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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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Answer:

a) It is moving at 43.15\frac{m}{s^{2}} when reaches the ground.

b) It is moving at 101.44\frac{m}{s^{2}} when reaches the ground.

Explanation:

Work energy theorem states that the total work on a body is equal its change in kinetic energy, this is:

W=K_f-K_i (1)

with W the total work, Ki the initial kinetic energy and Kf the final kinetic energy. Kinetic energy is defined as:

K=\frac{mv^2}{2} (2)

with m the mass and v the velocity.

Using (2) on (1):

W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2} (3)

In both cases the total work while the objects are in the air is the work gravity field does on them. Work is force times the displacement, so in our case is weight (w=mg) of the object times displacement (d):

W=Fd=wd=mgd (4)

Using (4) on (3):

mgd=\frac{mv_f^2}{2}-\frac{mv_i^2}{2} (5)

That's the equation we're going to use on a) and b).

a) Because the branch started form rest initial velocity (vi) is equal zero, using this and solving (5) for final velocity:

v_f=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*95}

v_f=43.15\frac{m}{s^{2}}

b) In this case the final velocity of the boulder is instantly zero when it reaches its maximum height, another important thing to note is that in this case work is negative because weight is opposing boulder movement, so we should use -mgd:

-mgd=-\frac{mv_i^2}{2}

Solving for initial velocity (when the boulder left the volcano):

v_i=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*525}

v_i=101.44 \frac{m}{s^{2}}

3 0
3 years ago
Which is not true about the pH scale?
pochemuha

Answer:

The pH scale was invented by Svante Arrhenius is false.

Explanation :

because the pH scale was invented by Soren Sorensen

4 0
3 years ago
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5. A 5.0 kg object accelerates uniformly from rest for 5.0 s and reaches a final velocity of 20.0 m/s. At 3.0 s, what is the obj
irina [24]

Answer:

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Explanation:

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4 years ago
(Plzzzz help!!!) (50 points!!!)
stellarik [79]

Answer:

Write the following Quantitiesin scientific notation.

a. 10130 Pa to 2 decimal place

b. 978.15m * s-2 to one decimal place

c 0.000001256 A to3 decimal place​

Add your answer and earn points.

Answer

5.0/5

2

kobenhavn

Expert

5.5K answers

43M people helped

Answer: a.

b.  

c.  

Explanation:

Scientific notation is defined as the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

For example : 5000 is written as

a. 10130 Pa to 2 decimal place is written as

b.  to 1 decimal place is written as

c.   to 3 decimal places is written as

Explanation:

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Calculate the acceleration of a 1000 kg car if the motor provides a small thrust of 1000 N and the static and dynamic friction c
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Explanation :

It is given that,

Mass of the car, m = 1000 kg              

Force applied by the motor, F_A=1000\ N

The static and dynamic friction coefficient is, \mu=0.5

Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,

F_A-\mu mg=ma

\dfrac{F_A-\mu mg}{m}=a

a=\dfrac{1000-0.5(1000)(9.81)}{1000}

a=-3.905\ m/s^2

So, the acceleration of the car is -3.905\ m/s^2. Hence, this is the required solution.

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