NEED HELP NOW DUE TODAY Which of these statements is most likely correct about Newton's law on gravity? (2 points)

What is the average velocity of an arrow that travels 12 m [E] in 0.15 s?

sergij07 [2.7K]

Velocity is defined as Distance divided by Time.
In other words, V = D/T.

Now that we have our formula, we can solve.
Let's plug in the numbers we have.

We have 12m [East (direction not necessary when solving yet)] for our distance, and 0.15s as our time.

Divide the distance (12 /) by the time (0.15)
12 / 0.15 = 80.

Your velocity is 80 m/s [E]

I hope this helps!

6 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:

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

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

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:

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:

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

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

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

After travelling 2 cm:

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

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

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

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

2 years ago

State the term used to describe the turning force exerted by the man

ozzi

]A force called the effort force is applied at one point on the lever in order to move an object, known as the resistance force, located at some other point on the lever.

The way levers work is by multiplying the effort exerted by the user. Specifically, to lift and balance an object, the effort force the user applies multiplied by its distance to the fulcrum must equal the load force multiplied by its distance to the fulcrum. Consequently, the greater the distance between the effort force and the fulcrum, the heavier a load can be lifted with the same effort force.

8 0

1 year ago

A ball is kicked from level ground at an angle 60° with initial velocity 10 m/s. The distance the ball travels, in meters, is:

RUDIKE [14]

Answer:

<em>the ball travels a distance of 8.84 m</em>

Explanation:

Range: Range is defined as the horizontal distance from the point of projection to the point where the projectile hits the projection plane again.

R = (U²sin2∅)/g.............................. Equation 1

Where R = range, U = initial velocity, ∅ = angle of projection, g = acceleration due to gravity.

<em>Given: U = 10 m/s, ∅ = 60°</em>

<em>Constant: g = 9.8 m/s²</em>

Substituting these values into equation 1

R = [10²×sin(2×60)]/9.8

R = (100sin120)/9.8

R = 100×0.8660/9.8

R = 86.60/9.8

R = 8.84 m

<em>Therefore the ball travels a distance of 8.84 m</em>

4 0

2 years ago

What is the smallest particle that relains<br> the properties of an element

VLD [36.1K]

Answer:

atom

Explanation:

Smallest particle of an element that identifies that element is an atom.

8 0

1 year ago