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Leokris [45]
4 years ago
12

You find a bag labeled "10 kg of lead-210". Its contents now weigh 1.25 kg. How many half-life periods have there been since the

bag was weighed and labeled? Lead-210 is a radioisotope with a half-life of about 22 years.
Physics
1 answer:
zlopas [31]4 years ago
5 0

Answer : The number of half-life periods will be, 3

Explanation :  Given,

Initial amount of lead = 10 kg

Amount of lead after decay = 1.25 kg

Half-life = 22 years

Formula used :

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

where,

a = amount of reactant left after n-half lives

a_o = Initial amount of the reactant

n = number of half lives

t_{1/2} = half-life

Now put all the given values in the above formula, we get:

1.25=\frac{10}{2^n}

2^n=8

2^n=2^3

n=3

Thus, the number of half-life periods will be, 3

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True or false Scientific endeavor is driven by both simple curiosity as well as societal demands
ICE Princess25 [194]

Answer:

its true that Scientific endeavor is driven by both simple curiosity as well as societal demands.

Explanation:

When a scientist has a curiosity about something he carried out a research. and when their is a demand of something in society that time scientific research is carried out. Therefore its true that a scientific endeavor is driven by  simple curiosity or societal demand.

For example

in society, there is demand of a medicine which can completely kill the cancer  and a scientist has curiosity to know how to kill cancer cell. In this way a scientific endeavor for cancer medicine can be carried out by both simple curiosity as well as societal demands.

6 0
4 years ago
Jhon is 12 y o his train is 5 minutes late Jhon has an apple that wheigs half it s own mass calculate the mass of the sun​
pav-90 [236]
Given the following information we have 20 watermelons from mark and 10 fishes from kim therefore we add the longitude of Walmart to the latitude of sams club and end up with a total of 1,000 dish soaps then we convert that into inches which leaves us at 20,000,000 inches of cats then multiply that number to 10 giraffes and we get
1.989 × 10^30 kg and therefore the mass of the sun is 1.989 × 10^30 kg.
7 0
3 years ago
Please help, it's super easy!
Alexandra [31]

The moon's gravitational pull on Earth causes water to bulge on two sides of the Earth(#3)

https://scijinks.gov/tides/

5 0
3 years ago
Read 2 more answers
What is the equivalent resistance of the circuit?
Kaylis [27]

Answer:

80 Ω.

Explanation:

In this circuit the resistances are in series.The equivalent resistance of a series circuit is equal to the sum of the resistances. Req= 60 + 20 = 80 Ω.

8 0
3 years ago
A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit
Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

d_1=4.7\times 10^{10} m

v_i=9.5\times 10^4 m/s

d_2=6\times 10^{12} m

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

K_i+U_i=K_f+U_f

\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}

\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f

v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})

v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}

Using the formula

v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}

v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}

Where mass of sun=M=1.98\times 10^{30} kg

G=6.7\times 10^{-11}

v_f=58515.9 m/s

4 0
3 years ago
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