Ask question

Ask question

All categories

3 years ago

6

The half-life of a certain radioactive substance is 12 hours. There are 19 grams present initially. a. Express the amount of su

bstance remaining as a function of time t. b. When will there be 8 grams remaining?

1 answer:

neonofarm [45]3 years ago

4 0

Answer:

(a) $N=19\times e^{-\lambda t}$

(b) 15 hours

Explanation:

half life, T = 12 hours

No = 19 g

(a) Let N be the amount remaining after time t.

Let λ be the decay constant.

$\lambda =\frac {0.6931}{T}$

The equation of radioactivity used here is given by

$N=N_{o}e^{-\lambda t}$

$N=19\times e^{-\lambda t}$

(b) N = 8 gram

Substitute the values in above equation

$\lambda =\frac {0.6931}{12}$

λ = 0.0577 per hour

So, $8=19\times e^{-0.577t}$

$e^{-0.0577t}=0.421$

Take natural log on both the sides

- 0.0577 t = - 0.865

t = 15 hours

You might be interested in

A sound wave traveling at 343 m/s is emitted by the foghorn of a tugboat. an echo is heard 2.80 s later. how far away is the ref

Nostrana [21]

The echo is heard 2.80 s later, this means this is the time the sound takes to travel to the reflecting object and then back to us. So, during this time, the sound wave has covered the distance L between us and the object twice:
$S=2L$
The speed of the sound wave is: $v=343 m/s$ , and since it is moving by uniform motion, we can find the distance covered by the wave using
$S=vt=(343 m/s)(2.80 s)=960 m$
And we said this corresponds to twice the distance between us and the reflecting object, so:
$L= \frac{S}{2} = \frac{960 m}{2} =480 m$
so, the object is 480 meters away.

3 0

3 years ago

You’ve had practice calculating the grams of hydrogen gas, but it is also possible to calculate the amount of oxygen gas produce

alekssr [168]

Answer:41.991ml

Explanation:

Equations: 2 H2O → 4H+ + 4e + O2 OXIDATION

2 H+ + 2e → H2 REDUCTION

Electrolysis is the chemical decomposition of compounds when electricity is made to pass through a molten compound or solution.

from the oxidation reaction:

1moles of oxygen requires 4moles of electrons to be discharged at the product

F=96500C/mol

Quantity of charge Q=It

=60*60*0.201A

Q=723.6C

Mole=Q/(F*mole ratio of electron)

Mole= 723.6/(4*96500)

Mole=((1809)/(965000))

M=0.0018746114

M1/M2=V1/V2

1/0.00187=22.4dm^3/V2

V2=22.4*0.00187

V2=0.04199129534dm^3

41.99129534ml

5 0

3 years ago

Help again please !!!

rusak2 [61]

D I think .... don’t be mad if I’m wrong

5 0

3 years ago

There is a naturally occurring vertical electric field near the Earth’s surface that points toward the ground. In fair weather c

trasher [3.6K]

Answer:

$\frac{F}{W} = 9.37 \times 10^{-4}$

Explanation:

Radius of the pollen is given as

$r = 12.0 \mu m$

Volume of the pollen is given as

$V = \frac{4}{3}\pi r^3$

$V = \frac{4}{3}\pi (12\mu m)^3$

$V = 7.24 \times 10^{-15} m^3$

mass of the pollen is given as

$m = \rho V$

$m = 7.24 \times 10^{-12}$

so weight of the pollen is given as

$W = mg$

$W = (7.24 \times 10^{-12})(9.81)$

$W = 7.1 \times 10^{-11}$

Now electric force on the pollen is given

$F = qE$

$F = (-0.700\times 10^{-15})(95)$

$F = 6.65 \times 10^{-14} N$

now ratio of electric force and weight is given as

$\frac{F}{W} = \frac{6.65 \times 10^{-14}}{7.1 \times 10^{-11}}$

$\frac{F}{W} = 9.37 \times 10^{-4}$

7 0

3 years ago

A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 hz. the sound he hears reflected back from t

Mashutka [201]

As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "Doppler's effect."

Now the general formula of the Doppler's effect is:
$f = (\frac{g + v_{r}}{g + v_{s}})f_{o}$ -- (A)

Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.

Where,
g = Speed of sound = 340m/s.
$v_{r}$ = Velocity of the receiver/observer relative to the medium = ?.
$v_{s}$ = Velocity of the source with respect to medium = 0 m/s.
$f_{o}$ = Frequency emitted from source = 400 Hz.
$f$ = Observed frequency = 408Hz.

Plug-in the above values in the equation (A), you would get:

$408 = ( \frac{340 + v_{r}}{340 + 0})*400$

$\frac{408}{400} = \frac{340 + v_{r}}{340}$

Solving above would give you,
$v_{r}$ = 6.8 m/s

The correct answer = 6.8m/s

7 0

3 years ago