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Mrrafil [7]
3 years ago
7

During nuclear decay, a new isotope is created. How is the nucleus of the new isotope different from the parent if the parent is

otope has undergone alpha decay?
Physics
2 answers:
Veseljchak [2.6K]3 years ago
6 0

Answer:

the difference in parent nuclei and daughter nuclei is that daughter nuclei is having smaller mass number and by 4 and smaller atomic number by 2

_z^AX ----> _{z-2}^{A-4}Y + _2^4He

Explanation:

Alpha nucleus is the helium nuclei which is having 2 protons and 2 neutrons

So we will have

\alpha = _2^4He

now we know that in all radioactive decay mass number and atomic number will remain constant on product and reactant side

So we have

X ------> Y + \alpha

now by mass number conservation and atomic number conservation we have

_z^AX ----> _{z-2}^{A-4}Y + _2^4He

so the difference in parent nuclei and daughter nuclei is that daughter nuclei is having smaller mass number and by 4 and smaller atomic number by 2

STALIN [3.7K]3 years ago
3 0

Answer:

Alpha decay will produce a daughter nucleus with more protons and beta decay will produce a daughter nucleus with fewer protons than the parent nucleus has.

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A loudspeaker has a circular opening with a radius of 0.1158 m. The electrical power needed to operate the speaker is 27.0 W. Th
Vilka [71]

Answer:

electrical power is converted by the speaker into sound power 17.25 %

Explanation:

given data

radius = 0.1158 m

power needed = 27.0 W

average sound intensity = 25.6 W/m²

solution

we know sound intensity that is

I = \frac{power}{area}   ...................1

power = intensity × area

Power = 25.6 × 4 ×π×r²

Power = 25.6 × 4 ×π×0.1158²

power = 4.313 W

so here electric power % into sound power

= \frac{4.313}{25} × 100

= 17.25 %

3 0
3 years ago
Carey and Marcy are performing a lab to determine the speed of sound in air with the apparatus that was demonstrated in the vide
andre [41]

Answer:

the frequency is the fundamental and distance is    L = ¼ λ

Explanation:

This problem is a phenomenon of resonance between the frequency of the tuning fork and the tube with one end open and the other end closed, in this case at the closed end you have a node and the open end a belly, so the wavelength is the basis is

          λ = 4 L

In this case L = 19.4 cm = 0.194 m

let's use the relationship between wave speed and wavelength frequency and

           v = λ f

where the frequency is f = 440 Hz

           v = 4 L f

let's calculate

           v = 4 0.194 440

           v = 341.44 m / s

so the frequency is the fundamental and distance is

                  L = ¼ λ

5 0
4 years ago
What do you know about muscular fitness activities?
Ulleksa [173]

Answer:

Examples of muscle-strengthening activities include:

lifting weights.

working with resistance bands.

heavy gardening, such as digging and shovelling.

climbing stairs.

hill walking.

cycling.

dance.

push-ups, sit-ups and squats.

Explanation:

3 0
3 years ago
the amount of surface area of the block contact with the surface is 2.03*10^-2*m2 what is the average pressure exerted on the su
CaHeK987 [17]

Complete question:

A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa

Answer:

The average pressure exerted on the surface by the block is 9655.17 Pa

Explanation:

Given;

density of the lead, ρ =  1.13 x 10⁴ kg/m³

mass of the lead block, m = 20 kg

surface area of the area of the block, A = 2.03 x 10⁻² m²

Determine the force exerted on the surface by the block due to its weight;

F = mg

F = 20 x 9.8

F = 196 N

Determine the pressure exerted on the surface by the block

P = F / A

where;

P is the pressure

P = 196 / (2.03 x 10⁻²)

P = 9655.17 N/m²

P = 9655.17 Pa

Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa

6 0
3 years ago
A truck with a mass of 1330 kg and moving with a speed of 15.0 m/s rear-ends a 805 kg car stopped at an intersection. The collis
Dovator [93]

Answer:

The Speed of the vehicles is 9.34m/s

Explanation:

For an elastic collision the two bodies move with similar velocities after collision

Given

M1=1330kg

V1=15m/s

M2=805kg

V2=0(the car is parked on neutral)

The formula is

M1V1+M2V2=(M1+M2)V

1330*15+805*0=(1330+805)V

19950+0=2135V

2135V=19950

divide both sides by 2135

V=19950/2135

V=9.34m/s

8 0
3 years ago
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