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3 years ago

Which change occurs when an atom undergoes alpha decay?

Physics

2 answers:

3241004551 [841]3 years ago

6 0

Answer:

its A on edge

Explanation:

did it now 2020

oee [108]3 years ago

3 0

In an alpha decay, an atom emits an alpha particle. An alpha particle consists of 2 protons and 2 neutrons: this means that during this kind of decay, the original atom loses 2 protons and 2 neutrons from its nucleus.

This also means that the atomic number Z of the element (the atomic number is the number of protons in the nucleus) decreases by 2 units in the process, while the mass number A (the mass number is the sum of the number of protons and neutrons) decreases by 4 units.

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If a molecule has four hybrid sp3 orbitals, it can be concluded that the molecule has a

Lilit [14]

If a molecule is found to have four hybrid sp3 then we say that it has a <span>tetrahedral molecular geometry. thsi is the correct term to define this type of molecules. So your option would be C. Hope this is useful</span>

7 0

3 years ago

Read 2 more answers

A 75-kilogram hockey player is skating across the ice at a speed of 6.0 meters per second. What is the magnitude of the average

saveliy_v [14]

Answer:

(3) 690 N

Explanation:

Force: This is the product of mass and acceleration of a body. The S.I unit of force is Newton(N).

The formula for force is given as,

F = ma..................... Equation 1

Where F = force, m = mass, a = acceleration.

Also,

a = (v-u)/t................... Equation 2

Where v = Final velocity, u = initial velocity, t = time.

Given: u = 6.0 m/s, v = 0 m/s (bring to stop), t = 0.65 s.

Substitute into equation 2

a = (0-6)/0.65

a = -6/0.65

a = -9.23 m/s²

Also given: m = 75 kg

Substitute again into equation 1

F = 75(-9.23)

F = -692.25 N

The negative sign tells that the force oppose the motion of the player

F ≈ 690 N

Hence the right option is (3) 690 N

3 0

3 years ago

A 74 KG student starting from rest slides down 11.8 m high water slide how fast is it going at the bottom of the slide?

-Dominant- [34]

the answer is going to be 873.2

8 0

3 years ago

A comet of mass 1.20 × 10¹⁰kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.500 AU and

irina [24]

The eccentricity of its orbit is $U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$

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The length of the semi-major axis is calculated as follows:

where $, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$$

$M=1.99 \times 10^{30} \mathrm{~kg}=$ mass of sur

$m=1.20 \times 10^{10} \mathrm{~kg}$ - a mass of the comet

$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$