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Nikitich [7]
3 years ago
14

A pulley has an initial angular speed of 12.5 rad/s and a constant angular acceleration of 3.41 rad/s2. Through what angle does

the pulley turn in 5.26 s?
Physics
1 answer:
Novosadov [1.4K]3 years ago
4 0

Answer:

Angle turn pulley is 113 rad

Explanation:

given data

angular speed w = 12.5 rad/s

angular acceleration a 3.41 rad/s²

time t = 5.26 s

to find out

what angle does the pulley turn

solution

we get here angle by this equation that is here

angle  turn = w × t + 0.5 × a × t²     ......................1

put here value we get

angle  turn = w × t + 0.5 × a × t²

angle  turn = 12.5 × 5.26 + 0.5 × 3.41 × 5.26²

Angle turn =  65.75 + 47.17

Angle turn = 112.92 rad = 113 rad

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What are the products of linear electron flow during the light reactions of photosynthesis?
Katena32 [7]

Answer:

NADPH and ATP

Explanation:

In the clear stage the light that "hits" chlorophyll excites an electron to a higher energy level. In a series of reactions, energy is converted (throughout an electron transport process) into ATP and NADPH. Water breaks down in the process releasing oxygen as a secondary product of the reaction. ATP and NADPH are used to make the C-C bonds in the dark stage.

Photophosphorylation is the process of converting the energy of the electron excited by light into a pyrophosphate bond of an ADP molecule. This occurs when water electrons are excited by light in the presence of P680. The energy transfer is similar to the chemosmotic electron transport that occurs in the mitochondria.

Light energy causes the removal of an electron from a P680 molecule that is part of Photosystem II, the electron is transferred to an acceptor molecule (primary acceptor), and then passes downhill to Photosystem I through a conveyor chain of electrons The P680 requires an electron that is taken from the water by breaking it into H + ions and O-2 ions. These O-2 ions combine to form O2 that is released into the atmosphere.

The light acts on the P700 molecule of Photosystem I, causing an electron to be raised to a higher potential. This electron is accepted by a primary acceptor (different from the one associated with Photosystem II).

The electron goes through a series of redox reactions again, and finally combines with NADP + and H + to form NADPH, a carrier of H needed in the independent phase of light.

Electron of photosystem II replaces the excited electron of the P700 molecule.

There is therefore a continuous flow of electrons (non-cyclic) from water to NADPH, which is used for carbon fixation.

Cyclic electron flow occurs in some eukaryotes and in photosynthetic bacteria. NADPH does not occur, only ATP. This also occurs when the cell requires additional ATP, or when there is no NADP + to reduce it to NADPH.

In Photosystem II, the "pumping" of H ions into the thylakoids (from the stroma of the chloroplast) and the conversion of ADP + P to ATP is motorized by an electron gradient established in the thylakoid membrane.

7 0
3 years ago
3. Differentiate between: (a) area and volume​
Margarita [4]

Answer:

area is 2d volume is 3d

Explanation:

Area refers to the size of two-dimensional surface. Volume refers to the size of a three-dimensional space.

8 0
3 years ago
A boulder with a weight of 780 N is resting at the edge of a cliff that rises 123 m above the ground. What is the gravitational
-Dominant- [34]
Potential energy U = mgh

Given h = 123 m,
mg = F = 780 N

Then
U = (123)(780)
= 95940
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3 0
2 years ago
Which free body diagram is in equilibrium?
bearhunter [10]

A. because everything is balanced.

6 0
2 years ago
1.Suppose someone pulls a cart up a ramp a distance of 85cm along the ramp with a force of 15N.
Drupady [299]

1. 12.75 J

Assuming that the force applied is parallel to the ramp, so it is parallel to the displacement of the cart, the work done by the force is

W=Fd

where

F = 15 N is the magnitude of the force

d = 85 cm = 0.85 m is the displacement of the cart

Substituting in the formula, we get

W=(15 N)(0.85 m)=12.75 J


2. 10.6 N

In this part, the cart reaches the same vertical height as in part A. This means that the same work has been done (because the work done is equal to the gain in gravitational potential energy of the object: but if the vertical height reached is the same, then the gain in gravitational potential energy is the same, so the work done must be the same).

Therefore, the work done is

W=Fd=12.75 J

However, in this case the displacement is

d = 120 cm = 1.20 m

Therefore, the magnitude of the force in this case is

F=\frac{W}{d}=\frac{12.75 J}{1.20 m}=10.6 N

3 0
3 years ago
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