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IRISSAK [1]
3 years ago
13

A typical AA size rechargeable NiMH battery can store 1100-2100 mAh of electric charge. The small print on the battery in your h

and says 1390 mAh. Can you express this amount of electric charge using the usual coulomb units?
Physics
1 answer:
Anni [7]3 years ago
8 0

Answer:

The electric charge, q (in coulomb units) = 5004 C  

Given:

The charge stored as printed on NiMH battery, q = 1390 mAh  

Solution:

To express the amount of electric charge printed on the battery in milli-ampere-hour (mAh) in coulomb, we will do simple conversion of milli amperes in ampere and hours in seconds:

1 mA = 1\times 10^{-3}

1 hour = 60\times 60 = 3600 s

Also, we know that the rate of flow of charge is electric current, I:

I = \frac{q}{t}

⇒ q = [tex]I\times t[tex]                                   (1)

where

q = electric charge

I = current

t = time taken for flow of current

Using eqn(1), we get:

q = [tex]1390\times 10^{-3}\times 60\times 60[tex]

q = 5004 A-s = 5004 C

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Measuring the Volume of Solids. The volume of solids is expressed in cubic measurements, such as cubic centimeter or cubic meter.
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3 years ago
A 20 KeV electron emits two bremsstrahlung photons as it is being brought to rest in two successive decelerations. The wavelengt
Degger [83]

Answer:

λ₁ = 87.5 10⁻¹² m ,  λ₂ =  2.175 10⁻¹⁰ m,    E₂ = 5.8 10³ eV

Explanation:

In this case you can use the law of conservation of energy, all the energy of the electron is converted into energized emitted photons

Let's reduce to the SI system

          E₀ = 20 10³ eV (1.6 10⁻¹⁹ J / 1eV) = 3.2 10⁻¹⁵ J

          Δλ = 1.30 A = 0.13 nm = 0.13 10⁻⁹ m

          Ef = E₁ + E₂

         E₀ = Ef

         E₀ = E₁ + E₂

The energy can be found with the Planck equation

          E = h f

          c = λ f

          f = c / λ

          E = hc / λ

They indicate that the wavelength of the second photon is

 

           λ₂ =  λ₁ +0.130 10⁻⁹

We replace

           E₀ = hv / λ₁ + hc / ( λ₁ + 0.130 10⁺⁹)

           E₀ / hv = 1 / λ₁ + 1 / ( λ₁ + 0.13 10⁻⁹)

          3.2 10⁻¹⁵ / (6.63 10⁻³⁴ 3 10⁸) = ( λ₁ + 0.13 10⁻⁹ +  λ₁) /  λ₁ ( λ₁ + 0.13 10⁻⁹)

          1.6 10¹⁰ ( λ₁² +0.13 10⁻⁹  λ₁) = 2  λ₁ + 0.13 10⁻⁹

           λ₁² + 0.13 10⁻⁹  λ₁ = 1.25 10⁻¹⁰  λ₁ + 8.125 10⁻²¹

            λ₁² + 0.005 10⁻⁹  λ₁ = 8.125 10⁻²¹

            λ₁² + 5 10⁻¹²  λ₁ - 8.125 10⁻²¹ = 0

Let's solve the second degree equation

            λ₁ = [-5 10⁻¹² ±√((5 10⁻¹²)² + 4 8.125 10⁻²¹)] / 2

    λ₁ = [-5 10⁻¹² ±√(25 10⁻²⁴ +32.5 10⁻²¹)] / 2 = [-5 10⁻¹² ±√ (32525 10⁻²⁴)] / 2

             λ₁ = [-5 10⁻¹² ± 180 10⁻¹²] / 2

            λ₁ = 87.5 10⁻¹² m

             λ₂ = -92.5 10⁻¹² m

We take the positive wavelength

The wavelength of the photons is

            λ₁ = 87.5 10⁻¹² m

            λ₂ =  λ₁ + 0.13 10⁻⁹

             λ₂ = 87.5 10⁻¹² + 0.13 10⁻⁹

             λ₂ = 0.2175 10⁻⁹ m = 2.175 10⁻¹⁰ m

The energy after the first deceleration is

            E₂ = E₀ –E₁

            E₂ = E₀ –hc / λ₁

            E₂ = 3.2 10⁻¹⁵ - 6.63 10⁺³⁴ 3 10⁸ / 87.5 10⁻¹²

            E₂ = 3.2 10⁻¹⁵ - 2.27 10⁻¹⁵

             E₂ = 0.93 10⁻¹⁵ J

             E₂ = 0.93 10⁻¹⁵ J (1 eV / 1.6 10⁻¹⁹ J)

             E₂ = 5.8 10³ eV

7 0
3 years ago
In science, Bob learns that the energy of a wave is directly proportional to the square of the waves amplitude. If the energy of
KIM [24]
The energy of a wave is directly proportional to the square of the waves amplitude. Therefore, E = A² where A is the amplitude. This therefore means when the amplitude of a wave is doubled the energy will be quadrupled, when the amplitude is tripled the energy increases by a nine fold and so on.
Thus, in this case if the energy is 4J, then the amplitude will be  √4 = 2 .
  


4 0
3 years ago
A ball rolls off the end of a horizontal table that is 4 meters off the ground. It is measured that the ball lands 3 meters away
ElenaW [278]

Answer:

The speed at which the ball rolled off the end of the table is 3.3 m/s

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the table.

The position vector of the ball can be calculated as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

r = position vector.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity.

When the ball reaches the ground, its position will be:

r final = (3, -4)

Then:

3 = x0 + v0x · t

-4 = y0 + v0y · t + 1/2 · g · t²

Since the origin of the frame of reference is located at the edge of the table, x0 and y0 = 0. v0y is also 0 ( see the initial velocity vector in the figure to elucidate why). Then:

3 m = v0x · t

-4 m = 1/2 · g · t²

We can solve for "t" in the equation of the y-component and use it in the equation of the x-component to obtain v0x:

-4 m = 1/2 · g · t²

-4 m = -1/2 · 9.8 m/s² · t²

8 m / 9.8 m/s² = t²

t = 0.9 s

Then:

3 m = v0x · 0.9s

3 m/ 0.9 s = v0x

v0x = 3.3 m/s

The speed at which the ball roll off the end of the table is 3.3 m/s

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