What is the definition of physical change

1 answer:

jeka943 years ago

7 0

They are used to separate mixtures into their component compounds but can no usually be used to deprecate compounds into chemical elements or simpler compounds

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In Figure 10-1, if the force exerted on a 3.0-kg backpack that is initally at rest is 20.0 N and the distance it acts over is 0.

pshichka [43]

I'm going to assume this is over a horizontal distance. You know from Newton's Laws that F=ma --> a = F/m. You also know from your equations of linear motion that v^2=v0^2+2ad. Combining these two equations gives you v^2=v0^2+2(F/m)d. We can plug in the given values to get v^2=0^2+2(20/3)0.25. Solving for v we get v=1.82 m/s!

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3 years ago

What's something that I've done in my life that is related to physics

Allisa [31]

That's a very difficult question to answer, because you give us
no information regarding what you have done in your life.

We can only assume that you have most likely breathed on occasion,
floated on your back in the ocean, lake or pool, maybe fallen off of a
ladder or out of bed, felt the warmth of the sun on your cheek, seen
a rainbow after a rainshower, heard the sound of thunder during a
summer storm, taken a trip in an airplane, and waited for a cup of
hot chocolate to cool off. The richness of any of these experiences
is greatly enhanced when you understand some of the Physics involved.

6 0

3 years ago

A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$