ANSWER IN LESS THAN A MINUTE!! EASYY!

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

If you are pushing on a box with a force of 20 N and there is a 7 N force on the box due to sliding friction, what is the net fo

vovangra [49]

Answer:

13 N

Explanation:

The Net Force of an object should be the difference between the forces applied to the object if the object is not in equilibrium. This object is not in equilibrium so therefore by finding the difference between the forces, you will find your answer. 20 N - 7 N = 13 N.

6 0

3 years ago

Read 2 more answers

Apollo 14 astronaut Alan B. Shepard Jr. used an improvised six-iron to strike two golf balls while on the Fra Mauro region of th

Illusion [34]

Complete Question

Apollo 14 astronaut Alan B. Shepard Jr. used an improvised six-iron to strike two golf balls while on the Fra Mauro region of the moon’s surface, making what some consider the longest golf drive in history. Assume one of the golf balls was struck with initial velocity v0 = 32.75 m/s at an angle θ = 32° above the horizontal. The gravitational acceleration on the moon’s surface is approximately 1/6 that on the earth’s surface. Use a Cartesian coordinate system with the origin at the ball's initial position.

Randomized Variables

vo 32.75 m/s

theta 32 degrees

What horizontal distance, R in meters, did this golf ball travel before returning to the lunar surface?

Answer:

The horizontal distance is $R = 590.2 \ m$

Explanation:

From the question we are told that

The initial velocity is $v_o = 32.75 \ m/s$

The angle is $\theta = 26^o$

The gravitational acceleration of the moon is $g_m = \frac{1}{6} * 9.8 = 1.633 m/s^2$

Generally the distance traveled is mathematically represented as

$R = \frac{v_o^2 sin 2(\theta)}{g_m}$

=> $R = \frac{32.75^2 sin 2(32)}{1.633}$

=> $R = 590.2 \ m$

8 0

3 years ago

How does fertilizer run off pollution?

velikii [3]

When it rains, the growth aids in the soil drift

5 0

3 years ago

A fan spins at 6.0 rev/s. You turn it off, and it slows at 1.0 rev/s2. What is the angular displacement before it stops

Komok [63]

Answer:

Angular displacement before it stops = 18 rev

Explanation:

Given:

Speed of fan w(i) = 6 rev/s

Speed of fan (Slow) ∝ = 1 rev/s

Final speed of fan w(f) = 0 rev/s

Find:

Angular displacement before it stops

Computation:

w(f)² = w(i) + 2∝θ

0² = 6² + 2(1)θ

0 = 36 + 2θ

2θ = -36

Angular displacement before it stops = -36 / 2

θ = -18

Angular displacement before it stops = 18 rev

4 0

3 years ago

ANSWER IN LESS THAN A MINUTE!! EASYY!​

ANSWER IN LESS THAN A MINUTE!! EASYY!