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3 years ago

Downward force of magnitude 5 N isexerted on the book by the force ofthetablegravityinertia .

1 answer:

4 0

Downward force of magnitude 5 N is exerted on the book by the force of <em>gravity</em>. We call that the "weight" of the book.

At the same time, upward force of 5 N is exerted on the book by the table. This one is called the "normal force".

Since the vertical forces on the book are 'balanced' (add up to zero), the book just lays there on the table, and does not accelerate.

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Nonmetals in the periodic table have a negative oxidation number. Which statement best explains why?

Tema [17]

They have a negative oxidation number because nonmetals gain electrons, thus making them negatively charged ions (anions). Metals become cations and have a positive charge because they lose electrons.

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4 years ago

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

3 0

3 years ago

For a sine wave depicting simple harmonic motion, the smaller the amplitude of the wave, the smaller the of the pendulum from th

stiks02 [169]

Displacement , shorter

7 0

4 years ago

Read 2 more answers

A proton, traveling with a velocity of 4.6 × 106 m/s due east, experiences a magnetic force that has a maximum magnitude of 6.0

Kobotan [32]

Answer:

$B = 0.0815 T$

since the direction of force is along south so the magnetic field must be upwards

Explanation:

Force on moving charge in magnetic field is given as

$F = q(\vec v \times \vec B)$

here we know this is force on proton

so we have

$q = 1.6 \times 10^{-19} C$

$F = 6.0 \times 10^{-14} (-\hat j) N$

also we know that the velocity of charge is

$v = 4.6 \times 10^6 \hat i m/s$

now from above formula we have

$(6.0 \times 10^{-14}) = (1.6 \times 10^{-19})(4.6 \times 10^6)B$

$B = 0.0815 T$

since the direction of force is along south so the magnetic field must be upwards

6 0

4 years ago

An oxygen molecule consists of two oxygen nuclei, each of massm= 2.7×10−26kg,separated by a distance 1.2×10−10m, and surrounded

fenix001 [56]

Answer:

a) I = 1,944 10⁻⁴⁶ Kg m²

, b) I = 7,915 10⁻⁵¹ Kg m²

Explanation:

a) The moment of inertia of point masses is

I = m r²

The nuclei have a very small size (10-15 m) so we can consider them punctual.

The distance to the center of mass passing through the middle of the two nuclei of equal mass

r = 0.6 10⁻¹⁰ m

The moment of total inertia

I = I₁ + I₂ = 2 I₀

I = 2 2.7 10⁻²⁶ (0.6 10⁻¹⁰)²

I = 1,944 10⁻⁴⁶ Kg m²

The kinetic energy of the rotation is

w = h / 2π

K = ½ I w²

K = ½ 1,944 10⁻⁴⁶ (h / 2π)² = ½ 1.944 10⁻⁴⁶ (6.63 10⁻³⁴ / 2π)²

K = 2.16 10⁻¹¹⁴ J (1eV / 1.6 10⁻¹⁹ J)

K = 2.16 10⁻⁹⁵ eV

B) the moment of inertia of the electron in the orbit, we can calculate it with the parallel axis theorem

I = $I_{cm}$ + m R²

I = $m_{e}$ r² + $m_{e}$ R²

Where R we calculate it by Pythagoras

R² = (0.6 10⁻¹⁰)² + (0.5 10⁻¹⁰) 2

R = √ (0.61 10⁻²⁰)

R = 0.78 10⁻¹⁰ m

I = 9.1 10⁻³¹ (0.5 10⁻¹⁰)² + 9.1 10⁻³¹ (0.78 10⁻¹⁰)²

I = (2,275 +5.54) 10⁻⁵¹

I = 7,915 10⁻⁵¹ Kg m²

The rotation energy of the electron

K = ½ I w²

If the angular velocity is the electrons outside the core, its kinetic energy is much lower by an order 10⁵, but the angular velocity of the electrons is much higher.

7 0

3 years ago