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LuckyWell [14K]
3 years ago
13

The rotating drum in a clothes-dryer has a radius of 0.31 m. If the acceleration at the rim of the drum is 27 m/s2, what is the

tangential speed of the rim?
Physics
1 answer:
olga nikolaevna [1]3 years ago
3 0

Answer:

v = 2.89 m / s

Explanation:

This is a kinematics exercise, the centripetal acceleration is

         a = v² / r

where a is the acceleration, v is the velocity and r the radius

let's clear

       v = √a r

let's calculate

       v = √ (27 0.31)

       

       v = 2.89 m / s

this is the speed of the drum which is constant

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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
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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

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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
A nonconducting solid sphere of radius 8.40 cm has a uniform volume charge density. The magnitude of the electric field at 16.8
mina [271]

Answer:

The sphere's volume charge density is 2.58 μC/m³.

Explanation:

Given that,

Radius of sphere R= 8.40 cm

Electric field E= 2.04\times10^{3}\ N/C

Distance r= 16.8 cm

We need to calculate the sphere's volume charge density

Using Gauss's law

\int{\vec{E}\cdot\vec{da}}=\dfrac{Q_{enc}}{\epsilon_{0}}

E\times 4\pi r^2=\dfrac{1}{\epsilon_{0}}\times\dfrac{4}{3}\piR^3\rho

E=\dfrac{\rho R^3}{3\epsilon_{0}r^2}

\rho=\dfrac{3\times E\times\epsilon_{0}r^2}{R^3}

Put the value into the formula

\rho=\dfrac{3\times2.04\times10^{3}\times8.85\times10^{-12}\times(16.8\times10^{-2})^2}{(8.40\times10^{-2})^3}

\rho=2.58\times10^{-6}\ C/m^3

\rho=2.58\ \mu C/m^3

Hence, The sphere's volume charge density is 2.58 μC/m³.

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While using a digital radiography system, suppose a radiographer uses exposure factors of 10 mAs and 70 kVp with an 8:1 grid for
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Answer:

b. 12.5 mAs, 70 kVp

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Given that the lead content in the 12:1 grid, is higher than the lead content in 8:1 grid and that 12:1 grid needs more mAs to compensate, and provides a higher image contrast, the amount of extra mAs is given by the Grid Conversion Factors, GCF, as follows;

The GCF for G.R. 8:1 = 4

The GCF for G.R. 12:1 = 5

Therefore, given that the mAs used by the radiographer for 8:1 Grid Ratio is 10 mAs, the mAs required for a G.R. of 12:1 in order to maintain the same exposure is given as follows;

mAs for G.R. of 12:1 = 10 mAs × 5/4 = 12.5 mAs

Therefore the new exposure factors are;

12.5 mAs, 70 kVp

5 0
3 years ago
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