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Vladimir [108]
3 years ago
12

Radium-226 emits alpha (α), beta (β) and gamma (γ) radiation.

Physics
1 answer:
Furkat [3]3 years ago
3 0

Answer:

Explanation:

alpha

Alpha Radiation (α): A large, unstable nucleus decays to produce a smaller, more stable nucleus and an alpha particle (identical to a helium nucleus, ⁴₂He or ⁴₂α).

It has a very high ionizing energy and low penetrating power. It can be stopped by paper skin

Beta Radiation (β): A neutron in an unstable nucleus decays, forming a proton and emitting a beta (β) particle (identical to an electron, ⁰₋₁e or ⁰₋₁b) and resulting in a more stable nucleus.

It has high ionizing energy and penetrating power. It can be stopped by aluminium sheet

Gamma Radiation (γ): An unstable nucleus releases energy in the form of a high energy photon (no mass)to become more stable; this often accompanies other forms of radioactivity.

It has very high penetrating power and very low ionizing energy. It can be stopped by lead block.

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Sound waves cannot travel through a medium

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Three uniform spheres of radius 2R, R, and 3R are placed in a line, in the order given, so their centers are lined up and the sp
kolezko [41]

Answer:

x = 2.33 R from the center of mass of the smallest sphere.

Explanation:

Due to the symmetry of the spheres, the center of mass of any of them, is located just in the center of the sphere.

If we align the centers of the spheres with the x-axis, the center of mass of any of them will have only coordinates on the x-axis, so the center of  mass of the system will have coordinates on the x-axis only also.

By definition, the x-coordinate of the center of mass of a set of discrete masses m₁, m₂, m₃, can be calculated as follows:

Xcm = \frac{m1*x1+m2*x2+m3*x3}{m1+m2+3}

In this case, we need to get the coordinates of the center of mass of each sphere:

If we place the spheres in such a way that the center of the first sphere has the x-coordinate equal to its radius (so it is just touching the origin), we will have:

x₁ = 2*R

For the second sphere, the center will be located at a distance equal to the diameter of  the first sphere plus its own radius, as follows:

x₂ = 4*R + R = 5*R

Finally, for the third sphere, the center will be located at a distance equal to the diameter of  the first sphere, plus the diameter of the second sphere,  plus its own radius, as follows:

x₃ = 4*R + 2*R + 3*R = 9*R

We can calculate the mass of each sphere (assuming that all are from the same material, with a constant density), as the product of the density and the volume:

m = ρ*V

For a sphere, the volume can be calculated as follows:

\frac{4}{3} *\pi *(r)^{3}

So, we can calculate the masses of the spheres, as follows:

m₁ = ρ*\frac{4}{3} *\pi *(2r)^{3}

m₂ = ρ*\frac{4}{3} *\pi *(r)^{3}

m₃ = ρ*\frac{4}{3} *\pi *(3r)^{3}

The total mass can be calculated as follows:

M= ρ*\frac{4}{3} *\pi * (8*r³ + r³ + 27*r³) =ρ*\frac{4}{3} *\pi * 36*r³

Replacing by the values, and simplifying common terms, we can calculate the x-coordinate of the center of mass of the system as follows:

Xcm = \frac{m1*x1+m2*x2+m3*x3}{m1+m2+3}

Xcm = \frac{(8*R^{3} *2*R)+(R^{3}*(5*R))+27*R^{3}*(9*R))}{36*R^{3} }=\frac{264*R^{4} x}{36*R^{3}} = 7.33 R

As the x-coordinate of the center fof mass of the entire system is located at 7.33*R from the origin, and the center of mass of the smallest sphere is located at 5*R from the origin, the center of mass of the system is located at a distance d:

d = 7.33*R - 5*R = 2.33 R

4 0
3 years ago
A bowling ball is far from uniform. Lightweight bowling balls are made of a relatively low-density core surrounded by a thin she
Leni [432]

Answer:

Part a)

I = 17.4 \times 10^{-3} kg m^2

Part b)

I = 14.9 \times 10^{-3} kg m^2

Explanation:

Part a)

Moment of inertia of the core of the ball

I_1 = \frac{2}{5}m_1r_1^2

I_1 = \frac{2}{5}(1.6)((\frac{0.196}{2})^2)

I_1 = 6.14 \times 10^{-3} kg m^2

now the moment of inertia for thin shell

I_2 = \frac{2}{3} m_2r_2^2

I_2 = \frac{2}{3}(1.6)((\frac{0.206}{2})^2)

I_2 = 11.3 \times 10^{-3} kg m^2

now total inertia of the ball is given as

I = I_1 + I_2

I = 17.4 \times 10^{-3} kg m^2

Part b)

Moment of inertia of uniform ball of mass 3.2 kg

I = \frac{2}{5} mr^2

I = \frac{2}{5}(3.2)((\frac{0.216}{2})^2)

I = 14.9 \times 10^{-3} kg m^2

4 0
3 years ago
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