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liq [111]
3 years ago
13

Jayden was given a marshmallow and a syringe in class to experiment with. She placed the marshmallow in the syringe and sealed t

he end. When she depressed the plunger of the syringe what happened to the marshmallow and why? A) The increased pressure, pushed the molecules closer together, and caused the marshmallow to shrink. B) The decreased pressure, pushed the molecules closer together, and caused the marshmallow to shrink. C) The increased pressure, allowed the molecules to spread apart, and caused the marshmallow to grow larger. D) The decreased pressure, allowed the molecules to spread apart, and caused the marshmallow to grow larger.
Physics
1 answer:
Tju [1.3M]3 years ago
3 0
<h2>Answer:</h2>

The correct option is A.

A) The increased pressure, pushed the molecules closer together, and caused the marshmallow to shrink.

<h2>Explanation:</h2>

Jayden experimented, she placed the marshmallow in the syringe and sealed the end. When she depressed the plunger of the syringe, the pressure increased and pushed the molecules closer together and causes the marshmallow to shrink.

<h2 />
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A river flows due south with a speed of 2.0 m/s .You steer a motorboat across the river; your velocity relative to the water is
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Answer:

a) v_m =\sqrt{v^2_x + v^2_y} = \sqrt{(2m/s)^2 +(4.8 m/s)^2}= 5.2 m/s

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c)t= \frac{w}{v_m}= \frac{600m}{4.8 m/s}= 125 s

d) Y = 2 m/s * 125 s = 250m

So it would be 250 to the South

Explanation:

Part a

For this case the figure attached shows the illustration for the problem.

We know that v_y = 2 m/s represent the velocity of the river to the south.

We have the velocity of the motorboard relative to the water and on this case is V_x= 4.8 m/s

And we want to find the velocity of the motord board relative to the Earth v_m

And we can find this velocity from the Pythagorean Theorem.

v_m =\sqrt{v^2_x + v^2_y} = \sqrt{(2m/s)^2 +(4.8 m/s)^2}= 5.2 m/s

Part b

We can find the direction with the following formula:

\theta= tan^{-1} \frac{v_y}{v_x} = tan^{-1} (\frac{2}{4.8})= 22.62 degrees and on this case to the South of the East.

Part c

For this case we can use the following definition

D = Vt

The distance would be D = w = 600 m and the velocity V = 4.8m/s and if we solve for t we got:

t= \frac{w}{v_m}= \frac{600m}{4.8 m/s}= 125 s

Part d

For this case we can use the same definition but now using the y compnent we have:

Y = v_y t

And replacing we got:

Y = 2 m/s * 125 s = 250m

So it would be 250 to the South

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