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aliina [53]
3 years ago
12

How is tidal energy obtained

Physics
1 answer:
yKpoI14uk [10]3 years ago
7 0
Under water turbans that are placed at the above to middle of the ocean they are used to capture kinetic motion
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Someone please help with this question
Alex787 [66]
The first one might be faunal succsession and the 2nd one might be metamorphic rock
5 0
3 years ago
Given the isotope 2Fes, which has an actual mass of 55.934939 u: a) b) Determine the mass defect of the nucleus in atomic mass u
SSSSS [86.1K]

Answer:

Mass defect of each iron-56 nuclei:

The binding energy per nucleon of Iron-56 is approximately 8.6 MeV.

Explanation:

According to the physics constants table on Chemistry Libretexts:

  • Proton rest mass: \rm 1.0072765\;amu;
  • Neutron rest mass: \rm 1.0086649\; amu.
  • Speed of light in vacuum: \rm 2.99792458\times 10^{8}\;m\cdot s^{-1}.
  • Charge on an electron: \rm 1.6021765\times 10^{-19}\;C.

<h3>a)</h3>

The mass defect of a nucleus is equal to the sum of the mass of its parts (protons and, in most cases, neutrons) minus the mass of the nucleus.

The atomic number of iron is 26. There are 26 protons in each iron-56 nucleus. The mass number 56 indicates that there are 56 nucleons (neutrons and protons) in each iron-56 nucleus. The other 56 - 26 = 30 particles are neutrons.

The mass of protons and neutrons in each iron-56 nucleus will be:

\rm 26 \times 1.0072765 + 30 \times 1.0086649 = 56.464736\;amu.

According to this question, the mass of an iron-56 nucleus is equal to 55.934939 amu. The mass defect will be

\rm 56.464736 - 55.934939 = 0.514197\;amu.

<h3>b)</h3>

By the mass-energy equivalence,

E = m\cdot c^{2}.

Refer to this equation, the speed of light in vacuum c^{2} is the conversion factor between mass m and energy E. The value of c is usually given only in SI units \rm m\cdot s^{-1}. Accordingly, the value of c^{2} will be in the SI unit \rm m^{2}\cdot s^{-2} = J\cdot kg^{-1}.

Convert million electron-volts to joules.

One electron-volt is equal to the electrical work done moving an electron across a potential difference of one volt.  

\begin{aligned}\rm 1 MeV&= \rm 10^{6}\; eV\\ &= \rm (10^{6}\times 1.6021765\times 10^{-19}\;C)\times 1\; V\\&=\rm 1.6021765\times 10^{-19}\;J\end{aligned}.

Convert the unit of c^{2} from \rm m^{2}\cdot s^{-2} = J\cdot kg^{-1} to the desired \rm MeV \cdot amu^{-1}:

\begin{aligned}c^{2} &= \rm {\left(2.99792458\times 10^{8}\;m\cdot s^{-1}\right)}^{2}\\&=\rm 8.987551787\times 10^{16}\; m^{2}\cdot s^{-2}\\ &= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\\&= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\times \frac{1\;MeV}{1.6021765\times 10^{-13}\;J}\times \frac{1\times 10^{-3}\;kg}{6.022142\times 10^{23}\;amu}\\&\approx \rm 931.602164\;MeV\cdot amu^{-1}\end{aligned}.

Total binding energy in each iron-56 nucleus:

\begin{aligned}E &= m\cdot c^{2}\\&= \rm 0.514197\;amu \times 9.31602164\;MeV\cdot amu^{-1} \\&=\rm 479.027038\; MeV \end{aligned}.

Again, the mass number 56 indicates that there are 56 nucleons in each iron-56 nucleus. The binding energy per nucleon of iron-56 \mathrm{^{56}Fe} will be:

\displaystyle \rm \frac{479.027038\; MeV}{56} \approx 8.6\; MeV.

6 0
3 years ago
A metal ion (X) with a charge of 4+ is attracted to a nonmetal ion (Z) with a
Keith_Richards [23]

Answer:

A) X3Z4

Explanation:

8 0
3 years ago
Tính chỉ số ampe kế thay ampe kế bằng vô kế có điện trở rất lớn thì vô kế chỉ bao nhiêu
ser-zykov [4K]

Answer:

I don't understand it.....

8 0
2 years ago
A proton moving at 3.0 × 10^4 m/s is projected at an angle of 30° above a horizontal plane. If an electric field of 400 N/C is a
GuDViN [60]

Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}

T = 7.83 X10⁻⁷ s

6 0
3 years ago
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