1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Tatiana [17]
3 years ago
11

A radioactive sample has a count rate of 800 counts per minute. One hour later, the count rate has fallen to 100 counts per minu

te. What is the half-life of the sample?
Physics
1 answer:
katovenus [111]3 years ago
4 0
Radioactivity equation states:

N = No e^(-0.693t/t1/2) --- Where t = time, t1/2 = half-life, N = count after time t, No = Initial count

Using the values given;

100 = 800 e^(-0.693*1/t1/2)
100/800 = e^(-0.693t/t1/2)
0.125 = e^(-0.693/t1/2)
Taking natural logs on both sides;
log 0.125 = -0.693/t1/2 * log (e)
-0.9031 = -0.693/t1/2 * 0.4323
t1/2 = (-0.693*0.4323)/(-0.9031) = 0.3317 hours

Therefore, half-life of the sample is approximately 0.3317 hours

You might be interested in
By what factor must the amplitude of a sound wave be decreased in order to decrease the intensity by a factor of 2?
alina1380 [7]
I believe that the answer to the question asked  above is the following

sound intensity = sound power / (4 pi R2<span>)
</span>
so if you decrease the intensity by a factor of 2 the sound wave will also decrease by a factor of 2.


Hope my answer would be a great help for you.    If you have more questions feel free to ask here at Brainly.
4 0
3 years ago
Read 2 more answers
A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the
kirill115 [55]

Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂  : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F  : Force ( N)

d  : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the  bar

mm = 1.5 kg :  mass of the  monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey  (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances  from forces the point 2

d₁₂ = (3-0.6) m = 2.4m  : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium  of moments at the point  2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

5 0
3 years ago
A 50 W light bulb is plugged into a standard
wolverine [178]

Answer:

$1.26

Explanation:

Power =energy/ time

energy =powerxtime

energy =50x31x24=37200

=37.2kwh

1kwh =3.39

37.2kwh=3.39x37.2=126.108cent

=$1.26

8 0
2 years ago
A 10-ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 ft/sec, how fast is the b
netineya [11]

Answer: The bottom of the ladder is moving at 3.464ft/sec

Explanation:

The question defines a right angle triangle. Therefore using pythagorean

h^2 + l^2 = 10^2 = 100 ...eq1

dh/dt = -2ft/sec

dl/ dt = ?

Taking derivatives of time in eq 1 on both sides

2hdh/dt + 2ldl/dt = 0 ....eq2

Putting l = 5ft in eq2

h^ + 5^2 = 100

h^2 = 25 = 100

h Sqrt(75)

h = 8.66 ft

Put h = 8.66ft in eq2

2 × 8.66 × (-2) + 2 ×5 dl/dt

dl/dt = 17.32 / 5

dl/dt = 3.464ft/sec

7 0
3 years ago
Displacement vectors of 4 km north, 2 km south, 5 km north, and 5 km south combine to a total displacement of
valentinak56 [21]

south = -(north)

Displacement = (4 km north) + (2 km south) + (5 km north) + (5 km south)

Displacement = (4 km north) - (2 km north) + (5 km north) - (5 km north)

Displacement = (4 - 2 + 5 - 5) km north

<u>Displacement = 2 km north </u>

6 0
3 years ago
Other questions:
  • Consider an element with energy levels E 0 and E ∗ and degeneracies of those energy levels g 0 and g ∗ , respectively. Determine
    10·1 answer
  • Can you use scalars and vectors to describe a home run?
    13·1 answer
  • Which measurement is a good indicator of physiological age​
    8·2 answers
  • a 20N mass is supported by two ropes. what is the tension in each rope? how woould i work this problem if i know the two angles
    8·1 answer
  • A particle is constrained to move along a straight line through O.
    6·1 answer
  • While traveling on a dirt road, the bottom of a car hits a sharp rock anda small hole develops at the bottom of its gas tank. If
    11·1 answer
  • You and your 3 friends are driving down the road. All the wheels go flat at the exact same time, that's not good. You have to us
    15·1 answer
  • A pendulum is in a spacecraft to measure
    15·1 answer
  • PHYSICS 50 POINTS PLEASE HELP
    10·2 answers
  • Can someone answer these question for me?​
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!