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kodGreya [7K]
3 years ago
9

Much of the energy released in catabolic reactions is captured in ATP for use in other reactions. When the phosphate is transfer

red directly from an organic molecule to ADP in order to produce ATP without an energized membrane, what has occurred?
a. substrate-level phosphorylation
b. oxidative phosphorylation
c. noncyclic photophosphorylation
d. cyclic photophosphorylation
Physics
1 answer:
Charra [1.4K]3 years ago
6 0

Answer:

A substrate-level phosphorylation

Explanation:

Large amount of energy released in catabolic reactions is captured in ATP for use in other reactions. The conversion of ADP into ATP With transfer of phosphate from ADP without an energized membrane. This process is called  substrate-level phosphorylation.

substrate-level phosphorylation is metabolic reaction in which ADP is converted to ATP by direct transfer of Phosphoryl Group (PO3) to ADP.

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In 2007, michael carter (u.s.) set a world record in the shot put with a throw of 24.77 m. what was the initial speed of the sho
Serga [27]

The initial speed of the shot is 15.02 m/s.

The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.

Pl refer to the attached diagram.

Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.

Write an expression for R.

R=u_xt=(ucos \theta)t

Therefore,

t=\frac{R}{ucos\theta} .......(1)

In the time t, the net displacement of the shotput is y in the downward direction.

Use the equation of motion,

y=u_yt-\frac{1}{2}gt^2=(usin\theta) t-\frac{1}{2}gt^2

Substitute the value of t from equation (1).

y=(ucos\theta)(\frac{R}{ucos\theta} )-\frac{1}{2} g(\frac{R}{ucos\theta} )^2\\ =Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

y=Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )\\ (-2.10m)=(24.77 m)(tan38.0^o)-\frac{(9.8 m/s^2)(24.77m)^2}{2u^2(cos38.0^o)^2} \\ u^2=225.71(m/s)^2\\ u=15.02m/s

The shot put was thrown with a speed 15.02 m/s.




7 0
3 years ago
The fluid in a grdulated cylinder should be read at the _____ of the meniscus.
galina1969 [7]
<span>The fluid in a graduated cylinder should be read at the BOTTOM of the meniscus.</span>
5 0
3 years ago
Read 2 more answers
Does light have to have a medium to pass through?
skad [1K]

Answer:

Light does not require any medium to travel because light is a transverse wave

hope it helps

7 0
2 years ago
An oceanographer is studying how the ion concentration in seawater depends on depth. She makes a measurement by lowering into th
Black_prince [1.1K]

Answer:

a)  R = ρ₀ L /π(r_b² - R_a²) , b)  ρ₀ = V / I    π (r_b² - R_a²) / L

Explanation:

a) The resistance of a material is given by

          R = ρ l / A

where ρ is the resistivity, l is the length and A is the area

the length is l = L and the resistivity is ρ = ρ₀

the area is the area of ​​the cylindrical shell

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we substitute

         R = ρ₀ L /π(r_b² - R_a²)

b) The potential difference is related to current and resistance by ohm's law

         V = i R

         

we subsist the expression of resistance

          V = I ρ₀ L /π (r_b² - R_a²)

           ρ₀ = V / I    π (r_b² - R_a²) / L

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3 years ago
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I'm guessing that you're supposed to enter whatever the expression becomes
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Just taking a wild guess here . . . . .

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But that's just a wild guess.  As I pointed out, you cut off
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6 0
3 years ago
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