a student viewing a cell with a microscope observe a cell wall, a cell membrane,and a nucleus. The presence of these structures indicates that the student is looking at a cell from a

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Answer 1

Answer: The answer should be Plant Cell

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3. You shoot an arrow straight up into the air with a velocity of 25 m/s.a. What is the velocity of the arrow at its peak?
b. How high did the arrow go?
c. How long is the arrow in the air?

Answers

a) The velocity of the arrow at its peak is zero

b) The maximum height of the arrow is 31.9 m

c) The time of flight is 5.10 s

Explanation:

The motion of the arrow fired straight upward into the air is a uniformly accelerated motion, with constant acceleration $g=9.8 m/s^2$ (acceleration of gravity) towards the ground.

The initial velocity of the arrow when it is fired is upward: since the acceleration is downward, this means that as the arrow moves upward, its velocity decreases in magnitude.

Eventually, at some point, the velocity of the arrow will become zero, and then it will change direction (downward) and will start increasing in magnitude. The moment when the velocity raches zero corresponds to the peak of the trajectory of the arrow: therefore, at the peak the velocity is zero.

Since the motion of the arrow is a uniformly accelerated motion, we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the arrow, we have:

u = 25 m/s

v = 0 (when the arrow reaches the peak)

$a=g=-9.8 m/s^2$ (negative because downward)

s is the maximum height reached by the arrow

And solving for s,

$s=(v^2-u^2)/(2a)=(0-(25)^2)/(2(-9.8))=31.9 m$

In order to find the time it takes for the arrow to reach the maximum height, we use the following suvat equation:

$v=u+at$

where here we have:

v = 0 is the final velocity at the peak

u = 25 m/s

$a=g=-9.8 m/s^2$

And solving for t,

$t=-(u)/(g)=-(25)/(-9.8)=2.55 s$

This is the time the arrow takes to reach the top of the trajectory: therefore, the total time of flight is twice this value,

$T=2t=2(2.55)=5.10 s$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

what is required to bring about a phase change 1.an increase or decrease in pressure 2.an increase or decrease in energy 3.an increase in energy only 3.a decrease in energy only

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Answer: An increase or decrease in energy!

Explanation: I just took the quiz, I got it wrong with the increase or decrease of pressure. This is the right answer nowww.

The period of a 261 Hertz sound wave is

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That would be the result of 1/261

At what age is the difference between the average height boys and girls

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Hey There!

The age that determines difference between the average height boys and girls, is 5 to 8.

Have A Brainly Day :)

How long does it take for the speed of light and sound to travel around the world

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-- Light travels straight, not around in a circle. But if it did, it would cover
a distance equal to the length of the equator in about 0.13 second.

-- At the speed of sound (in air at standard temperature and pressure),
it would take about 32.6 hours to cover the same distance.

a person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500 m/s2. What is the velocity of the stroller after it has traveled 4.75 m ?

Answers

Answer:

Approximately $2.18 m\cdot s^(-1)$ .

Explanation:

Consider one of the equations for constant acceleration ("SUVAT" equations)

$v^(2) - u^(2) = 2 \; a \cdot x$ ,

where

$v$ is the final velocity of the object,
$u$ is the initial velocity of the object,
$a$ is the acceleration of the object, and
$x$ is the distance that the object had traveled while its velocity changed from $u$ to $v$ .

Note that unlike other SUVAT equations, this one does not ask for the time required for the speed of the object to change from $u$ to $v$ . Since in this problem, time isn't given, this time-less equation would particular useful.

Here

$v$ the final velocity needs to be found.
$u = 0$ for the stroller started from rest.
$a =\rm 0.500 \;m \cdot s^(-2)$ is the acceleration of the stroller, and
$x = \rm 4.75\; m$ is the distance that the stroller traveled while its velocity changed from $u$ to $v$ .

Rearrange the equation to isolate the unknown, $v$ :

$v^(2) = u^(2) + 2 \; a \cdot x$ .

Make sure that all units are standard, so that the unit of the output will also be standard. Apply the equation:

$v = \sqrt{u^(2) + 2 \; a \cdot x} = √(0^2 + 2 * 0.500 * 4.75 )\approx \rm 2.18\; m\cdot s^(-1)$ .

Hence the final velocity will be approximately $\rm 2.18 m\cdot s^(-1)$ .

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