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masya89 [10]
3 years ago
10

Find the half-life of a radioactive material if after 1 year 99.67% of the initial amount remains. (Round your answer to one dec

imal place.)
Physics
1 answer:
Svetradugi [14.3K]3 years ago
3 0

Answer:

209.68 years

Explanation:

Let T be the half life.

t = 1 year

N = 99.67 % of No = 0.9967 No

Use the law of radioactivity

N = No x e ^(- λ t)

Where, λ is decay constant.

λ = 0.6931 / T

So,

0.9967 No = No x e^(- λ t)

0.9967 = e^(- λ t)

e^( λ t) = 1 / 0.9967 = 1.0033

 λ t = 3.3 x 10^-3

(0.6931 x 1) / T = 3.3 x 10^-3

T = 209.68 years

Thus, the halflife is 209.68 years.

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maw [93]

You've already told us the speed in ft/s .  It's right there in the question.  You said that light travels about  982,080,000 ft/s.

We don't know how accurate that number is, but for purposes of THIS question, that's the number we're going with.

In scientific notation, it's written . . . <em>9.8208 x 10⁸ ft/s .</em>

We don't know where you were going with the number of seconds in a year.  But to answer the question that you eventually asked, it turned out that we don't even need it.

6 0
3 years ago
Calculate the maximum absolute uncertainty for R if:
erma4kov [3.2K]

Answer:

 ΔR = 5 s

Explanation:

The absolute uncertainty or error in an expression is

       ΔR = | \frac{dR}{dB} | ΔB + | \frac{dR}{dA} | ΔA

the absolute value guarantees to take the unfavorable case, that is, the maximum error.

We look for the derivatives

       \frac{dR}{dB} = 1

       \frac{dR}{dA} = -1

we substitute

       ΔR = 1 ΔB + 1 ΔA

       

of the data

       ΔB = 3 s

       ΔA = 2 s

         

       ΔR = 3 + 2

       ΔR = 5 s

7 0
3 years ago
HELP 10 POINTS <br> Which set of balloons would exhibit a greater electric force and why?
vovikov84 [41]

Answer:

The second one because there is less <em><u>distance between them.</u></em>

7 0
3 years ago
The molecular weight of oxygen gas (o2) is 32 while that of hydrogen gas (h2) is 2.
anzhelika [568]

use the formula: v^2=(3kT)/m

Where:

<em>v is the velocity of a molecule</em>

<em>k is the Boltzmann constant (1.38064852e-23 J/K)</em>

<em>T is the temperature of the molecule in the air</em>

<em>m is the mass of the molecule</em>

For an H2 molecule at 20.0°C (293 K):

v^2 = 3 × 1.38e-23 J/K × 293 K / (2.00 u × 1.66e-27 kg/u)

v^2 = 3.65e+6 m^2/s^2

v = 1.91e+3 m/s

For an O2 molecule at same temp.:

v^2 = 3 × 1.38e-23 J/K × 293 K / (32.00 u × 1.66e-27 kg/u)

v^2 = 2.28e+5 m^2/s^2

v = 478 m/s

Therefore, the ratio of H2:O2 velocities is:

1.91e+3 / 478 = 4.00

7 0
3 years ago
A water skier lets go of the tow rope upon leaving the end ofa
ANTONII [103]

Answer:

1.35m

Explanation:

At the highest point of the jump, the vertical speed of the skier should be 0. So the 13m/s speed is horizontal, this speed stays the same from the jumping point to the highest point. The 14m/s speed at jumping point is the combination of both vertical and horizontal speeds.

The vertical speed at the jumping point can be computed:

v_v^2 + v_h^2 = v^2

v_v^2 + 13^2 = 14^2

v_v^2 = 196 - 169 = 27

v_v = \sqrt{27} = 5.2 m/s

When the skier jumps to the its potential energy is converted to kinetic energy:

E_p = E_k

mgh = mv_v^2/2

where m is the skier mass and h is the vertical distance traveled, v_v is the vertical velocity at jumping point, and h is the highest point.

Let g = 10m/s2

We can divide both sides of the equation by m:

gh = v_v^2/2

h = \frac{v_v^2}{2g} = \frac{27}{2*10} = 1.35 m

3 0
3 years ago
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