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Harrizon [31]
3 years ago
5

An alpha particle (which has two protons) is sent directly toward a target nucleus with 90 protons. the alpha particle has a kin

etic energy of 0.45 pj when far away. what is the least center-to-center distance that the alpha particle will be from the target nucleus, assuming that the nucleus does not move?
Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
5 0
It does not move because of the sun the sun has no energy.
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The radioactive element radium (Ra) decays by a process known asalpha decay,in which the nucleus emits a helium nucleus. (Factoi
tamaranim1 [39]

<u>Answer:</u> The energy released in the decay process is 4.6800\times 10^{11}J

<u>Explanation:</u>

The equation for the alpha decay of Ra-226 follows:

_{88}^{226}\textrm{Ra}\rightarrow _{2}^{4}\textrm{He}+_{86}^{222}\textrm{Rn}

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

\Delta m=(m_{Rn}+m_{He})-(m_{Ra})

We know that:

m_{Rn}=222.0176u\\m_{Ra}=226.0254u\\m_{He}=4.0026u

Putting values in above equation, we get:

\Delta m=(222.0176+4.0026)-(226.0254)=-0.0052g=-5.2\times 10^{-6}kg

(Conversion factor: 1 kg = 1000 g )

To calculate the energy released, we use Einstein equation, which is:

E=\Delta mc^2

E=(-5.2\times 10^{-6}kg)\times (3\times 10^8m/s)^2

E=-4.6800\times 10^{11}J

Hence, the energy released in the decay process is 4.6800\times 10^{11}J

8 0
3 years ago
Which type of image is formed on a cinema screen?
Taya2010 [7]

Answer:

A real image is formed on a cinema screen.

Explanation:

It is a type of image that is formed when light rays coming from an object meet each other in actual after reflection or refraction. A real image can be easily obtained on the screen. It is always inverted

5 0
3 years ago
In which of the following scenarios is the total momentum of the system conserved?
leonid [27]

Answer:

The total momentum of a system is conserved only when the system is closed.

Explanation:

7 0
2 years ago
An object of mass m travels along the parabola yequalsx squared with a constant speed of 5 ​units/sec. What is the force on the
Dmitriy789 [7]

Explanation:

The object is moving along the parabola y = x² and is at the point (√2, 2).  Because the object is changing directions, it has a centripetal acceleration towards the center of the circle of curvature.

First, we need to find the radius of curvature.  This is given by the equation:

R = [1 + (y')²]^(³/₂) / |y"|

y' = 2x and y" = 2:

R = [1 + (2x)²]^(³/₂) / |2|

R = (1 + 4x²)^(³/₂) / 2

At x = √2:

R = (1 + 4(√2)²)^(³/₂) / 2

R = (9)^(³/₂) / 2

R = 27 / 2

R = 13.5

So the centripetal force is:

F = m v² / r

F = m (5)² / 13.5

F = 1.85 m

7 0
3 years ago
Anyone need a girl huh
Bess [88]

Answer:

no no no no

Explanation:

3 0
3 years ago
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