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IceJOKER [234]
3 years ago
7

Suppose one organ in the organ system failed to work properly, How would the organ system be affected? Would it still be able to

function? Explain your reasoning
Physics
1 answer:
scoray [572]3 years ago
8 0

The organ system failing would affect the rest of the body. You would probablly not live long. The human anatomy is very important. It won work properly. It will affect the rest of your body and probably land you in the hospital. This is why one failing organ system can be fatal.

You might be interested in
A teacher told a learner to react benzene (CH) with chlorine (Cl₂) to
kolezko [41]

The minimum quantity of benzene, C₆H₆ needed for the reaction is 106.67 g

<h3>How to determine the theoretical yield of chlorobenzene, C₆H₅Cl</h3>

From the question given, the following data were obtained

  • Actual yield = 100 g
  • Percentage yield = 65%
  • Theoretical yield =?

Percentage yield = (Actual / Theoretical) × 100

65% = 100 / Theoretical

0.65 = Actual / Theoretical

Cross multiply

0.65 × Theoretical = 100

Divide both sides by 0.65

Theoretical = 100 / 0.65

Theoretical yield = 153.85 g

<h3>How to determine the mass of benzene, C₆H₆ needed</h3>

Balanced equation

C₆H₆ + Cl₂ → C₆H₅Cl + HCl

Molar mass of C₆H₆ = 78 g/mol

Mass of C₆H₆ from the balanced equation = 1 × 78 = 78 g

Molar mass of C₆H₅Cl = 112.5 g

Mass of C₆H₅Cl from the balanced equation = 1 × 112.5 = 112.5 g

SUMMARY

From the balanced equation above,

112.5 g of C₆H₅Cl were obtained from 78 g of C₆H₆

Therefore,

153.85 g of C₆H₅Cl will be produced from = (153.85 × 78) / 112.5 = 106.67 g of C₆H₆

Thus, the minimum amount of benzene, C₆H₆ needed for the reaction is 106.67 g

Learn more about stoichiometry:

brainly.com/question/16735180

#SPJ1

3 0
2 years ago
What would be the first thing you would do if your clothes caught fire while working in a laboratory? select one of the options
Ksivusya [100]
<span>c. run towards a source of water to extinguish the fire </span>
3 0
3 years ago
How is the Sun related to nuclear electromagnetic and heat energy
Snezhnost [94]

-- The source of most of the energy that radiates from the sun is nuclear energy.

-- Most of the energy that radiates from the sun is electromagnetic energy.

-- Heat energy is part of the electromagnetic energy that radiates from the sun.
Other parts include radio, microwave, visible light, ultraviolet, and X-ray energy.

5 0
3 years ago
A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.?a.What is the speed of the mountain lio
Arisa [49]

Answer:

What is the speed of the mountain lion as it leaves the ground?

9.98m/s

At what angle does it leave the ground?

50.16°

Explanation:

This is going to be long, so if you want to see how it was solved refer to the attached solution. If you want to know the step by step process, read on.

To solve this, you will need use two kinematic equations and SOHCAHTOA:

d = v_it + \dfrac{1}{2}at^{2}\\\\vf = vi + at

With these formulas, we can derive formulas for everything you need:

Things you need to remember:

  • A projectile at an angle has a x-component (horizontal movement) and y-component (vertical movement), which is the reason why it creates an angle.
  • Treat them separately.
  • At maximum height, the vertical final velocity is always 0 m/s going up. And initial vertical velocity is 0 m/s going down.
  • Horizontal movement is not influenced by gravity.
  • acceleration due to gravity (a) on Earth is constant at 9.8m/s

First we need to take your given:

10.0 m long (horizontal) and maximum height of 3.0m (vertical).

d_x=10.0m\\d_y=3.0m

What your problem is looking for is the initial velocity and the angle it left the ground.

Vi = ?     Θ =?

Vi here is the diagonal movement and do solve this, we need both the horizontal velocity and the vertical velocity.

Let's deal with the vertical components first:

We can use the second kinematic equation given to solve for the vertical initial velocity but we are missing time. So we use the first kinematic equation to derive a formula for time.

d_y=V_i_yt+\dfrac{1}{2}at^{2}

Since it is at maximum height at this point, we can assume that the lion is already making its way down so the initial vertical velocity would be 0 m/s. So we can reduce the formula:

d_y=0+\dfrac{1}{2}at^{2}

d_y=\dfrac{1}{2}at^{2}

From here we can derive the formula of time:

t=\sqrt{\dfrac{2d_y}{a}}

Now we just plug in what we know:

t=\sqrt{\dfrac{(2)(3.0m}{9.8m/s^2}}\\t=0.782s

Now that we know the time it takes to get from the highest point to the ground. The time going up is equal to the time going down, so we can use this time to solve for the intial scenario of going up.

vf_y=vi_y+at

Remember that going up the vertical final velocity is 0m/s, and remember that gravity is always moving downwards so it is negative.

0m/s=vi_y+-9.8m/s^{2}(0.782s)\\-vi_y=-9.8m/s^{2}(0.782s)\\-vi_y=-7.66m/s\\vi_y=7.66m/s

So we have our first initial vertical velocity:

Viy = 7.66m/s

Next we solve for the horizontal velocity. We use the same kinematic formula but replace it with x components. Remember that gravity has no influence horizontally so a = 0:

d_x=V_i_xt+\dfrac{1}{2}0m/s^{2}(t^{2})\\d_x=V_i_xt

But horizontally, it considers the time of flight, from the time it was released and the time it hits the ground. Also, like mentioned earlier the time going up is the same as going down, so if we combine them the total time in flight will be twice the time.

T= 2t

T = 2 (0.782s)

<em>T = 1.564s</em>

<em>So we use this in our formula:</em>

<em>d_x=V_i_xT\\\\10.0m=Vi_x(1.564s)\\\\\dfrac{10.0m}{1.564s}=V_i_x\\\\6.39m/s=V_i_x</em>

Vix=6.39m/s

Now we have the horizontal and the vertical component, we can solve for the diagonal initial velocity, or the velocity the mountain lion leapt and the angle, by creating a right triangles, using vectors (see attached)

To get the diagonal, you just use the Pythagorean theorem:

c²=a²+b²

Using it in the context of our problem:

Vi^{2}=Viy^2+Vix^2\\Vi^2=(7.66m/s)^2+(6.39m/s)^2\\\sqrt{Vi}=\sqrt{(7.66m/s)^2+(6.39m/s)^2}\\\\Vi=9.98m/s

The lion leapt at 9.98m/s

Using SOHCAHTOA, we know that we can TOA to solve for the angle, because we have the opposite and adjacent side:

Tan\theta=\dfrac{O}{A}\\\\Tan\theta=\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{7.66m/s}{6.39m/s}\\\\\theta=50.17

The lion leapt at an angle of 50.16°.

6 0
3 years ago
How far from the castle wall does the launched rock hit the ground?
lina2011 [118]
King Arthur's knights use a catapult to launch a rock from their vantage point on top of the castle wall, 14 m above the moat. The rock is launched at a speed of 27 m/s and an angle of 32degrees above the horizontal.
4 0
3 years ago
Read 2 more answers
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