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777dan777 [17]
3 years ago
12

IM SOO CONFUSED PLS HELP!! The mass of the nucleus is approximately EQUAL to the mass number multiplied by ____ Atomic Mass unit

.
a. 1
b. 10
c. 100
d. 1000​
Physics
1 answer:
nevsk [136]3 years ago
5 0

Answer:

option a.

Explanation:

We can think of an atom as a nucleus (where the protons and neutrons are) and some electrons orbiting it.

We also know that the mass of an electron is a lot smaller than the mass of a proton or the mass of an electron.

So, if all the protons and electrons of an atom are in the nucleus, we know that most of the mass of an atom is in the nucleus of that atom.

Then we define the mass number, which is the total number of protons and neutrons in an atom. Such that the mass of a proton (or a neutron) is almost equal to 1u

Then if we define A as the total number of protons and neutrons, and each one of these weights about 1u

(where u = atomic mass unit)

Then the weight of the nucleus is about A times 1u, or:

A*1u = A atomic mass units.

Then the correct option is:

The mass of the nucleus is approximately EQUAL to the mass number multiplied by __1__ Atomic Mass unit.

option a.

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Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

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Answer:

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