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LekaFEV [45]
3 years ago
10

Using the videos and websites provided, or other online resources you find, and analyze the events of the Biosphere 2 story and

history.
Biosphere 2 Research Resources

A: Arizona's Beautiful Biodome Failure
B: Biosphere 2: 25 Years Scientists Lived Inside
C: TED Talk about life in Biosphere 2
D: Biosphere 2: Not Such a Bust
E: Biosphere 2: An American Space Odyssey


In the chart below:
Decide which events, components, or aspects contributed to the success or failure of Biosphere 2. You must include a minimum of 5 events, components, aspects of the project.

Provide evidence of these successes or failures from your research.

Include the citation of the resource where you found this information.


Biosphere event, component, or aspect (ex: one Biospherian needing to leave temporarily)
Min. of 5 required
Success (S) or Failure (F)?
Evidence of success or failure from the provided resources
Citation (Letter)
Physics
2 answers:
Andru [333]3 years ago
4 0
Your answer has to be B
Likurg_2 [28]3 years ago
4 0
Get an app called SOCRATIC works like a charm
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7. The energy source used to produce most of the electrical energy in the United States is ____________________.
Fudgin [204]

Fossil fuels like coal, natural gas and oil.

Hopes this helps!

4 0
3 years ago
Read 2 more answers
~~~NEED HELP ASAP~~~<br>Please solve each section and show all work for each section.
anastassius [24]

Explanation:

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass m_A as

x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)

y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)

Substituting (2) into (1), we get

\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)

where f_N= \mu_kN, the frictional force on m_A. Set this aside for now and let's look at the forces on m_B

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>

Let the x-axis be (+) up along the inclined plane. We can write the forces on m_B as

x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)

y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)

From (5), we can solve for <em>N</em> as

N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension<em> </em><em>T</em><em> </em> is given by

T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)

Substituting (7) into (4) we get

m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba

Collecting similar terms together, we get

(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag

or

a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)

Putting in the numbers, we find that a = 1.4\:\text{m/s}. To find the tension <em>T</em>, put the value for the acceleration into (7) and we'll get T = 21.3\:\text{N}. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get N = 50.9\:\text{N}

8 0
3 years ago
A simple experiment to measure the speed of sound doesn't involve a stopwatch. You can fill up along tube with water and put a t
Serhud [2]

Answer:

Explanation:

In order to answer this problem you have to know the depth of the column, we say R, this information is important because allows you to compute some harmonic of the tube. With this information you can compute the depth of the colum of air, by taking tino account that the new depth is R-L.

To find the fundamental mode you use:

f_n=\frac{nv_s}{4L}

n: mode of the sound

vs: sound speed

L: length of the column of air in the tube.

A) The fundamental mode id obtained for n=1:

f_1=\frac{v_s}{4L}

B) For the 3rd harmonic you have:

f_3=\frac{3v_s}{4L}

C) For the 2nd harmonic:

f_2=\frac{2v_s}{4L}

7 0
2 years ago
If you drop a rock off a cliff and it hits the ground 3 seconds later how tall is that cliff
skelet666 [1.2K]

Answer:

Basic kinematics, negating drag and assuming ideal conditions, we use the equation:

d=vi*t+1/2*a*t^2

Since vi is 0 (we know this because you’re dropping it, not throwing it)…

…and the only acceleration acting on it is gravity, a=9.8 m/s^2…

…we get

d=1/2(9.8)(5)^2

Explanation:

Some quick mental math tells us that this is about 125 m.

Plugging it in, we find it to be 122.5 m.

6 0
3 years ago
A 2,537-kg truck moving at 14 m/s strikes a car waiting at a traffic light, hooking bumpers. The two continue to move together a
slavikrds [6]

Answer:

1902.75 kg

Explanation:

From Law of conservation of momentum,

m₁u₁ + m₂u₂ = V (m₁ + m₂).................... Equation 1

make m₂ the subject of the equation,

m₂ = (m₁V - m₁u₁)/(u₂-V)..................... Equation 2

Where m₁ = mass of the truck, m₂ = mass of the car, u₁ initial velocity of the truck, u₂ = initial velocity of the car V = common velocity

Given: m₁ = 2537 kg, u₁ = 14, V= 8 m/s, u₂ = 0 m/s ( as the car was at rest waiting at a traffic light)

Substituting into equation 2.

m₂ =[2537(8) - 2537(14)]/(0-8)

m₂ = (20296-35518)/-8

m₂ = -15222/-8

m₂ = 1902.75 kg.

Thus the mass of the car = 1902.75 kg

5 0
3 years ago
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