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user100 [1]
4 years ago
13

7. If silicon-32 has a decay chain with the following steps:

Physics
1 answer:
Westkost [7]4 years ago
7 0
<h3>Answer:</h3>

Aluminium-28

<h3>Explanation:</h3>

We are given;

Silicon-32

it undergoes a decay chain starting with alpha decay then followed by beta decay.

  • We need to know that when a radioactive isotope undergoes alpha decay its mass number decreases by 4 while the atomic number decreases by 2.
  • When its a beta decay the mass number remains unchanged while atomic number increases by 1.
  • In this case;

Silicon-32 has an atomic number 14 and mass number 32.

#1. Alpha decay

  • Si-32 undergoes alpha decay to form a radioisotope with a mass number 28 and atomic number 12.

That is; ³²₁₄Si → ²⁸₁₂X + ⁴₂He

#2. Beta decay

  • The radioisotope formed then undergoes beta decay to form a radioisotope with a mass number 28 and atomic number 13.

That is;

²⁸₁₂X → ²⁸₁₃Y + ⁴₂He

  • Therefore, the radioisotope formed after the two decay at the end of decay chain is Al-28.

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A train moving west with an initial velocity of 20 m/s accelerates at 4 m/s2 for 10 seconds. During this time, the train
Luden [163]

Answer:400m

Explanation: x=20⋅10+ 4.10^2/2                                                                                                        =400m

​

4 0
3 years ago
While on the moon, the Apollo astronauts enjoyed the effects of a gravity much smaller than that on
Ainat [17]

Answer:

1.628 \frac{m}{sec^{2} }

Explanation:

Anywhere in the universe, In a closed system, <u>Conservation of energy</u> is applicable.

In this case

Neil is initially on the surface of moon and has a velocity of 1.51 \frac{m}{sec} in upward direction.

⇒He has Kinetic energy= K_{i} = \frac{1}{2} m{v^{2} } J

But with respect to the surface of the moon,

where m=mass of moon

           v=velocity of Neil

He has Potential energy= P_{i}=0 J

At the highest point of his jump, his velocity =0

⇒ Kinetic energy=K_{f}=0 J

His Potential energy with respect to the surface of moon=P_{f}=m \times g\times h

where m=mass of moon

           g= gravitational acceleration on moon

           h=height from moon's surface

By Conservation Energy Principle

K_{i}+P_{i}=K_{f}+P_{f}

K_{i}+0=0+P_{f}

\frac{1}{2} m{v^{2} } = m \times g\times h

\frac{v^{2} }{2} =  g\times h

\frac{1.5^{2} }{2} J= g\times 0.7 m

⇒ g = \frac{1.14}{0.7} = 1.628 \frac{m}{sec^{2} }

8 0
3 years ago
A 10.00 kg mass is moving to the right with a velocity of 14.0 m/s. A 12.0 kg mass is moving to the left with a velocity of 8.00
Basile [38]

Answer:

2 m/s

Explanation:

From the conservation of momentum, the initial momentum of the system must be equal to the final momentum of the system.

Let the 10.00 kg mass be m_1 and the 12.0 kg mass be m_2. When they collide and stick, they have a combined mass of m_1+m_2.

Momentum is given by p=mv. Set up the following equation:

\displaystyle m_1v_1+m_2v_2=(m_1+m_2)v_f, where v_f is the desired final velocity of the masses.

Call the right direction positive. To indicate the 12.0 kg object is travelling left, its velocity should be substitute as -8.00 m/s.

Solving yields:

10\cdot 14 + 12\cdot (-8)=(10+12)v_f\\\implies v_f=\boxed{2 \text{ m/s}}

4 0
2 years ago
Two mechanical devices typically used in laboratories to accurately measure small objects or distances are the _____ and _____.
Lera25 [3.4K]
Micrometers and calipers

3 0
3 years ago
A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equ
valentinak56 [21]

Explanation:

The given data is as follows.

                    mass = 0.20 kg

              displacement = 2.6 cm

              Kinetic energy = 1.4 J

       Spring potential energy = 2.2 J

Now, we will calculate the total energy present present as follows.

         Total energy = Kinetic energy + spring potential energy

                           = 1.4 J + 2.2 J

                            = 3.6 Joules

As maximum kinetic energy of the object will be equal to the total energy.

So,      K.E = Total energy

                = 3.6 J

Also, we know that

                  K.E = \frac{1}{2}mv^{2}_{m}

or,                   v = \sqrt{\frac{2K.E}{m}}

                        = \sqrt{2 \times 3.6 J}{0.2 kg}

                        = \sqrt{36}

                        = 6 m/s

thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.

4 0
3 years ago
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