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3 years ago

If the mass of 293nv is 293.15 amu and that of 295nv is 295.30 amu , what is the atomic weight of nv?

1 answer:

5 0

Honestly, I am quite confused with what Nv stands for because there is no element with that symbol. However, I still get the concept of finding the average molecular mass of an element. Let's just assume that nv stands for a specific type of element and it has two isotopes: nv-293 and nv-295. Isotopes have the same number of protons but differ in mass number (protons+neutrons).

To find the average atomic weight, just multiply the individual weights with the respective composition of the isotope. Since there are only two isotopes, they constitute 50% each. So, the average atomic weight is

(50%)(293.15 amu) + (50%)(<span>295.30 amu) = 294.225 amu

Hence, the atomic weight of nv is 294.225 atomic mass units.</span>

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3 years ago

The distance between two particles is 2 centimeters. If the distance is increased to 4 centimeters, the force will be ?

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Answer:

The new force is 1/4 of the previous force.

Explanation:

Given

$Initial\ Distance = 2cm$ ---- $r_1$

$New\ Distance = 4cm$ --- $r_2$

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Determine the new force

Let the two particles be q1 and q2.

The initial force F1 is:

$F_1 = \frac{kq_1q_2}{r_1^2}$ --- Coulomb's law

Substitute 2 for r1

$F_1 = \frac{kq_1q_2}{2^2}$

$F_1 = \frac{kq_1q_2}{4}$

The new force (F2) is

$F_2 = \frac{kq_1q_2}{r_2^2}$

Substitute 4 for r2

$F_2 = \frac{kq_1q_2}{4^2}$

$F_2 = \frac{kq_1q_2}{4*4}$

$F_2 = \frac{1}{4}*\frac{kq_1q_2}{4}$

Substitute $F_1 = \frac{kq_1q_2}{4}$

$F_2 = \frac{1}{4}*F_1$

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2 years ago

An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. Find the average mag

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Answer:

The induced emf in the loop is $7.35\times 10^{-4}\ V$

Explanation:

Given that,

Length of the wire, L = 1.22 m

It changes its shape is changed from square to circular. Then the side of square be its circumference, 4a = L

4a = 1.22

a = 0.305 m

Area of square, $A=a^2=(0.305)^2=0.0930\ m^2$

Circumference of the loop,

$C=2\pi r=L\\\\r=\dfrac{L}{2\pi}\\\\r=\dfrac{1.22}{2\pi}=0.194\ m$

Area of circle,

$A'=\pi r^2\\A'=\pi (0.194)^2\\\\A'=0.118\ m^2$

The induced emf is given by :

$\epsilon=\dfrac{\d\phi}{dt}\\\\\epsilon=\dfrac{\d(BA)}{dt}\\\\\epsilon=B\dfrac{A'-A}{t}\\\\\epsilon=0.125 \times \dfrac{0.118-0.0930}{4.25}\\\\\epsilon=7.35\times 10^{-4}\ V$

So, the induced emf in the loop is $7.35\times 10^{-4}\ V$

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