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ZanzabumX [31]
3 years ago
14

Refer to Figure 13-2 and determine how much parent material will be left after five half-lives

Physics
2 answers:
Kamila [148]3 years ago
4 0

Answer: 3.125g

Explanation: This is a type of radioactive decay and all the radioactive process follow first order kinetics.

Half life is the time required to decompose half of the reactants. thus after every half life the reactant will decompose to half of its present value.

Now, to calculate the number of half lives, we use the formula:

a=\frac{a_o}{2^n}

where,

a = amount of reactant left after n-half lives = ?

a_o =Let Initial amount of the reactant = 100 g

n = number of half lives

a=\frac{100}{2^5}

a=3.125g

Tomtit [17]3 years ago
4 0
After five half-lives, there will be

           (1/2) · (1/2) · (1/2) · (1/2) · (1/2)  =  1/32  =  3.125%

of the original material remaining.
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Part A<br>Does the number of particles change as the substance changes its state?​
vovangra [49]

No, the number of particles does not change as the substance changes its state.

<h3><u>Explanation: </u></h3>

Change of state from one phase to another is achieved by providing or absorbing heat or pressure. For instance, liquid water if heated becomes vapour steam and if cooled becomes solid ice. Vapour can be compressed to form liquid water again and thus change of state is a reversible action.

The "chemical composition of the matter remains the same" irrespective of its state. Unless a chemical change is carried out, no change occurs with the number of particles. Phase change only affects the "arrangement of molecules", its structure and its motion.

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3 years ago
Does coffee contain corn? What other food in cafes are corn?
professor190 [17]
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4 0
3 years ago
Betelgeuse is how many times larger than the sun
Aleksandr-060686 [28]
Betelgeuse is one of the largest known stars and is probably at least the size of the orbits of Mars or Jupiter around the sun. That's a diameter about 700 times the size of the Sun or 600 million miles. For a star it has a rather low surface temperature (6000 F compared to the Sun's 10,000 F).
3 0
3 years ago
The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de
spin [16.1K]

Answer:

the spring constant k = 5.409*10^4 \ N/m

the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

Explanation:

From Hooke's Law

F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m

Thus; the spring constant k = 5.409*10^4 \ N/m

The amplitude is decreasing 37% during one period of the motion

e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}

b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) }    }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s

Therefore; the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

5 0
3 years ago
PRACTICE ANOTHER A cube of wood having an edge dimension of 19.7 cm and a density of 647 kg/m3 floats on water. (a) What is the
dedylja [7]

CHECK COMPLETE QUESTION

The cube of wood having an edge dimension of 19.7cm and a density of 647kg/m3 floats on water

(a) What is the distance from the horizontal top surface of the cube to the water level answer in cm

(b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface answer in kg

Answer:

a)6.29cm

b)2.78 kg

Explanation:

Given:

Let us calculate the volume first, we were given dimension as 19.7cm=0.197m

Volume is (0.197 meters)³ = 0.00764m³

Then we can calculate the mass as;

Given mass is 647 kg/m³ x 0.00774m³ = 4.947kg

The weight = mass × acceleration due to gravity

weight = 4.947 x 9.8 N/kg = 48.44N

By Floating we can say the the buoyancy force has to equal the weight (48.44 N) which has

which is equal to the weight of volume of the displaced water. Or the mass, the calculation is the same.

We know that density of fresh water at 20ºC is 998 kg/m³

Then we can calculate the volume of displaced water as

4.947 kg / 998 kg/m³ = 0.00496 m³

We know that the displaced water has a shape of a rectangular solid with 0.197 meters on the two horizontal dimensions, and h as the height to the surface then

V = 0.197²h = 0.00496

0.00496= 0.197²h

h = 0.1278 meters or 12.78 cm

Then the the distance exposed, would be 19.7–12.78 = 6.29 cm

b) ifthe cube is fully submerged, the volume of the displaced water is 0.00774m³

mass of displaced water is 0.00774m³ x 998 kg/m³ = 7.724 kg

Added mass is the mass of the displaced water – mass of block

= 7.724–4.947 = 2.78 kg

5 0
3 years ago
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