a block of mass 3kg slides along a horizontal surface that has neglitjble friction except dor one section what is the magntiude of acerage frictional force exerted on the block by rough section of surface

Answer:

1st: Radiation

2nd: Conduction

3rd: Convection

Explanation:

I actually learned this before in school. Yay

(1) Both the girl and the boy have the same nonzero angular displacement.

Hence the correct option is A.

(2) The girl has greater linear speed.

Hence the correct option is B.

(1) For a given elapsed time interval, both the girl and the boy will have the same angular displacement. The angular displacement is determined by the angle swept out by the riders as the merry-go-round rotates. Since both riders are on the same merry-go-round and are moving with it at the same rate, they will both have the same angular displacement.

Therefore, Both the girl and the boy have the same nonzero angular displacement.

Hence the correct option is A.

(2) The linear speed of a rider depends on their distance from the center of the merry-go-round. The linear speed is given by the formula:

v = ω * r

Where:

v is the linear speed

ω is the angular speed (which is constant for the merry-go-round)

r is the distance from the center of the merry-go-round

Since the girl is near the outer edge of the merry-go-round, she has a greater distance from the center compared to the boy. As a result, the girl will have a greater linear speed than the boy.

Therefore, The girl has greater linear speed.

Hence the correct option is B.

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Answer:

The density of the 1,000,000 kg iceberg = The density of the 10 g iceberg

Explanation:

The given quantities of iceberg that will be the basis of the comparison of the densities of the iceberg;

The mass of the iceberg = 1,000,000 kg

The mass of the ice cube taken from the iceberg = 10 g

The density of a substance, is the measure of the mass of the substance per volume of the substance, it is a constant for the substance

[tex]Density = \dfrac{Mass}{Volume}[/tex]

For the iceberg, the density of the iceberg is the property that indicates the volume occupied by a given mass of the iceberg

Because of density, we can estimate that the volume occupied by a 1,000 kg iceberg is 10 times the volume occupied by a 100 kg iceberg

We can write

The density of the 100 kg iceberg = 100 kg/(x m³)

The density of the 1,000 kg iceberg = 1,000 kg/(10 × x m³) = 100 kg/(x m³)

Therefore, the density of the 100 kg iceberg = The density of the 1,000 kg iceberg

Similarly,

The density of the 1,000,000 kg iceberg = The density of the 10 g iceberg because the 10 g iceberg was obtained from the 1,000,000 kg iceberg, and therefore, they have the same density.

When light waves fall on a thin film of oil or soap bubble, they reflect back from both the top and bottom surfaces. Therefore, the minimum thickness of the soap bubble film is 386.7 nm.

The light waves interfere with one another and either enhance or cancel each other out, depending on their relative phase at the time of reflection. When two light waves reinforce each other, the interference is constructive. In this question, we have to find the minimum thickness of a soap bubble film with refractive index 1.50 that results in constructive interference in reflected light if this film is illuminated by a beam of light of wavelength 580 nm.So, we know that the condition for constructive interference is given by2t=nλ/1.5where, t is the thickness of the soap film, λ is the wavelength of the light and n is the order of interference. To find the minimum thickness, we need to consider the first order of interference, i.e., n=1.Substituting the values in the above equation, we get2t= (1 × 580 × 10-9) /1.5= 0.3867 × 10-6 m= 386.7 nm.

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The potential difference across the 2 Ω resistor is 2 V.

The potential difference across the 2 Ω resistor is equal to the current flowing through it multiplied by the resistance of the resistor. The current flowing through the circuit is 1 A, and the resistance of the 2 Ω resistor is 2 Ω.

Therefore, the potential difference across the 2 Ω resistor is:

= 1 A * 2 Ω = 2 V.

V = I * R

V = 1 A * 2 Ω

V = 2 V

Therefore, the potential difference across the 2 Ω resistor is 2 V.

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The resistance of the circuit is 6.00 kΩ when resistor connected in series.

Resistance: It is the opposition to the flow of electric current. It is a measure of how much the resistor opposes the flow of electricity through it. Series: In a series circuit, the current that flows through each of the components is the same and the voltage across the circuit is the sum of the individual voltage drops. When we connect multiple resistors in series, we connect them end to end to create a single path for current flow. The total resistance of the series circuit is equal to the sum of the individual resistances. In this problem, three resistors are connected in series: a 5.00 * 10² Ω, a 2.00 kΩ, and a 3.50 kΩ resistor. We need to find the total resistance of the circuit. First, we need to convert the 5.00 * 10² Ω into kΩ by dividing by 1000. \frac{5.00 * 10² Ω }{ 1000 }= 0.5 kΩ

Now we can add up the resistances in kΩ to find the total resistance in kΩ.R(total) = 0.5 kΩ + 2.00 kΩ + 3.50 kΩR

(total) = 6.00 kΩ .Therefore, the resistance of the circuit is 6.00 kΩ.

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Answer:

The body systems that work together to maintain the energy Sierra needs are;

The digestive system, the respiratory system, and the circulatory system

Explanation:

Cellular respiration in the body cells require oxygen to produce energy which are used by the muscles and other body cells. Carbon dioxide is also produced and is the build up of carbon dioxide has to be removed from the body as the by product of cellular respiration which is toxic at the cell level

Therefore, the body systems that work together to maintain the energy Sierra needs are;

The digestive system; Takes in the energy containing food and brakes them into chemicals that are transported to the cells for cellular respiration

The respiratory system; Takes in oxygen and removes carbon dioxide from the blood from and to the atmosphere

The circulatory system; Supplies food and oxygen from the digestive and respiratory system to the cells and transports produced carbon dioxide from the cells to the lungs from where it is passed out of the body by th respiratory system.

The resultant magnitude, R, and angle, [tex]$\theta$[/tex] , of vectors a and b can be determined using the analytical method.

Given that the magnitude of vector a is 7.1 and the magnitude of vector b is 6.0, we can calculate the resultant as follows:

[tex]$$R = \sqrt{a^2 + b^2} = \sqrt{(7.1)^2 + (6.0)^2} \approx 9.17.$$[/tex]

To find the angle [tex]$\theta$[/tex], we can use the inverse tangent function:

[tex]$$\theta = \arctan\left(\frac{b}{a}\right) = \arctan\left(\frac{6.0}{7.1}\right) \approx 40.57^\circ.$$[/tex]

Therefore, the resultant magnitude is approximately 9.17 and the angle with respect to vector a is approximately [tex]$40.57^\circ$[/tex].

The magnitude of the resultant vector can be found using the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In this case, the hypotenuse represents the resultant vector R, and the other two sides represent vectors a and b. By substituting the given magnitudes, we can calculate R as approximately 9.17.

To determine the angle [tex]$\theta$[/tex], we use the inverse tangent function. The ratio [tex]$\frac{b}{a}$[/tex] represents the slope of the right triangle formed by vectors a and b. By taking the arctan of this ratio, we find that [tex]$\theta$[/tex] is approximately [tex]$40.57^\circ$[/tex]. This angle indicates the direction in which the resultant vector points, with respect to vector a.

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The angular speed ω[tex]_{final}[/tex]when the small object is directly below the axis is 0. This means that the system comes to rest in that position.

To solve this problem, we can use the principle of conservation of angular momentum. When the small object is directly below the axis, the total angular momentum of the system remains constant.

The angular momentum L of an object is given by the formula:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.

The moment of inertia of a solid disk rotating about an axis through its center is given by:

I_disk = (1/2) × m × R²

where m is the mass of the disk and R is the radius.

Similarly, the moment of inertia of a point mass m located at the rim of the disk is given by:

I_object = m × R²

The total moment of inertia of the system, when the small object is glued to the rim, is the sum of the moments of inertia of the disk and the object:

[tex]I_{total}[/tex] = I_disk + I_object

[tex]I_{total}[/tex]= (1/2) × m × R² + m × R²

[tex]I_{total}[/tex]= (3/2) × m × R²

Now, at the initial position, the angular momentum of the system is given by:

I[tex]_{total}[/tex] × ω[tex]_{initial}[/tex]= L[tex]_{initial}[/tex]

Since the disk is released from rest, ω_initial is 0.

When the small object is directly below the axis, the moment of inertia becomes:

I[tex]_{final}[/tex]= I[tex]_{disk}[/tex]

The angular momentum at this position is:

L[tex]_{final}[/tex]= I[tex]_{final}[/tex] × ω[tex]_{final}[/tex]

Since angular momentum is conserved, we can equate the initial and final angular momentum:

L[tex]_{initial}[/tex] = L[tex]_{initial}[/tex]

I[tex]_{total}[/tex] × ω[tex]_{initial}[/tex]= I[tex]_{final}[/tex] × ω[tex]_{final}[/tex]

Substituting the expressions for the moments of inertia and simplifying:

[(3/2) × m × R²] * 0 = (1/2) × m × R² × ω_final

Simplifying further:

0 = (1/2) * ω[tex]_{final}[/tex]

Therefore, we find that the angular speed ω_final when the small object is directly below the axis is 0. This means that the system comes to rest in that position.

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Answer:

b.

Explanation:

Answer:

Attractive force and r = 0.399 m

Explanation:

One charge is positive and the other charge is negative. Opposite charges attract, so there has to be a force that attracts between them.

q1 = 10.0 x 10^-6 C

q2 = -3 x 10^-6 C

F = 1.7 N

Plug those values into Coulomb's Law:

[tex]F = k\frac{q1q2}{r^{2} } \\1.7 = \frac{(9x10^{9})(10.0 x 10^{-6})(-3 x 10^{-6})}{r^{2} }[/tex]

Solve for r

r = 0.399 m

The adiabatic process refers to a thermodynamic process in which there is no heat transfer involved between the system and the surroundings. During an adiabatic process, the change in the internal energy of the system is achieved by the transfer of energy from or to the system, which results in a change in temperature.

Argon gas is initially at a pressure of 100 kPa and a temperature of 300 K and expands adiabatically from 0.01 m3 to 0.027 m3 while doing work on a piston. The work done by the gas on the piston during the adiabatic expansion can be calculated using the formula for the work done by the gas: W = (γ / (γ - 1)) * p * (V2 - V1)where,γ = Cp / Cv is the ratio of specific heats of the gas.

Cp = Specific heat at constant pressure, Cv = Specific heat at constant volume, p = Initial pressure of the gasV1 = Initial volume of the gasV2 = Final volume of the gas.

Initial pressure, p1 = 100 kPa, Initial temperature, T1 = 300 K, Initial volume, V1 = 0.01 m3, Final volume, V2 = 0.027 m3. The specific heat at constant pressure and constant volume of argon gas is constant. The value of γ can be calculated as follows:γ = Cp / Cv = 1.67 / 1.40 = 1.193Therefore,γ / (γ - 1) = 3.77.

Work done by the gas can be calculated as W = 3.77 * 100 kPa * (0.027 m3 - 0.01 m3)W = 95.44 kJHence, the work done by the argon gas during the adiabatic expansion is 95.44 kJ.

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The specific heat of the copper pan is the lowest among the other given pans. Therefore, a copper pan should carter used that will heat up and cool down the fastest. Therefore, option (A) is correct.

Specific heat of a substance can be described as the heat energy required to raise the temperature of one unit mass of a material by 1 °C. The S.I. unit of the specific heat capacity of a material is J/g°C.

The thermal capacity of a substance is a physical characteristic of a substance so it depends upon the nature of the material.

The mathematical expression of specific heat can be represented as :

Q = m×C× ΔT        Where C is the specific heat.

The specific heat is an intensive characteristic of a material and does not depend upon the shape or size of the quantity.

As given the values of the specific heat of metals of the frying pans, the highest specific heat means that the pan will take more heat to increase just one-degree temperature. As copper has the lowest specific heat so it can easily heat up in comparison to other pans.

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Using the theorem of calculus the derived derivative of the function found is f(x) = ∫₀ˣ 5 + sec(5t) dt is f'(x) = -x^5 + sec(5t).

Using the first part of the Fundamental Theorem of Calculus, we can find the derivative of the function f(x) = ∫[0, x] (5 + sec(5t)) dt.

Let F(x) be the antiderivative of the integrand 5 + sec(5t) with respect to t. By evaluating the integral at the upper limit x and subtracting the value at the lower limit 0, we obtain F(x) - F(0).

To find the derivative of f(x), we differentiate both sides of the equation with respect to x. Using the chain rule, we have:

f'(x) = (d/dx)(F(x) - F(0))

Since F(0) is a constant, its derivative is zero. Therefore, the equation simplifies to:

f'(x) = d/dx (F(x)) = F'(x)

The derivative of F(x) is the original integrand, 5 + sec(5t). Therefore, the derivative of the function f(x) is:

f'(x) = 5 + sec(5t)

Hence, the derivative of f(x) is 5 + sec(5t).

The derivative of the function f(x) = ∫₀ˣ 5 + sec(5t) dt is f'(x) = -x^5 + sec(5t).

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Answer: The resistance in the circuit is 250 ohms

Explanation:

According to Ohm's law:

[tex]V=IR[/tex]

where V = voltage  = 55 V

I = current in Amperes = 0.22 A

R = Resistance = ?

Putting in the values we get:

[tex]55V=0.22A\times R[/tex]

[tex]R=250ohm[/tex]

Thus the resistance in the circuit is 250 ohms

The magnetic field strength needed on the square loop to create a maximum torque of 320 N·m is approximately 43.24 N/A.

To calculate the magnetic field strength needed to create a maximum torque on a square loop, we can use the formula:

Torque (T) = N × B × A × sinθ

Where:

T = Torque

N = Number of turns in the loop

B = Magnetic field strength

A = Area of the loop

θ = Angle between the magnetic field and the plane of the loop

In this case, we are given:

N = 185 turns

A = (20.0 cm)² = 0.04 m² (since the loop is square)

T = 320 N·m

θ = 90 degrees (since the torque is maximum)

Rearranging the formula, we can solve for B:

B = T / (N × A × sinθ)

Substituting the given values:

B = 320 N·m / (185 × 0.04 m² × sin(90°))

B = 320 N·m / (7.4 m²)

B ≈ 43.24 N/A

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The question is -

Calculate the magnetic field strength needed on a 185-turn square loop 20.0 cm on a side to create a maximum torque of 320 N·m, if the loop is carrying 21 A.

T?

Solution :

As the charge is stationary, hence

[tex]$F_m= qvB \sin \theta$[/tex]

[tex]$F_m=0$[/tex]

Hence, no torque at all.

When the charge is moving in positive x direction and the field will be in the negative y direction outside the bar, then :

[tex]$F = q(V \hat i \times B(- \hat j))$[/tex]

    [tex]$= -qV B (\hat i \times \hat j)$[/tex]

    [tex]$=qVB(- \hat k)$[/tex]

Hence, the force have direction [tex]$(- \hat k)$[/tex].

When instead of charge, an iron nail is used, then there will be induced magnetic field in the soft iron. The nature of the pole induced will be opposite near tot he bar. That is the north pole will be induced near the south pole and vice versa. That is why whichever be the pole of magnet closest to iron will be attracted by iron.

We are givenInitial velocity vector vA has a magnitude of 3.00 meters per second and points 20.0o north of east, while final velocity vector vB has a magnitude of 6.00 meters per second and points 40.0o south of east. We need to find the magnitude and the direction of the change in velocity vector Δv (which is the vector subtraction of the two vectors:

final velocity vector minus initial velocity vector).Let's solve the given problem:From the above figure, the direction of Δv is at an angle θ to the east of south:

[tex]θ = θ2 - θ1= 40.0 - (-20.0)= 60.0o[/tex]

Magnitude of the Δv: Let's use the Pythagorean theorem to find the magnitude of Δv. We have:[tex]$$|\Delta \vec{v}| = \sqrt{|\vec{v}_B|^2+|\vec{v}_A|^2-2|\vec{v}_A||\vec{v}_B|\cos(\theta)}$$[/tex]  

Putting the given values in the above equation, we get

[tex]$$|\Delta \vec{v}| = \sqrt{(6.00)^2+(3.00)^2-2(6.00)(3.00)\cos(60.0)}$$$$|\Delta \vec{v}| = 3.10\ \text{m/s}$$[/tex]

So, the magnitude of the Δv is 3.10 m/s.Therefore, the magnitude and the direction of the change in velocity vector Δv is 3.10 m/s at an angle of 60.0o to the east of south.

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Hopkins Memorial Forest in western Massachusetts has been collecting meteorological and environmental data for more than 100 years. Scientists at the forest have observed that in the past few years, the sulfate content in water samples collected from Birch Brook has averaged 7.48 mg/L with a standard deviation of 1.60 mg/L.

Meteorological and environmental data are crucial for understanding the state of the natural environment. Hopkins Memorial Forest in western Massachusetts has been collecting this data for over a century. The data collected in the forest can provide valuable insights into how environmental factors such as climate change, pollution, and other factors affect the local ecosystem.They have found that the sulfate content in these samples has averaged 7.48 mg/L with a standard deviation of 1.60 mg/L.

This information is useful for understanding how sulfate levels in the water are changing over time, and whether this could have any implications for the local ecosystem.The scientists at Hopkins Memorial Forest in western Massachusetts have a unique dataset that can provide valuable insights into how environmental factors affect ecosystems over time. By continuing to collect meteorological and environmental data, they will be able to gain a better understanding of how the environment is changing and how these changes could affect the local ecosystem in the future.

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Answer:

d = 1.07 mile

Explanation:

The rationale for this method is that the speed of light is much greater than the speed of sound, the definition of speed in uniform motion is  

v = d / t  

d = v t  

the speed of sound is worth  

v = 343 m / s  

Therefore, the speed of sound must be multiplied by time to do this, all the units must be in the same system, as the distance in miles is requested  

v = 343 m/s (1mile/1609 m) (3600s/1 h) = 343 (2.24) = 767.4 mile/h  

v = 343 m / s (1 mile / 1609 m) = 0.213, mile/ s  

If the measured time is t = 5s we multiply it by the speed  

we substitute  

d = 0.213 5

d = 1.07 mile

If you want to calculate the speed, this method in general is not widely used, since you must know the distance where the lightning occurred, which is relatively complicated.

The ratio of the periods, t1/t2, for the two masses m1 and m2 in simple harmonic motion (SHM) on springs with the same spring constant k is 1:2.

In SHM, the period of oscillation (T) is given by the equation:

T = 2π√(m/k)

where T is the period, m is the mass, and k is the spring constant.

Let's calculate the periods for the two masses:

For mass m1:

T1 = 2π√(m1/k)

For mass m2 (which is 4 times m1):

T2 = 2π√(m2/k)

To find the ratio of the periods, we divide T1 by T2:

t1/t2 = (T1)/(T2)

Substituting the expressions for T1 and T2:

t1/t2 = (2π√(m1/k))/(2π√(m2/k))

Canceling out the common factors of 2π, we get:

t1/t2 = (√(m1/k))/(√(m2/k))

Since m2 = 4m1, we can substitute this value:

t1/t2 = (√(m1/k))/(√((4m1)/k))

Simplifying further:

t1/t2 = (√(m1/k))/(2√(m1/k))

The square root terms cancel out, resulting in:

t1/t2 = 1/2

The ratio of the periods, t1/t2, for the two masses m1 and m2 that oscillate in SHM on springs with the same spring constant k is 1:2.

This means that the period of oscillation for the larger mass (m2) is twice the period of the smaller mass (m1).

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Answer: The maximum displacement the spring is stretched or compressed from its equilibrium position is its AMPLITUDE.

Explanation:

In simple harmonic motion, the restoring force which pulls the oscillating body back towards its rest position is proportional in magnitude to the displacement of the body from the rest position.

The simple harmonic motion in terms of MASS AND SPRING, simple pendulum and loaded test tube is the motion or movement of a particle in a to and fro movement along a straight line under the influence of force.

Mass and spring: This means when a string of suspended mass, M, with initial level of the spring is at rest, the spring will start moving upward and downward due to the imbalance of the suspended mass.

The maximum displacement as the spring is stretched or compressed from its equilibrium position is its AMPLITUDE. This is measured in units of meter.

Answer:

since the charges are of different nature it's a attractive force

Explanation:

magnitude of force=

9*10^9*5*10^-6*3*10^-6/0.04

= 3.375N answer

The train covers a distance of 100 meters in the given time.

To find the distance covered by the train in the given time, we can use the equations of motion.

S = ut + (1/2)at²

The equation S = ut + (1/2)at² is derived from the basic equations of motion. The first term (ut) represents the distance covered in the initial velocity u multiplied by time t. The second term (1/2)at² represents the distance covered due to the acceleration a during time t.

The initial velocity (u) of the train is 10 m/s, and the acceleration (a) is 4 m/s². We are given that this acceleration is attained after 5 seconds, so the time (t) is also 5 seconds. We need to find the distance covered (S).

Substituting the given values:

S = (10 m/s)(5 s) + (1/2)(4 m/s²)(5 s)²

S = 50 m + (1/2)(4 m/s²)(25 s²)

S = 50 m + 50 m

S = 100 m

It's important to note that the given problem assumes a constant acceleration throughout the entire time interval.

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Answer: the average velocity should be 13.422

Explanation:

tell me if i'm wrong please because i want to know if my calculations are reliable

Answer:

I think its A

Explanation:

Not 100 percent sure tho but they do go up and down in big movements.

(a) The magnitude of the impulse experienced by the soccer ball during the bounce is 0.6923 kg·m/s, (b) the magnitude of the constant force acting on the soccer ball during the bounce is 13.846 N and (c) the soccer ball transferred approximately 12.026 joules of energy to the environment during the bounce.

What is energy and how is it measured?

Energy is a fundamental concept in physics that refers to the ability of a system to do work or cause a change. It is a scalar quantity and is associated with various forms, such as kinetic energy, potential energy, thermal energy, and others.

The SI unit of energy is the joule (J).

Part (a):

The magnitude of the impulse (J) experienced by the soccer ball during the bounce can be calculated using the equation:

J = Δp,

where Δp is the change in momentum.

The change in momentum is given by:

Δp = m * Δv,

where m is the mass of the soccer ball and Δv is the change in velocity.

The initial velocity of the soccer ball is zero as it is dropped from rest. The final velocity can be calculated using the equation for final velocity in free fall:

v² = u² + 2gh,

where v is the final velocity, u is the initial velocity (zero in this case), g is the acceleration due to gravity, and h is the height.

Calculating the final velocity:

v²  = 0² + 2 * 9.8 m/s² * 2.12 m,

v ≈ 6.02 m/s.

Substituting the values into the equation for change in momentum:

Δp = m * (v - u),

Δp = 0.115 kg * (6.02 m/s - 0 m/s).

Calculating:

Δp ≈ 0.6923 kg·m/s.

Therefore, the magnitude of the impulse experienced by the soccer ball during the bounce is approximately 0.6923 kg·m/s.

Part (b):

The magnitude of the constant force (F) acting on the soccer ball can be calculated using the equation:

F = Δp / Δt,

where Δp is the change in momentum and Δt is the time interval.

Given that the soccer ball was in contact with the ground for Δt = 0.05 s, we can substitute the values into the equation:

F = 0.6923 kg·m/s / 0.05 s.

Calculating:

F = 13.846 N.

Therefore, the magnitude of the constant force acting on the soccer ball during the bounce is 13.846 N.

Part (c):

The energy transferred to the environment during the bounce can be calculated as the work done by the force of the ball on the ground.

The work done is given by:

W = F * d,

where F is the magnitude of the force and d is the distance over which the force acts.

In this case, the force acts over the distance between the initial and final heights, which is h₁ - h₂.

Substituting the values:

W = 13.846 N * (3.05 m - 2.12 m).

Calculating:

W ≈ 12.026 J.

Therefore, the soccer ball transferred approximately 12.026 joules of energy to the environment during the bounce.

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Given that the electric potential at a distance of 6 m from a certain point charge is 240 V relative to infinity. The electric potential (relative to infinity) at a distance of 2 m from the same charge is 80 V.

Electric potential (relative to infinity) at a distance of 2 m from the same charge can be calculated as follows: By using the formula of electric potential, the electric potential at any point of space due to a point charge is given by;

V = kq / r

where,V = Electric potential due to point charge

q = Point charge

k = Coulomb's constant = 9 × 10^9 Nm^2C^-2

r = Distance between the charge and point where electric potential is to be calculated. Hence,Electric potential at a distance of 6 m from point charge q,V = kq / r1 = 9 × 10^9 × q / 6 ............(1)

Electric potential at a distance of 2 m from point charge q,

V = kq / r2 = 9 × 10^9 × q / 2 ............(2)

Divide equation (1) by equation (2), we get,

240 / V = 6 / 2V = 240 / (6 / 2)

By solving the above equation, we get

V = 80 V

Therefore, the electric potential (relative to infinity) at a distance of 2 m from the same charge is 80 V.

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The sample is approximately 216 years old. In a radioactive decay process, the parent isotope gradually transforms into daughter isotopes over time.

The half-life of an isotope is the time it takes for half of the parent isotope to decay into the daughter isotope. In this case, if the sample contains 25% parent isotope and 75% daughter isotopes, it means that half of the parent isotope has decayed, resulting in the current ratio. Since the half-life of the parent isotope is 72 years, we can determine the age of the sample by calculating the number of half-lives that have occurred. Each half-life represents a reduction of 50% in the parent isotope.

Starting with 100% parent isotope, after one half-life (72 years), it reduces to 50% parent and 50% daughter isotopes. After the second half-life (another 72 years), it reduces to 25% parent and 75% daughter isotopes, which matches the given ratio in the sample. Therefore, two half-lives have occurred, resulting in an age of approximately 144 years. To find the total age of the sample, we multiply the half-life by the number of half-lives. In this case, 72 years (half-life) multiplied by 2 (number of half-lives) gives us an approximate age of 144 years. Therefore, the sample is approximately 144 years old.

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