Ask question

Ask question

All categories

3 years ago

12

How much time will have passed by the time a sample of an isotope has decayed to 6.25% of its original mass if the half-life is

15 days? 45 days 60 days 75 days 30 days

1 answer:

aleksandrvk [35]3 years ago

5 0

Answer:

60 days.

Explanation:

Let the original mass (N₀) = 1 g

Amount remaining (N) = 6.25% of its original mass

= 6.25% × 1

= 6.25/100 × 1

= 0.0625 g

Half life (t½) = 15 days

Time (t) =?

Next, we shall determine the rate of decay. This can be obtained as follow:

Decay constant (K) = 0.693/ half life

K = 0.693 / t½

Half life (t½) = 15 days

Decay constant (K) =?

K = 0.693 / t½

K = 0.693 / 15

K = 0.0462 / day

Finally, we shall determine the time taken for the sample of the isotope to decay to 6.25% of its original mass.

This can be obtained as follow:

Original amount (N₀) = 1 g

Amount remaining (N) = 0.0625 g

Decay constant (K) = 0.0462 / day

Time (t) =?

Log (N₀/N) = Kt/2.3

Log (1/0.0625) = 0.0462 × t / 2.3

Log 16 = 0.0462 × t / 2.3

1.2041 = 0.0462 × t /2.3

Cross multiply

0.0462 × t = 1.2041 × 2.3

Divide both side by 0.0462

t = (1.2041 × 2.3)/0.0462

t = 59.9 ≈ 60 days

Therefore, the time taken for the sample of the isotope to decay to 6.25% of its original mass is 60 days

You might be interested in

A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.

SVETLANKA909090 [29]

A) Force of the wall on the ladder: 186.3 N

B) Normal force of the ground on the ladder: 725.2 N

C) Minimum value of the coefficient of friction: 0.257

D) Minimum absolute value of the coefficient of friction: 0.332

Explanation:

a)

The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:

$W=mg$ : weight of the ladder, with m = 20 kg (mass) and $g=9.8 m/s^2$ (acceleration of gravity)

$W_M=Mg$ : weight of the person, with M = 54 kg (mass)

$N_1$ : normal reaction exerted by the wall on the ladder

$N_2$ : normal reaction exerted by the floor on the ladder

$F_f = \mu N_2$ : force of friction between the floor and the ladder, with $\mu$ (coefficient of friction)

Also we have:

L = 4.1 m (length of the ladder)

d = 3.0 m (distance of the man from point A)

Taking the equilibrium of moments about point A:

$W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}$

where

$Wsin 21^{\circ}$ is the component of the weight of the ladder perpendicular to the ladder

$W_M sin 21^{\circ}$ is the component of the weight of the man perpendicular to the ladder

$N_1 sin 69^{\circ}$ is the component of the normal force perpendicular to the ladder

And solving for $N_1$ , we find the force exerted by the wall on the ladder:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N$

B)

Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of $N_2$ .

We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.

Therefore, we have:

$\sum F_y = 0\\N_2 - W - W_M =0$

And substituting and solving for N2, we find:

$N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N$

C)

Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.

The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.

Therefore, we can write:

$\sum F_x = 0\\F_f - N_1 = 0$

And re-writing the equation,

$\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257$

So, the minimum value of the coefficient of friction is 0.257.

D)

Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.

From part C), we saw that the coefficient of friction can be written as

$\mu = \frac{N_1}{N_2}$

This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}$

We see that this quantity is maximum when d is maximum, so when

d = L

Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)$

And substituting, we get

$N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N$

And therefore, the minimum coefficient of friction in order for the ladder not to slip is

$\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332$

Learn more about torques and equilibrium:

brainly.com/question/5352966

#LearnwithBrainly

7 0

3 years ago

A lamp consumes 1000J of ekectrical energy in 10s. Calculate its power.

Mice21 [21]

You do 1000 divide it by 10 which equals 100 W

4 0

3 years ago

a car accelerates from rest at 3.6m/s^2. how much time does it need to attain a speed of 7 m/s? awnser in units of s.

Ivahew [28]

Initial velocity: 0
final velocity: 7 m/s
a = 3.6
t = ?
x = ?

(7-0)/3.6 = t
t = 1.94 s

6 0

3 years ago

Read 2 more answers

. What is the relationship between potential energy, kinetic energy, and speed as the skater moves down and up the U-shaped ramp

kykrilka [37]

Top of the U ramp: potential energy is the highest, while kinetic energy is the lowest

Bottom of the U ramp(aka the curve part): potential energy is the lowest and the kinetic energy is the highest

THEREFORE, PE and KE have an INVERSE RELATIONSHIP.

6 0

3 years ago

Read 2 more answers

A golf ball is struck with a five iron on level ground. it lands 100.0 m away 4.60 s later. what was the magnitude and direction

finlep [7]

consider the motion in x-direction

$v_{ox}$ = initial velocity in x-direction = ?

X = horizontal distance traveled = 100 m

$a_{x}$ = acceleration along x-direction = 0 m/s²

t = time of travel = 4.60 sec

Using the equation

X = $v_{ox}$ t + (0.5) $a_{x}$ t²

100 = $v_{ox}$ (4.60)

$v_{ox}$ = 21.7 m/s

consider the motion along y-direction

$v_{oy}$ = initial velocity in y-direction = ?

Y = vertical displacement = 0 m

$a_{y}$ = acceleration along x-direction = - 9.8 m/s²

t = time of travel = 4.60 sec

Using the equation

Y = $v_{oy}$ t + (0.5) $a_{y}$ t²

0 = $v_{oy}$ (4.60) + (0.5) (- 9.8) (4.60)²

$v_{oy}$ = 22.54 m/s

initial velocity is given as

$v_{o}$ = sqrt(( $v_{ox}$ )² + ( $v_{oy}$ )²)

$v_{o}$ = sqrt((21.7)² + (22.54)²) = 31.3 m/s

direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg

6 0

3 years ago