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maksim [4K]
4 years ago
11

Electrons are important to electric current because they are able to

Physics
1 answer:
Agata [3.3K]4 years ago
4 0
Electrons are important to the electric current because they are able to move from one atom to another. When an atom loses an electron, it becomes positively charged and when an atom gains an electron, it becomes negatively charged.
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What does a tachometer measure?
nlexa [21]
A tachometer measures, for example it measures the MPH(Miles Per Hour) in a car.  So like it measures your speed, so you could be going 100 MPH.
6 0
4 years ago
5. the combustion of a single molecule of methane produces about 10 ev of energy. a methane molecule has a mass of 16 amu. the f
Alchen [17]

The mass of methane will produce as much energy as a single gram of uranium-235 1,287 kilograms of methane. Option D is correct.

<h3>What is uranium?</h3>

Uranium, with the atomic number 92 and the symbol U in the periodic table, is a weakly radioactive element. One of the heavy metals that may be used as a concentrated energy source is uranium.

Given data;

⇒One mole of U-235 = 235 grams

=> 1 gram of U-235 = 1/235 moles

⇒1 mole of U-235 = 6.023 x 10²³ atoms

The no of atoms is;

=> 1/235 moles of U-235

⇒N = 6.023 x 10²³/235 atoms

⇒N=25.6 x 10²⁰ atoms

If one atom of U-235 gives is 189 x 10⁶ eV energy,25.6 x 10²⁰ atoms of U-235 gives;

=> 25.6 x 10²⁰ atoms of U-235 gives;

E = 25.6 x 10²⁰ x 189 x 10⁶

E= 4.8 x 10²⁹ eV energy

One methane molecule produces;

E = 10 eV of energy

=> To produce 4.8 x 10²⁹ eV energy, no. of molecules required;

= 4.8 x 10²⁸

6.032 x 10²³ molecules of methane = 16 gms

=> 4.8 x 10²⁸ molecules of methane = 16 x 4.8 x 10²⁸/ 6.032 x 10²³ gms

m= 12.75 x 10⁵ gms

m= 1275 kilograms of methane

m≅ 1287 kilograms of methane

1 gram of U-235 has the same amount of energy as 1287 kilograms of methane.

Hence option D is correct,

To learn more about uranium refer to the link;

brainly.com/question/9099776

#SPJ1

7 0
2 years ago
The electrons that produce the picture in a
vredina [299]
Given:
u = 10⁵ m/s, the entrance velocity
v = 2.5 x 10⁶ m/s, the exit velocity
s = 1.6 cm = 0.016 m, distance traveled

Let a = the acceleration.
Then
u² + 2as = v²
(10⁵ m/s)² + 2*(a m/s²)*(0.016 m) = (2.5 x 10⁶ m/s)²
0.032a = 6.25 x 10¹² - 10¹⁰ = 6.24 x 10¹²
a = 1.95 x 10¹⁴ m/s²

Answer: 1.95 x 10¹⁴ m/s²

4 0
3 years ago
In an incompressible three-dimensional flow field, the velocity components are given by u = ax + byz; υ = cy + dxz. Determine th
Ksivusya [100]

An incompressible flow field F in a 3D cartesian grid with components u,v,w:

F = u + v + w

where u,v,w are functions of x,y,z

Must satisfy:

∇·F = du/dx + dv/dy + dw/dz = 0

We have a field F defined:

F = u+v+w, u = ax+byz, v = cy+dxz

du/dx = a, dv/dy = c

Recall ∇·F = 0:

∇·F = du/dx + dv/dy + dw/dz = 0

a + c + dw/dz = 0

dw/dz = -a-c

Solve for w by separation of variables:

w = ∫(-a-c)dz

w = -az - cz + f(x,y)

f(x,y) is some undetermined function of x and y

The question states that w is not a function of x and y, therefore f(x,y) = 0...

w = -az - cz

8 0
3 years ago
In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10
sergij07 [2.7K]

Answer:

A)\Phi=83.84\times 10^{-9}

B)\Phi=0 Wb

C)emf=5.4090\times 10^{-4}V

Explanation:

Given that:

  • no. of turns i the coil, n=200
  • area of the coil, a=13.1 \times 10^{-4}\,m^2
  • time interval of rotation, t=3.1\times 10^{-2}\,s
  • intensity of magnetic field, B=6.4\times 10^{-5}\,T

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

\Phi=B.a\,cos \theta..................................(1)

\theta is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}

\Phi=83.84\times 10^{-9} Wb

(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴  \theta=90^{\circ}

From eq. (1)

\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}

\Phi=0 Wb

(C)

According to the Faraday's Law we have:

emf=n\frac{B.a}{t}

emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}

emf=5.4090\times 10^{-4}V

7 0
3 years ago
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