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2 years ago

A radioactive decay is illustrated. Which radioactive emission is a product of this decay?

Physics

1 answer:

hram777 [196]2 years ago

5 0

The radioactive decay is shown in the image attached to this answer

Answer:

Alpha

Explanation:

An alpha decay is said to have occurred when the mass number of the daughter nucleus decreases by four units and the atomic number of the daughter nucleus decreases by two units.

If we look at the equation from left to right, we will notice that the mass number of the daughter nucleus decreased by four units and the atomic of the daughter nucleus number by two units compared to that of the parent nucleus leading to the emission of an alpha particle.

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You walk 10 miles north, 20 miles south, and 5 miles north. Taking north to be the positive y direction, define a position vecto

gulaghasi [49]

Considering north as positive Y axis

First 10 miles north means a positive 10 displacement along y direction

Second 20 miles south means a negative 20 displacement along y direction

Third 5 miles south means a positive 5 displacement along y direction

So total displacement = 10 - 20 + 5 = -5 miles displacement

So position vector of the final position = - 5 j

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3 years ago

Write a hypothesis for Part I of the lab, which is about the

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2 years ago

Jenna made an electric circuit as seen in the picture. She placed a thermometer near her light bulb. After the light bulb was tu

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The thermometer had most likely gotten a very high number on it from the heat of the light bulb. The longer he light bulb was on, the higher the number went on the thermometer.

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2 years ago

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A hiker leaves her camp and walks 3.5 km in a direction of 55° south of west to the lake. after a short rest at the lake, she hi

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Let's choose east as positive x-direction and south as positive y-direction. We can resolve the two displacement along these two axes:

- Displacement 1 (3.5 km, $55^{\circ}$ south of west

$d_{1x}=-(3.5 km)( cos 55^{\circ})=-2.01 km$

$d_{1y}=(3.5 km)( sin 55^{\circ})=2.87 km$

- Displacement 2 (2.7 km, $16^{\circ}$ east of south

$d_{2x}=(2.7 km)( sin 16^{\circ})=0.74 km$

$d_{1y}=(2.7 km)( cos 16^{\circ})=2.60 km$

So, the total components on the two directions are

$d_x = -2.01 km+0.74 km=-1.27 km$

$d_y=2.87 km+2.60 km=5.47 km$

And the magnitude of the hiker's resultant displacement is

$d=\sqrt{(1.27 km)^2+(5.47 km)^2}=5.6 km$

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3 years ago

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How many types Of Ornital does the atomic orbital have ?

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Answer:

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2 years ago

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