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AnnyKZ [126]
2 years ago
15

What is the mass moment of inertia of a 20kg sphere with a radius of 0.2m about a point on the sphere's perimeter

Physics
1 answer:
Kobotan [32]2 years ago
7 0

Answer:

I = M R^2 is the moment of inertia about a point that is a distance R from the center of mass (uniform distributed mass).

The moment  of inertia about the center of a sphere is 2 / 5 M R^2.

By the parallel axis theorem the moment of inertia about a point on the rim of the sphere is  I = 2/5 M R^2 + M R^2 = 7/5 M R^2

I = 7/5 * 20 kg * .2^2 m = 1.12 kg m^2

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Americium-241 is a radioactive substance used in smoke detectors. The half life of americium is 432 years. If a smoke detector i
romanna [79]

Answer:

0.5 g

Explanation:

The half-life of a radioactive isotope is the time it takes for a certain amount of that isotope to halve.

In this case, we have an isotope of Americium-241, which has a half-life of

\tau = 432 y

This means that after 1 half-life (432 years), there will be an amount of americium equal to half its initial amount.

In this problem, the initial amount of americium is

m_0 = 1 g

Therefore, after 432 years (1 half-life), the amount of americium left will be half:

m(432y) = \frac{m_0}{2}=0.5 g

7 0
3 years ago
A 1.31 kg object is attached to a horizontal spring of force constant 2.70 N/cm and is started oscillating by pulling it 6.20 cm
goldfiish [28.3K]

Answer:

The answer is below

Explanation:

a) The change in energy is the difference between the final energy and the initial energy.

ΔE (energy change) = Ef (final energy) - Ei (initial energy)

\Delta E=\frac{1}{2}kA_f^2 -\frac{1}{2}kA_i^2\\\\k=force\ constant=2.7\ N/cm=270\ N/m, A_f=final\ dispalacment= 3.7\ cm=0.037\ m,\\ A_i=initial \ displacement = 6.2\ cm=0.062\ m\\\\Hence:\\\\\Delta E=\frac{1}{2}(270)(0.037)^2 -\frac{1}{2}(270)(0.062)^2\\\\\Delta E=-0.334 \ J

The negative sign shows that energy is lost to the environment. Hence 0.334 J is lost to the environment.

b) According to the law of conservation of energy, energy cannot be created or destroyed but transformed from one form to another.

The oscillating object loses energy due to wind resistance, friction between the spring and the object. Given that the air is frictionless, hence the energy loss is due to friction which is converted to heat.

6 0
3 years ago
The magnitude J of the current density in a certain wire with a circular cross section of radius R = 2.11 mm is given by J = (3.
oksano4ka [1.4K]

Answer:

i = 2.84 \times 10^{-3} A

Explanation:

As we know that current density is ratio of current and area of the crossection

now we have

J = \frac{di}{dA}

so the current through the wire is given as

i = \int J dA

now we have

i = \int_{0.921R}^R J dA

here we have

J = (3.25 \times 10^8)r^2

now plug in the values in above equation

i = \int_{0.921R}^R (3.25 \times 10^8)r^2 2\pi r dr

now we have

i = \int_{0.921R}^R 2\pi (3.25 \times 10^8)r^3 dr

i = (2.04 \times 10^9) \frac{r^4}{4}

now plug in both limits as mentioned

i = (2.04 \times 10^9)(\frac{R^4}{4} - \frac{(0.921R)^4}{4})

i = (2.04\times 10^9)(0.07 R^4)

here R = 2.11 mm

i = (2.04 \times 10^9)(0.07 (2.11 \times 10^{-3})^4)

i = 2.84 \times 10^{-3} A

8 0
3 years ago
Someone please answer please don’t take my points.
Anni [7]

Answer:

bro today my teacher will teach this answer only when I will be getting the answer I will copy and paste

Explanation:

wait for sometimes

5 0
3 years ago
The potential difference between two parallel conducting plates in vacuum is 165 V. An alpha particle with mass of 6.50×10-27 kg
shepuryov [24]

Answer:

kinetic energy (K.E) = 5.28 ×10⁻¹⁷            

Explanation:

Given:

Mass of  α particle (m) = 6.50 × 10⁻²⁷ kg

Charge of  α particle (q) = 3.20 × 10⁻¹⁹ C

Potential difference ΔV = 165 V

Find:

kinetic energy (K.E)

Computation:

kinetic energy (K.E) = (ΔV)(q)

kinetic energy (K.E) = (165)(3.20×10⁻¹⁹)

kinetic energy (K.E) = 528 (10⁻¹⁹)

kinetic energy (K.E) = 5.28 ×10⁻¹⁷              

4 0
3 years ago
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