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3 years ago

In the oxygen cycle, oxygen is

Physics

1 answer:

dlinn [17]3 years ago

7 0

Answer:

4.) added to the atmosphere by photosynthesis and removed by cellular respiration and combustion.

Explanation:

The atmospheric cycle is a biogeochemical cycle and is a transition between the oxidation state in ions, oxides, and molecules. The largest reservoir on the earth for oxygen island or lithosphere.

The crust and mantle together make up 99.5% of oxygen by volume. The rest is found in the air that 21%, 22% in the biosphere, and 33% in the hydrosphere.

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If an object is thrown upward at 128 feet per second from a height of 76 feet, its height S after t seconds is given by the foll

faltersainse [42]

<h2>a) Average velocity in first 4 seconds is 64 ft/s upward</h2><h2>b) Average velocity in second 4 seconds is 63.5 ft/s downward</h2>

Explanation:

a) Given S(t) = 76 + 128t − 16t²

s(0) = 76 + 128 x 0 − 16 x 0² = 76 ft

s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

Displacement in 4 seconds = 332 - 76 = 256 ft

Time = 4 - 0 = 4 s

$\texttt{Velocity = }\frac{256}{4}=64ft/s$

Average velocity in first 4 seconds is 64 ft/s upward

a) Given S(t) = 76 + 128t − 16t²

s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

s(8) = 76 + 128 x 8 − 16 x 8² = 78 ft

Displacement in 4 seconds = 78 - 332 = -254 ft

Time = 4 - 0 = 4 s

$\texttt{Velocity = }\frac{-254}{4}=-63.5ft/s$

Average velocity in second 4 seconds is 63.5 ft/s downward

3 0

3 years ago

The transfer of energy when particles of a fluid move from one place to another is called

siniylev [52]

Answer:

Answer below!!!!

Explanation:

Convection is the transfer of thermal energy by particles moving through a fluid.

Hope I Helped!!!

;)

5 0

3 years ago

A tree loses water to the air by the process of transpiration at the rate of 110 g/h. This water is replaced by the upward flow

Ksivusya [100]

Answer:

The answer is 1.87nm/s.

Explanation:

The $110g/hr$ water loss must be replaced by $110g/hr$ of sap. 110g of sap corresponds to a volume of

$110g \div \dfrac{1040*10^3g}{1*10^6cm^3} = 106cm^3$

thus rate of sap replacement is

$106cm^3/hr = 106*10^{-6}m^3/3600s = 2.94*10^{-8}m^3/s$

The volume of sap in the vessel of length $x$ is

$V = Ax$ ,

where $A$ is the cross sectional area of the vessel.

For 2000 such vessels, the volume is

$V = 2000Ax$

taking the derivative of both sides we get:

$\dfrac{dV}{dt} = 2000A \dfrac{dx}{dt}$

on the left-hand-side $\dfrac{dx}{dt}$ is the velocity $v$ of the sap, and on right-hand-side $\dfrac{dV}{dt} = 2.94*10^{-8}m^3/s$ ; therefore,

$2.94*10^{-8}m^3/s=2000Av$

and since the cross-sectional area is

$A = \pi (\dfrac{100*10^{-3}m}{2} )^2 = 7.85*10^{-3}m^2$ ;

therefore,

$2.94*10^{-8}m^3/s =2000(7.85*10^{-3}m^2)v$

solving for $v$ we get:

$v = \dfrac{2.94*10^{-8}m^3/s}{2000(7.85*10^{-3}m^2)}$

$\boxed{v =1.875*10^{-9}m/s = 1.875nm/s}$

which is the upward speed of the sap in each vessel.

8 0

4 years ago

A ball is dropped off a building and falls past a window that is 2.2m long. If it takes .28s for the ball to cross the window wh

Sidana [21]

0.616 is the distance from the top
of the building to the top of the window.

8 0

2 years ago

You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int

uysha [10]

In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
$Q(t) = Q_0 (1-e^{-t/\tau})$
where t is the time, $Q_0 = CV$ is the maximum charge on the capacitor at voltage V, and $\tau = RC$ is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using $R=39.1 \Omega$ and $C= 65.7 mF=65.7\cdot 10^{-3}F$ , the time constant of the circuit is:
$\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s$

2) To find the charge on the capacitor at time $t=1.95 \tau$ , we must find before the maximum charge on the capacitor, which is
$Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C$
And then, the charge at time $t=1.95 \tau$ is equal to
$Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C$

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be $Q_0 = 0.59 C$ . We can see this also from the previous formule, by using $t=\infty$ :
$Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C$

4 0

3 years ago