Which does not contain a lens?

An object in projectile motion will follow which path? a]curved up from the ground b]curved down toward the ground c]straight do

Lesechka [4]

It's B because when you throw something it doesn't go up it slowly descends downward

7 0

2 years ago

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The sound waves from a noisy jet travel from the air into water. Which property of the wave will not change?

notka56 [123]

The frequency of the wave will not change. Since the change in medium doesn't affect the source of the waves, the frequency of those waves do not change.

Hope this helps! :)

8 0

2 years ago

A scientist pours 0.120 L of solution A into an Erlenmeyer flask. She adds 2.345 L of solution B. How many significant figures a

nika2105 [10]

0.120L + 2.345L = 2.465L = 4 significant figures in the answer

6 0

2 years ago

Make the following conversion. 0.0097 mg = _____ g

Cloud [144]

Your answer should be 9.7 :)

8 0

3 years ago

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A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,