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polet [3.4K]
3 years ago
14

Difference between universal law of gravitation and acceleration due to gravity​

Physics
1 answer:
marishachu [46]3 years ago
7 0

Explanation:

The acceleration on an object due to the gravity of any massive body is represented by g (small g). The force of attraction between any two unit masses separated by unit distance is called universal gravitational constant denoted by G(capital g). The relation between G and g is not proportional. That means they are independent entities.

G and g

In physics, G and g can be related mathematically as –

\(g=\frac{GM}{R^{2}}\)

Where,

1=g is the acceleration due to the gravity of any massive body measured in m/s2.

2=G is the universal gravitational constant measured in Nm2/kg2.

3=R is the radius of the massive body measured in km.

4=M is the mass of the massive body measured in Kg.

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What is the net charge of a metal ball if there are 21,749 extra electrons in it?
pickupchik [31]

Answer:

Q=3.47\times 10^{-15}\ C

Explanation:

Given that,

Number of extra electrons, n = 21749

We need to find the net charge on the metal ball. Let Q is the net charge.

We know that the charge on an electron is q=1.6\times 10^{-19}\ C

To find the net charge if there are n number of extra electrons is :

Q = n × q

Q=21749\times 1.6\times 10^{-19}\ C

Q=3.47\times 10^{-15}\ C

So, the net charge on the metal ball is 3.47\times 10^{-15}\ C. Hence, this is the required solution.

6 0
3 years ago
Two positive charges of 20 micro coulomb and 100 micro coulomb and the distance between them is 150cm.What will be the electrica
Roman55 [17]

Answer:

0.8 N

Explanation:

From coulomb's law,

Formula:

F = kqq'/r²........................ Equation 1

Where F = Force of repulsion, k = coulomb's constant, q = first positive charge, q' = second positive charge, r = distance between the charge.

Given: q = 20 μC = 20×10⁻⁶ C, q' = 100 μC = 100×10⁻⁶ C, r = 150 cm = 1.5 m.

Constant:  k = 9×10⁹ Nm²/C²

Substitute these values into equation 1

F = (20×10⁻⁶ )( 100×10⁻⁶)(9×10⁹)/1.5²

F = 1800×10⁻³/2.25

F = 1.8/2.25

F = 0.8 N

3 0
3 years ago
6, P 14 are consecutive terms in an AP<br>find the value of P.​
xenn [34]

In an arithmetic progression, consecutive terms differ by the same value.

So, we have

P-6 = 14-P

which reflects the fact that the difference between P and 6 must be the same than the one between P and 14.

The equation solves to

2P=20\iff P=10

And in fact, if you start with

6, 10, 14

every pair of consecutive terms differ by 4.

6 0
3 years ago
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
When finding the upper bound of the density, you put what number in the denominator?
lilavasa [31]
Answer: The result of "the upper bound of the density"  does not go on the denominator. 
So simplified, no. The answer is no.
3 0
3 years ago
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