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3 years ago

9

What is the vacuole in a plant cell?

2 answers:

Marina CMI [18]3 years ago

8 0

The vacuole functions to maintain the proper pressure within the plant cells to provide structure and support for the growing plant.

arsen [322]3 years ago

4 0

The central vacuole<span> is a </span>cellular<span> organelle found in</span>plant cells<span>. It is often the largest organelle in the </span>cell<span>. It is surrounded by a membrane and functions to hold materials and wastes. It also functions to maintain the proper pressure within the </span>plant cells<span> to provide structure and support for the growing</span>plant<span>.</span>

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You are driving due north on i-81 to come to jmu with a speed of 10 m/s, suddenly you realize you forgot your book. You make a u

Readme [11.4K]

first we make a U turn and travel towards home in t = 20 s

so the distance of home from initial position is

$d_1 = v*t_1$

$d_1 = 10*20 = 200 m$

Now after picking up the book we travel back with speed 20 m/s

so again after t = 20 s the displacement is given as

$d_2 = v*t = 20*20 = 400 m$

so the net displacement is given as

$\vec d = \vec d_2 - \vec d_1$

$\vec d = 400 - 200 = 200 m$

so it will be displaced by total displacement 200 m

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3 years ago

A 60 kg runner accelerates during a race at 3m/s2. what force is exerted on the earth with each step

marusya05 [52]

Answer:

180 Newton(N)

Explanation:

force =mass *acceleration

=60 * 3

=180 kgm/s^2

=180 N

6 0

3 years ago

A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the

irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

$x=V_{o} cos \theta t$ (1)

$y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2}$ (2)

Where:

$x$ is the horizontal distance between the cannon and the ball

$V_{o}=23 m/s$ is the cannonball initial velocity

$\theta=0\°$ since the cannonball was shoot horizontally

$t$ is the time

$y=0$ is the final height of the cannonball

$y_{o}=1.96 m$ is the initial height of the cannonball

$g=9.8 m/s^{2}$ is the acceleration due gravity

Isolating $t$ from (2):

$t=\sqrt{-\frac{2(y-y_{o})}{g}}$ (3)

$t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}}$ (4)

$t=0.63 s$ (5)

Substituting (5) in (1):

$x=(23 m/s) cos(0\°) 0.63 s$ (6)

Finally:

$x=14.49 m$

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3 years ago

Once again, move the balloon to the right and let it go. Note how fast the balloon moves. Next, brush the balloon against the en

lilavasa [31]

Answer:

Yes. The balloon moves faster when it has more electrons and the sweater has fewer electrons

Explanation:

From Plato. Hope this helps!

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3 years ago

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A 150-kg object takes 1.5 minutes to travel a 2,500-meter straight path. It begins the trip traveling 120 m/s and decelerates to

podryga [215]

I believe it is -1.11 m/s^2. I will let you know if its correct

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4 years ago

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