The relationship between the decay constant (λ) and the half-life (t1⁄2) is given by the equation t(1/2) = 0.693/λ. Use the equation for N to derive this relationship.
Hint: when t = t(1/2), N/No = 0.5 (1 point)

A sample of carbon-14 initially consists of 5 × 1024 particles. Carbon-14 has a half-life of 5730 years.

a. What is the decay constant for carbon-14? (Answer in units of yr–1.) (1 point)
b. How many radioactive particles of the sample remain after 100 years? (1 point)
c. What percentage of the radioactive particles remains after 500 years? (1 point)
d. How many radioactive particles of the sample remain after 1000 years? (1 point)
e. How much time will it take for 50% of the particles to decay? (1 point)
f. How much time will it take for 99% of the particles to decay? (1 point)
g. How many half-lives will it take for 99% of the sample to decay? (1 point)
h. What is the initial decay rate of the sample? (Answer in decays/yr.) (1 point)
i. After 200 years, what is the decay rate of the sample? (1 point)
j. How long will it take for the decay rate to decrease to 1015 decays/year? (1 point)
k. How many half-lives have passed after the time you found in part (j)? (1 point)

Answers

Answer 1

Answer: A: 70/2=35

B: 35/2=17

C: 17.5/2=8.75

D: 8.75 of C-14 will be left

E: 5,730 years

F: 5,631

Explanation:

that’s all I got, hope I helped kinda


Related Questions

Calculate the electric force between two point charges of -54.0 μC and +36.0 μC when they are 2.00 cm apart.

**note -- the metric prefix "micro" = μ = x10 - 6 **


Is this attractive or repulsive?


Can someone help me please? Thank you!!!!

Answers

Answer:

43740 N and attractive force

Explanation:

Given that,

Charge 1, q₁ = -54.0 μC

Charge 2, q₂ = +36.0 μC

The distance between charges, r = 2 cm = 0.02 m

We need to find the force between charges. The formula for the force between charges is given by :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

Put all the values,

[tex]F=9\times 10^9\times \dfrac{54\times 10^{-6}\times 36\times 10^{-6}}{(0.02)^2}\\\\F=43740\ N[/tex]

As the charges are opposite, the force between them is attractive. Hence, the required force is 43740 N.

Two identical spheres a and b carry charges of +0.6 coulomb and -0.2 coulomb, respectively. If these spheres touch, the resulting charge on the first sphere will be...
A) + 0.8 C
B) + 0.2 C
C) - 0.3 C
D) + 0.4 C

Answers

The answer is D because when the positive charged side touches the negative charged side it nullifys part of the positively charged side, basically subtraction from my understanding?

Please urgent science question explaining needed

Answers

Answer:

what is the question? there isn't one

I think you forgot the question

which way would 2 negatively charged balloons naturally move? what would that do to the amount of potential energy stored in the field?

Answers

Answer:

gsg

Explanation:

The balloons with negative charges moves apart from each other by electrostatic force of repulsion. The potential energy in them increases as they move away.

What is force of repulsion?

The basic principle of charges is that, two like charges repel and unlike charge attracts. That is: two negative charged bodies or two positively charged bodies will repel each other and one negative charged body attracts a positively charged body.

In both cases attraction or repulsion the charged particles exerts a force with each other called the electrostatic force. The electrostatic force between two charges is directly proportional to the magnitude of charges and inversely proportional to the distance between them.

The balloons with negative charges moves apart from each other by electrostatic force of repulsion. The potential energy in them increases as they move away.

Find more on electrostatic repulsion:

https://brainly.com/question/24013868

#SPJ2

ik its a lot but can u help
tysm for the person who helped me

Answers

Answer:

It's option 1: they have the same density.

Bill is farsighted and has a near point located 125 cm from his eyes. Anne is also farsighted, but her near point is 75.0 cm from her eyes. Both have glasses that correct their vision to a normal near point (25.0 cm from the eyes), and both wear the glasses 2.0 cm from the eyes. Relative to the eyes, what is the closest object that can be seen clearly (a) by Anne when she wears Bill's glasses and (b) by Bill when he wears Anne's glasses

Answers

Answer:

a) The closest object Anne can see, while wearing Bill's glasses is at approximately 22.39 cm relative to her eyes

b) The closest object Bill can see while wearing Anne's glasses is at approximately 26.38 cm relative to his eyes

Explanation:

The point where Bill has a near point = 125 cm

The point where Anne has a near point = 75.0 cm

Their vision are both corrected to the normal near point = 25.0 cm

The distance of their glasses from the eye, d = 2.0 cm

The lens formula is presented as follows;

[tex]\dfrac{1}{f} =\dfrac{1}{d_0} +\dfrac{1}{d_i}[/tex]

The required distance of the of the object from Bill's glass, [tex]d_{oB}[/tex] = 25 cm - 2.0 cm = 23.0 cm

The distance of the of the image from Bill's glass, [tex]d_{iB}[/tex] = -(125 cm - 2.0 cm) = -123 cm

Therefore, for Bill, we have;

[tex]\dfrac{1}{f_B} =\dfrac{1}{d_{0B}} +\dfrac{1}{d_{iB}}[/tex]

Plugging in the values gives;

[tex]\dfrac{1}{f_B} =\dfrac{1}{23} -\dfrac{1}{123} = \dfrac{100}{2,829}[/tex]

Therefore;

[tex]f_B[/tex] = 2,829/100 cm = 28.29 cm

The required distance of the of the object from Anne's glass, [tex]d_{oA}[/tex] = 25 cm - 2.0 cm = 23.0 cm

The distance of the of the image from Anne's glass, [tex]d_{iA}[/tex] = -(75 cm - 2 cm) = -73 cm

The focal length for Anne is therefore;

[tex]\dfrac{1}{f_A} =\dfrac{1}{d_{0A}} +\dfrac{1}{d_{iA}}[/tex]

Plugging in the values gives;

[tex]\dfrac{1}{f_A} =\dfrac{1}{23} -\dfrac{1}{73} = \dfrac{50}{1,679}[/tex]

Therefore, we have;

[tex]f_A[/tex] = 1,679/50 cm = 33.58 cm

a) When Anne wears Bill's lasses, we have;

[tex]\dfrac{1}{f_B} =\dfrac{1}{d_{0A}} +\dfrac{1}{d_{iA}}[/tex]

Therefore, we get;

[tex]\dfrac{1}{28.29} =\dfrac{1}{d_{0A}} -\dfrac{1}{73}[/tex]

[tex]\dfrac{1}{d_{0A}} = \dfrac{1}{28.29} + \dfrac{1}{73} = \dfrac{10,129}{206,517}[/tex]

[tex]d_{0A}[/tex] ≈ 20.39 cm

The distance of the closest object Anne can see, from her eye, [tex]d_{oe}[/tex] =  [tex]d_{0A}[/tex] + d

∴ [tex]d_{oe}[/tex] ≈ 20.39 cm + 2.0 cm = 22.39 cm

The distance of the closest object Anne can see, from her eye, [tex]d_{oe}[/tex] ≈ 22.39 cm

b) The closest object that can be seen when Bill wears Anne's glasses, we have;

[tex]\dfrac{1}{f_A} =\dfrac{1}{d_{0B}} +\dfrac{1}{d_{iB}}[/tex]

Therefore;

[tex]\dfrac{1}{33.58} =\dfrac{1}{d_{0B}} -\dfrac{1}{123}[/tex]

[tex]\dfrac{1}{d_{0B}} = \dfrac{1}{33.58} +\dfrac{1}{123} = \dfrac{7,829}{206,517}[/tex]

∴ [tex]d_{oB}[/tex] ≈ 26.38 cm

The closest object Bill can see while wearing Anne's glasses,  [tex]d_{oB}[/tex] ≈ 26.38 cm.

Una rueda que gira a 300 r.P.M. Aumenta su velocidad bajo una aceleración angular de 6 2 s rad : calcula: a) La velocidad angular después de 10 s. b) El número de vueltas que da en ese tiempo. Respuestas: f = 873.0 r.P.M.  = 97.746 vueltas.

Answers

Answer:

a)   w = 873 rev,  b)  θ = 97.75 rev

Explanation:

This is a rotation kinematics exercise

         w = w₀ + α t

         θ = θ₀ + w₀ t + ½ α t²

let's start by reducing the magnitudes to the SI system

         w₀ = 300 rpm (2pi rad / 1 rev) (1 min / 60s) = 31.42 rad / s

          α = 6 rad / s²

a) let's look for the angular velocity

            w = 31.42 + 6 10

             w = 91.42 rad / s

b) θ₀ = 0

             θ = 0 + 31.42 to + ½ 6 10²

             θ = 614.2 rad

As they ask for the result in rpm and revolutions, let's carry out the reduction

         w = 91.42 rad / s (1 rev / 2pi rad) (60 s / 1min)

         w = 873 rev

         θ = 614.2 rad (1 rev / 2pi rad)

         θ = 97.75 rev

Monochromatic light of wavelength lambda, such as light from a laser, is directed through two narrow parallel slits that are a distance d apart. There is a viewing screen a very long distance R away from the slits that is used to observe the interference pattern that emerges from the light shining through the slits. The distance that light travels from one slit to the screen is r1 and the distance that light travels from the other slit to the screen is r2. The central bright fringe of the interference pattern results due to a path difference, r1 – r2, of zero. What path difference is required to produce the first order bright fringe that is adjacent to the central bright fringe?

Answers

Answer:

When r1 – r2 equal the wavelength of the monochromatic light

Explanation:

This is because to observe a bright fringe a constructive interference pattern of path difference, r1 – r2 = mλ is observed, where m is an integer and λ = wavelength of monochromatic light.

For bright fringes and thus constructive interference, r1 – r2 must be integral multiples of the wavelength, λ. When r1 – r2 , we have the central bright fringe and thus m = 0.

The first order bright fringe which is adjacent to the central bright fringe is obtained when m = 1,

So r1 – r2 = 1 × λ

r1 – r2 = λ

Thus, the path difference required to produce the first order bright fringe that is adjacent to the central bright fringe is when r1 – r2 = λ, the wavelength of the monochromatic light.

18 kilogram Mass Blokus addressed a level surface if the coefficient of static friction between the Block in the surface is 0.6 what is horizontal force is required to take gravity 10m/s2









Answers

Question: 18 kilogram Mass Block rest on level surface if the coefficient of static friction between the Block and the surface is 0.6 what  horizontal force is required to just move the blcok ( take gravity as 10m/s2 )

Answer:

108 N

Explanation:

From the question,

Applying

F' = mgμ................ Equation 1

Where F' = Frictional force = horizontal  force required to just move the block,  m = mass of the block, g = acceleration due to gravity, μ = coefficient of static friction.

From the question,

Given: m = 18 kg, μ = 0.6, g = 10 m/s²

Substitute these values into equation 1

F' = 18×0.6×10

F' = 108 N

which of the following is a vector physical quantity

a. speed
b. energy
c. mass
d. displacement

Answers

Answer:

D. displacement

Explanation:

Vector quantities have both magnitude and direction.

Scalar quantities have only magnitude.

The two scientists who gave us a better understanding of the universe are?

Answers

Answer:

I'm pretty sure it is Edwin Powell Hubble and Albert Einstein

Explanation:

pls send the answer..pls...​

Answers

Answer:

The fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C

Explanation:

Given;

initial temperature of the liquid, t₁ = 76.3  +/-  0.4⁰C

final temperature of the liquid, t₂ = 67.7  +/-  0.3⁰C

The change in temperature of the liquid is calculated as;

Δt = t₂  -  t₁

Δt = (67.7 - 76.3)  +/-  (0.3 - 0.4)

Δt = (-8.6)  +/-  (-0.1)

Δt = 8.6 +/- 0.1 ⁰C

Therefore, the fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C

How would you calculate the number of moles of oxygen you had?

Answers

Answer:

Explanation:

1 mole of gas contains 6.02x10^23 molecules. That is the Avogadro's number. To find the number of moles, you divide the number of molecules by the Avogadro's number.

The correct answer is C.

Answer:

Explanation:

ans is C. divided by Avogadro no.

A circuit has a voltage of 10 V and a current of 5 A. What must the resistance be?
0.5 ohms
50 ohms
2 ohms
15 ohms

Answers

Answer:

R=V/I

R= 2

Explanation:

R = 10V/5A

R = 2ohms

A 1 km long train passes through a 2km long tunnel .
How much time does it take to pass it completely
“GIVEN SPEED OF TRAIN = 1KM/Hr”

Answers

Answer:

If the tunnel is straight, it will have 2 hours to pass through it

Explanation:

Train speed distance time

1km/hr. 1km. 1hr. 1km/hr. 2km. 2hr

The primary coil of a transformer has 411 turns; the secondary coil has 1000 turns. An alternating current is sent through the primary coil. The emf in the primary is of amplitude 11.7 V. What is the emf amplitude in the secondary

Answers

Answer:

The emf amplitude in the secondary coil is 26.53 V

Explanation:

Given that,

The number of turns of the primary coil of a transformer = 441

The number of turns of the secondary coil of a transformer = 1000

The emf in the primary is of amplitude 11.7 V

We need to find the emf amplitude in the secondary. We know that,

[tex]\dfrac{E_1}{N_1}=\dfrac{E_2}{N_2}\\\\E_2=\dfrac{E_1N_2}{N_1}[/tex]

Put all the values,

[tex]E_2=\dfrac{11.7\times 1000}{441}\\\\E_2=26.53\ V[/tex]

So, the emf amplitude in the secondary coil is 26.53 V.

Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 138000 kg and a velocity of 0.288 m/s, and the second having a mass of 123000 kg and a velocity of -0.131 m/s. (The minus indicates direction of motion.) What is their final velocity

Answers

Answer:

their final velocity is 0.091 m/s

Explanation:

Given;

mass of the first train, m₁ = 138,000 kg

mass of the second train, m₂ = 123,000 kg

initial velocity of the first train, u₁ = 0.288 m/s

initial velocity of the seocnd train, u₂ = -0.131 m/s (opposite direction to the first)

Let their common final velocity after been coupled = v

Apply the principle of conservation of linear momentum;

m₁u₁  +  m₂u₂ = v(m₁  +  m₂)

(138,000 x 0.288)    +    (-0.131 x 123,000)   =   v(138,000 + 123,000)

39,744   -   16,113   =  v(261,000)

23,631 = v(261,000)

v = 23,631 / 261,000

v = 0.091 m/s

Therefore, their final velocity is 0.091 m/s

I WILL MARK YOU THE BRAINLIEST NO LINKS
An upward force applied by a fluid on an object in the fluid is the
Archimedes Principle
Hooke's Law
buoyant force
spring force

Answers

Answer:

buoyant force

Explanation:

just trust meeee

Answer:

C

Explanation:

How do we weigh planets? Explain your thinking

Answers

The weight (or the mass) of a planet is determined by its gravitational effect on other bodies. Newton's Law of Gravitation states that every bit of matter in the universe attracts every other with a gravitational force that is proportional to its mass.

An alpha particle (charge 2e) is sent at high speed toward an atomic nucleus. The electric force acting on the alpha particle is 79.8 N when it is 2.29 x 10(-14) m away from the nucleus. What is the charge on the atomic nucleus? (e = 1.60 x 10(-19) C, Kc = 9 x 10(9) N•m(2)/c(2)

Answers

Answer:

1.45×10⁻¹⁶ C

Explanation:

Applying,

F = kqq'/r²................ Equation 1

Where F = force acting on the alpha particle, k = coulomb's constant, q = charge on the alpha particle, q' = charge on the atomic nucleus, r = distance of seperation between the alpha particle and the atomic nucleus.

make q' the subject of the equation

q' = Fr²/(kq)................. Equation 2

From the question,

Given: F = 79.8 N, r = 2.29×10⁻¹⁴ m, q = 2e = (2×1.6×10⁻¹⁹) = 3.2×10⁻¹⁹ C, k = 9×10⁹ Nm²/C²

Substitute these values into equation 2

q' = [79.8×(2.29×10⁻¹⁴)²]/(3.2×10⁻¹⁹×9×10⁹)

q' = (4.18×10⁻²⁶)/(2.88×10⁻¹⁰)

q' = 1.45×10⁻¹⁶ C

Help fast please physics

Answers

Answer:

10

Explanation:

what do you mean by orbit?​

Answers

Answer:

An orbit is a regular, repeating path that one object in space takes around another one. An object in an orbit is called a satellite. A satellite can be natural, like Earth or the moon. ...Planets, comets, asteroids and other objects in the solar system orbit the sun

Explanation:

In physics, an orbit is the gravitationally curved trajectory of an object, such as the trajectory of a planet around a star or a natural satellite around a planet. Normally, orbit refers to a regularly repeating trajectory, although it may also refer to a non-repeating trajectory.

The switch in a circuit breaker opens when which of the following in the circuit becomes too high?
A) current
B) resistance
C) static charge
D) total charge

Answers

Answer:

b) resistance

Explanation:

since resistance is responsible for heating, if it increases the circuit breaker opens breaking the flow of current to the circuit and thus preventing short circuits.

A student applies a 10 N force to a wood block with a mass of 5 kg. The block is pushed across four different surfaces. The accelerations of the block are recorded. Which surface showed the least friction?

Answers

The complete question is as follows: A student is subjected to a reaction force of 10 N northward from a 5 kg block while pushing the block over a smooth, level surface. Ignoring friction, what is the acceleration of the block?

Answer: The acceleration of the block is [tex]2 m/s^{2}[/tex].

Explanation:

Given: Force = 10 N

Mass = 5 kg

It is known that force applied on an object is the product of mass and acceleration.

Mathematically, [tex]F = m \times a[/tex]

where,

F = force

m = mass

a = acceleration

Substitute the values into above formula as follows.

[tex]F = m \times a\\10 N = 5 kg \times a\\a = \frac{10}{5}\\= 2 m/s^{2}[/tex]

Thus, we can conclude that the acceleration of block is [tex]2 m/s^{2}[/tex].

which of these is a mechanical wave
1)Light .
2)Sound .
3)Gamma ray .
4)Electromagnetic .

Answers

Answer:

Sound waves

Explanation:

Mechanical Sound Waves - A sound wave moves through air by displacing air particles in a chain reaction. ... Sound energy, or energy associated with the vibrations created by a vibrating source, requires a medium to travel, which makes sound energy a mechanical wave.

(≚ᄌ≚)ℒℴѵℯ❤

29. A boy lifts a book with the weight of 30.0 N from the floor to the top of a table, a vertical distance of 2.0. meters. Compute the increase in the potential energy of the weight.
Remember to include your data, equation, and work when solving this problem.

Answers

Answer:

60J

Explanation:

Since the book's weight is 30N and weight = mass * gravity (around 10m/s²):

30 = X * 10 ===>   X = 3.

Using the potential energy formula we have Gravity * Height * Mass:

10 * 2 * 3 = 60J

The ratio of the number of pigs to the number of horse in a farm is 2 to 3 (2:3), if therected are 24 horses, what is the number of pigs?

Answers

Answer:

16 pigs

Explanation:

Let the horses be H.

Let the pigs be P

Let the total number of animals be x.

Given the following data;

Ratio of P to H = 2:3 = 2 + 3 = 5

Number of horses = 24

To find the number of pigs;

First of all, we would determine the total number of animals x.

From the ratio of horse, H:

3/5 * x = 24

Cross-multiplying, we have;

3x = 120

x = 120/3 = 40

Next, we find the number of pigs;

Pigs = 2/5 * 40

Pigs = 80/5 = 16 pigs

what is a Pulley why is it used​

Answers

Answer:

A simple pulley is a wheel with a rope that allows you to pull one end and have it lift whatever is on the other end. A modern, common example of this is a crane, often used in construction.

Explanation:

the magnetic field at a certain point next to a wire carrying 10.0 A of current is 4.20x10^-5 T. what is the field at that point when the current is 20 A

Answers

Answer:

B=8.40x10^-5 T

Explanation:

To find the magnetic field at the new current, first you need to find the distance. To do so, alter the original equation from finding the field to finding the distance.

r (distance) = 4 pi x 10^-7 / 2 pi multiplied by 10.0 / 4.20 x 10^-5

which equals 0.04762

Next, just plug the numbers back into the original equation, taking into account the requested changes.

which equals 8.40x10^-5

Also it was right on Acellus ;)

Hope this helps :)

2. Stars normally convert hydrogen into helium through nuclear fusion. That requires incredibly hot temperatures and high pressure. At the moment, the temperature of empty space is nowhere near warm enough to fuse any elements together. What type of temperature and pressure conditions needed to be present in the early universe in order to create the first hydrogen atoms

Answers

Answer:

About 4,000 K and 10⁻¹⁷ atm

Explanation:

The Big Bang theory states that the Big Bang which is the origin of the universe was about 13.75 billion years ago, and the temperature a few seconds later was 10³²K

The first element began forming at about 3 minutes after the Big Bang with a temperature of 10⁹ K, the nuclei of simple elements

The nuclei of hydrogen and helium began combine with electrons at a temperature of 3,000 K to 4,000 K to form the first neutral atoms. The pressure of the universe at that stage was 10⁻¹⁷ atmospheres

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