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LiRa [457]
3 years ago
7

The unit for atomic mass is A. Gram B. Amu C. Pound D. None of the above

Physics
2 answers:
Colt1911 [192]3 years ago
7 0

Answer:

The unit of atomic mass is Amu.

(B) is correct option.

Explanation:

Atomic mass :

The mass of a single atom of a chemical element.

An atom made by the proton, electron and neutron.

One atomic mass unit is explained by one by twelve of the mass of a single carbon - twelve atom.

We know that,

Gram is the unit of mass in C.G.S unit.

Amu is the unit of atomic mass.

Pound is the unit of mass in F.P.S units.

Hence, The unit of atomic mass is Amu.

DIA [1.3K]3 years ago
6 0

It's measured in Amu, so the answer is B.

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The other options except Option is not applicable since the gravitational force is a long range force, in which the satellite revolves very close to the surface of the Earth where the gravity is felt.The zero weight experienced by the astronaut in a satellite is due to the earth pulling along with satellite. Due to gravitational force of the Earth,the astronaut falls freely .

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You run from your friends house that is 1k away. You then walk home. What distance did you travel?
kolezko [41]

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Suppose you have a total charge qtot that you can split in any manner. Once split, the separation distance is fixed. How do you
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Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

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F = \frac{Kq1q2}{r^{2} }

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Product of Charges = q x (Q-q)

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Answer:

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T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

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But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

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R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

T_p=64*365.25\\T_p=23376days

4 0
3 years ago
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