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2 years ago

How does Math work in periodic Tables like protons neutrons and electrons

Physics

1 answer:

Brrunno [24]2 years ago

6 0

Answer:

The stability of isotopes is affected by the ratio of protons to neutrons and also by the presence of certain “magic” numbers of protons or neutrons that represent closed and filled quantum shells. These quantum shells correspond to the shell model of the nucleus. Filled shells confer unusual stability on a nuclide. A magic number for the atomic number (with filled shells) tends to increase the number of stable isotopes of an element.

Explanation:

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A turntable, with a mass of 1.5 kg and diameter of 20 cm, rotates at 70 rpm on frictionless bearings. Two 540 g blocks fall from

ivann1987 [24]

Answer:

The turntable's angular speed after the event is 28.687 revolutions per minute.

Explanation:

The system formed by the turntable and the two block are not under the effect of any external force, so we can apply the Principle of Conservation of Angular Momentum, which states that:

$I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}$ (1)

Where:

$I_{T}$ - Moment of inertia of the turntable, in kilogram-square meters.

$r$ - Distance of the block regarding the center of the turntable, in meters.

$m$ - Mass of the object, in kilograms.

$\omega_{o}$ - Initial angular speed of the turntable, in radians per second.

$\omega_{f}$ - Final angular speed of the turntable-objects system, in radians per second.

In addition, the momentum of inertia of the turntable is determined by following formula:

$I_{T} = \frac{1}{2}\cdot M\cdot r^{2}$ (2)

Where $M$ is the mass of the turntable, in kilograms.

If we know that $\omega_{o} \approx 7.330\,\frac{rad}{s}$ , $M = 1.5\,kg$ , $m = 0.54\,kg$ and $r = 0.1\,m$ , then the angular speed of the turntable after the event is:

$I_{T} = \frac{1}{2}\cdot M\cdot r^{2}$

$I_{T} = 7.5\times 10^{-3}\,kg\cdot m^{2}$

$I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}$

$\omega_{f} = \frac{I_{T}\cdot \omega_{o}}{2\cdot r^{2}\cdot m +I_{T}}$

$\omega_{T} = 3.004\,\frac{rad}{s}$ ( $28.687\,\frac{rev}{min}$ )

The turntable's angular speed after the event is 28.687 revolutions per minute.

3 0

3 years ago

A cyclist travels 36km in 3 hours. What is the cyclist’s average speed?

dybincka [34]

To determine what the cyclists average speed is, simply divide the distance the cyclist has travelled by the time the cyclist has traveled for.

Assuming that this is the average rate the cyclist is moving at it would be 12 km/hr.

4 0

3 years ago

The temperature of 5 pounds of water is 40 degrees. Btus are added until the temp of

Vikentia [17]

Answer:

69becuz u smell l i k e c h e e z e

5 0

4 years ago

A 15.0-kg object sitting at rest is struck elastically in a head-on collision with a 10.5-kg object initially moving at 3.0 m/s.

DIA [1.3K]

Answer:

The final velocity of the 15.0-kg object after the collision is 2.47 m/s in forward direction.

Explanation:

Given;

mass of the object, m₁ = 15 kg

initial velocity of this object, u₁ = 0

mass of the second object, m₂ = 10.5 kg

initial velocity of this object, u₂ = 3.0 m/s

let the final velocity of the first object = v₁

also, let the final velocity of the second object = v₂

Apply the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(15 x 0) + (10.5 x 3) = 15v₁ + 10.5v₂

31.5 = 15v₁ + 10.5v₂ ----- (1)

One directional velocity;

u₁ + v₁ = u₂ + v₂

0 + v₁ = 3 + v₂

v₂ = v₁ - 3 ------(2)

Substitute (2) into (1);

31.5 = 15v₁ + 10.5v₂

31.5 = 15v₁ + 10.5(v₁ - 3)

31.5 = 15v₁ + 10.5v₁ - 31.5

63 = 25.5v₁

v₁ = 63 / 25.5

v₁ = 2.47 m/s

Therefore, the final velocity of the 15.0-kg object after the collision is 2.47 m/s in forward direction.

7 0

3 years ago

When Carbon -14 decays into Nitrogen -14, what kind of decay is this?

Nutka1998 [239]

Answer:

me no speak d english

Explanation:

3 0

4 years ago