass is 5kg .after burning,the mass of the ash is 1kg .explain what may have happened to the other 4kg

BigorU [14]

The "missing" 4 kg may have escaped the scene of the fire in the
form of hot gases, and as particles of soot and ash that were seen
leaving in the form of "smoke".

5 0

3 years ago