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skad [1K]
3 years ago
11

A diagram of the carbon-oxygen cycle is shown below.

Physics
2 answers:
goldfiish [28.3K]3 years ago
8 0
B.plants is responsible of removing carbon dioxide
shutvik [7]3 years ago
8 0

Answer:its plants factories and animals

Explanation:

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On March 21, a stick casts the following shadow. What is the most likely time of day?
jenyasd209 [6]

Answer:

9:00 AM

Explanation:

I took the test and that was the answer

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3 years ago
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Which of the following is not a galilean moon?
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The answer is B) titan
7 0
3 years ago
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Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma
AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0
4 years ago
A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac
RoseWind [281]

Answer:

a) \Delta U_{g} = 12.945\,J, b) \Delta U_{k} = 12.945\,J, c) k = 2930.059\,\frac{N}{m}

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)

\Delta U_{g} = 12.945\,J

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

\Delta U_{k} = 12.945\,J

c) The spring constant of the gun is:

\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}

k = \frac{2\cdot \Delta U_{k}}{x^{2}}

k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}

k = 2930.059\,\frac{N}{m}

4 0
3 years ago
On a popular amusement park ride, the riders stand along the inside wall of a large cylinder with a diameter of 8.27 m. The cyli
____ [38]

Answer:

Explanation:

D = 8.27 m   ⇒  R = D / 2 = 8.27 m / 2 = 4.135 m

ω = 0.66 rev/sec = (0.66 rev/sec)*(2π rad/1 rev) = 4.1469 rad/s

We can apply the equation

Ff = W  ⇒  μ*N = m*g   <em>(I)</em>

then we have

N = Fc = m*ac = m*(ω²*R)

Returning to the equation <em>I</em>

<em />

μ*N = m*g    ⇒    μ*m*ω²*R = m*g   ⇒ μ = g / (ω²*R)

Finally

μ = (9.81 m/s²) / ((4.1469 rad/s)²*4.135 m) = 0.1379

3 0
3 years ago
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