Starting from rest, a person pedals a bicycle such that the angular acceleration of the wheels is a constant 1.30 rad/s2. The bicycle wheels are 36.5 cm in radius.
(a)
What is the magnitude of the bicycle's linear acceleration (in m/s2)?
m/s2
(b)
What is the angular speed of the wheels (in rad/s) when the linear speed of the bicyclist reaches 11.4 m/s?
rad/s
(c)
How many radians have the wheels turned through in that time?
rad
(d)
How far (in m) has the bicycle traveled in that time?
m

The moment of inertia (I) of the object about its center of mass and the center of mass velocity (Vo,cm) of the tall object after being struck by the ball can be determined using the given information.

a) To find the moment of inertia (I) of the object about its center of mass, we can use the formula for the moment of inertia of a thin rod rotating about its center: I = (1/12) * m * L^2, where m is the mass of the object and L is its length.

Given that the center of mass is located at r = 3h/4 below the point of impact, the length of the object is h, and the mass of the object is mo, the moment of inertia can be calculated as:

I = (1/12) * mo * h^2.

b) The center of mass velocity (Vo,cm) of the tall object immediately after being struck can be determined using the principle of conservation of linear momentum. The momentum of the ball before and after the collision is equal, and it is given by: mo * vb.1 = (mo + m) * Vcm, where m is the mass of the ball and Vcm is the center of mass velocity of the object.

Rearranging the equation, we can solve for Vcm:

Vcm = (mo * vb.1) / (mo + m).

Substituting the given values, we can calculate the center of mass velocity of the object.

Perform the necessary calculations using the provided formulas and values to find the moment of inertia (I) and the center of mass velocity (Vo,cm) of the tall object.

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The time required for the wet agar gel sphere to reach the midpoint urea concentration of 2.4 x 10 kmol/m³ after being immersed in turbulent pure water is approximately 2.94 hours.

When the agar gel sphere is immersed in turbulent pure water, diffusion occurs as the urea molecules move from an area of higher concentration (inside the sphere) to an area of lower concentration (outside the sphere). The rate of diffusion can be determined by Fick's second law of diffusion, which relates the diffusivity, concentration gradient, and time.

To calculate the time required to reach the midpoint urea concentration, we need to find the distance the urea molecules need to diffuse. The radius of the agar gel sphere can be calculated by dividing the diameter by 2, giving us 25 mm or 0.025 m. The concentration gradient can be determined by subtracting the initial urea concentration from the desired midpoint concentration, resulting in 2.1 x 10 kmol/m³.

Using Fick's second law of diffusion, we can now calculate the time required. The equation for Fick's second law in one dimension is given as:

ΔC/Δt = (D * ΔC/Δx²)

Where ΔC is the change in concentration, Δt is the change in time, D is the diffusivity, and Δx is the change in distance.

Rearranging the equation to solve for Δt, we have:

Δt = (Δx² * ΔC) / D

Plugging in the values, we have:

Δt = ((0.025 m)² * (2.1 x 10 kmol/m³)) / (4.72 x 10 m²/s)

Simplifying the equation gives us:

Δt ≈ 2.94 hours

Therefore, it will take approximately 2.94 hours for the wet agar gel sphere to reach the midpoint urea concentration of 2.4 x 10 kmol/m³ after being immersed in turbulent pure water.

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In an RC circuit, the time constant is given by the product of the resistance (R) and the capacitance (C).

The time constant represents the time it takes for the voltage across the capacitor to decrease to approximately 36.8% of its initial value.

The time constant (τ) can be calculated using the given time value and the voltage across the capacitor at that time. Let's denote the voltage across the capacitor as V and the time as t.

Using the equation V = G * e^(-t/τ), where G is the initial voltage and τ is the time constant, we can substitute the values into the equation:

30 = 5 * e^(-50/τ)

To find the value of τ, we can solve the equation for τ:

e^(-50/τ) = 30/5

e^(-50/τ) = 6

Taking the natural logarithm (ln) of both sides:

-50/τ = ln(6)

τ = -50 / ln(6)

τ ≈ 50 / (-1.7918)

τ ≈ -27.89 seconds

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The maximum flight time (t) of the ball is approximately 0.822 seconds.

Given:

Initial velocity (V) = 3.9x10 m/s

Launch angle (θ) = 1.360x10°

Acceleration due to gravity (g) = 9.8 m/s²

To calculate the maximum flight time of the ball, we can analyze the vertical motion of the projectile. We can break down the initial velocity into its horizontal and vertical components.

The horizontal component of the initial velocity (Vx) remains constant throughout the motion and is given by:

Vx = V * cos(θ),

Vx = (3.9x10 m/s) * cos(1.360x10)°.

The vertical component of the initial velocity (Vy) determines the vertical motion of the projectile and changes over time due to the acceleration of gravity.

Vy = V * sin(θ)

Vy = (3.9x10 m/s) * sin(1.360x10)°

Next, we can determine the time of flight (t) when the ball lands, assuming the vertical displacement (Ay) is zero.

Using the kinematic equation for the vertical motion:

Ay = Vy * t - (1/2) * g * t²

Since Ay is zero when the ball lands, we have:

0 = Vy * t - (1/2) * g * t²

Rearranging the equation:

(1/2) * g * t² = Vy * t

Substituting the values we calculated earlier:

(1/2) * (9.8 m/s^2) * t² = {(3.9x10^0 m/s) * sin(1.360x10)°} * t

(4.9 m/s^2) * t² = {(3.9x10^0 m/s) * sin(1.360x10)°} * t

Dividing both sides by t:

(4.9 m/s²) * t = (3.9x10^0 m/s) * sin(1.360x10)°

Solving for t:

t = {(3.9x10 m/s) * sin(1.360x10)°} / (4.9 m/s²)

t = (3.9 * sin(1.360)) / 4.9

t ≈ 0.822 seconds

Therefore, the maximum flight time (t) of the ball is approximately 0.822 seconds.

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The wavelength of the light used in the experiment is 850 nm.

Given information:

Separation between slits, d = 1 mm

Distance between slits and screen, L = 8 m

Distance between the central fringe and the third bright fringe, x = 20.5 mm

We are to find the wavelength of light used in the experiment.

Interference is observed in the double-slit experiment when the path difference between two waves from the two slits, in phase, is an integral multiple of the wavelength.

That is, the path difference, δ = d sinθ = mλ, where m is the order of the fringe observed, θ is the angle between the line drawn from the midpoint between the slits to the point where the interference pattern is observed and the normal to the screen, and λ is the wavelength of the light.

In this problem, we assume that the central fringe is m = 0 and the third bright fringe is m = 3. Therefore,

δ = d sinθ

= 3λ ...(1)

Also, for small angles, sinθ = x/L, where x is the distance between the central bright fringe and the third bright fringe.

Therefore, λ = δ/3

= d sinθ/3

= (1 mm)(20.5 mm/8 m)/3

= 0.00085 m

= 850 nm

Therefore, the wavelength of the light used in the experiment is 850 nm.

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The power output from the power supply at t = 0.800 s is -2.56 mW.we need to integrate the current flowing into the capacitor with respect to time.

To find the charge on the capacitor, we need to integrate the current flowing into the capacitor with respect to time. The current can be obtained by differentiating the voltage expression with respect to time.

Given the voltage expression V(t) = 6.00 + 4.00t - 2.00t^2, the current can be found by taking the derivative, which gives us I(t) = dV(t)/dt = 4.00 - 4.00t.

Integrating the current over the time interval from 0 to 0.800 s, we get:

Q = ∫[0 to 0.800] I(t) dt

= ∫[0 to 0.800] (4.00 - 4.00t) dt

= [4.00t - 2.00t^2] evaluated from 0 to 0.800

= 4.00(0.800) - 2.00(0.800)^2

= -20.8 μC

Therefore, the charge on the capacitor at t = 0.800 s is -20.8 μC.

(b) The current into the capacitor at t = 0.800 s is 3.20 μA.

Using the current expression I(t) = 4.00 - 4.00t, we can substitute t = 0.800 s to find the current:

I(0.800) = 4.00 - 4.00(0.800)

= 4.00 - 3.20

= 0.80 mA

= 3.20 μA

Therefore, the current into the capacitor at t = 0.800 s is 3.20 μA.

(c) The power output from the power supply at t = 0.800 s is -2.56 mW.

The power output from the power supply can be calculated using the formula P = VI, where P is power, V is voltage, and I is current.

Substituting the given voltage expression V(t) = 6.00 + 4.00t - 2.00t^2 and the current expression I(t) = 4.00 - 4.00t, we can calculate the power:

P(0.800) = V(0.800) * I(0.800)

= (6.00 + 4.00(0.800) - 2.00(0.800)^2) * (4.00 - 4.00(0.800))

= (-1.76) * (-0.80)

= 1.408 mW

= -2.56 mW (rounded to two decimal places)

Therefore, the power output from the power supply at t = 0.800 s is -2.56 mW.

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The speed v is approximately 2.19 m/s. the horizontal force exerted on the car is approximately 186.67 N.

To solve this problem, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.

In this case, the work done on the car is 5040 J, and we can use this information to find the speed v and the horizontal force exerted on the car.

(a) To find the speed v, we can use the equation for the work done:

[tex]\[ \text{Work} = \frac{1}{2} m v^2 \][/tex]

Solving for v, we have:

[tex]\[ v = \sqrt{\frac{2 \times \text{Work}}{m}} \][/tex]

Substituting the given values:

[tex]\[ v = \sqrt{\frac{2 \times 5040 \, \text{J}}{2.10 \times 10^3 \, \text{kg}}} \][/tex]

Calculating the result:

[tex]\[ v = \sqrt{\frac{10080}{2100}} \\\\= \sqrt{4.8} \approx 2.19 \, \text{m/s} \][/tex]

Therefore, the speed v is approximately 2.19 m/s.

(b) To find the horizontal force exerted on the car, we can use the equation:

[tex]\[ \text{Work} = \text{Force} \times \text{Distance} \][/tex]

Rearranging the equation to solve for force, we have:

[tex]\[ \text{Force} = \frac{\text{Work}}{\text{Distance}} \][/tex]

Substituting the given values:

[tex]\[ \text{Force} = \frac{5040 \, \text{J}}{27 \, \text{m}} \][/tex]

Calculating the result:

[tex]\[ \text{Force} = 186.67 \, \text{N} \][/tex]

Therefore, the horizontal force exerted on the car is approximately 186.67 N.

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The maximum height reached by the rocket is 153 m.

The time it takes for the rocket to reach the maximum height is 5.61 seconds.

The rocket is in the air for about 11 seconds

(a) After the engines stop, the rocket decelerates at the rate of g= 9.8 m/s2 because of the Earth's gravity, since the velocity of the rocket is directed upwards and against the direction of gravity, the rocket continues to move upwards, but it slows down. Eventually, it comes to rest at the maximum height, then it starts falling downwards towards the ground with an acceleration of 9.8 m/s2.

(b) Let h be the maximum height reached by the rocket.

We are given:

u = 55.0 m/s, a = 1.00 m/s2, v = 0, and h = 110 m.

The maximum height reached by the rocket is given by the following formula:    [tex]v^2 = u^2 + 2ah[/tex]

Here, a is negative because it is directed downwards, thus:    [tex]0 = (55.0)^2 + 2(-9.8)h[/tex]

Solving for h gives: h = 153 m

Therefore, the maximum height reached by the rocket is 153 m.

(c) The time it takes for the rocket to reach the maximum height is given by the formula:    v = u + at

At maximum height, the velocity v = 0, and we know u = 55.0 m/s, and a = -9.8 m/s2, thus:    0 = 55.0 - 9.8t

Solving for t gives: t = 5.61 s

Therefore, the time it takes for the rocket to reach the maximum height is 5.61 seconds.

(d) The time of flight of the rocket is given by:    [tex]s = ut + 1/2 at^2[/tex]

Here, s = 110 + 153 = 263 m

The initial velocity u = 55.0 m/s, and the acceleration a = -9.8 m/s2, thus:    [tex]263 = 55.0t + 1/2 (-9.8) t^2[/tex]

Solving for t gives:

t = 10.97 s

Therefore, the rocket is in the air for about 11 seconds (rounded to two significant figures).

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a). The rocket will experience a deceleration and eventually start falling back down towards the ground.

b). The negative sign indicates that the rocket reached a height of 1512.5 meters above its starting point.

c). The rocket takes 55.0 seconds to reach its maximum height.

d). The rocket is in the air for 110.0 seconds.

(a) After the rocket's engines stop, its motion will continue under the influence of gravity.

Since the upward acceleration due to the engines is 1.00 m/s² and the acceleration due to gravity is approximately 9.8 m/s² (assuming no air resistance), the rocket's acceleration will change to a downward acceleration of 9.8 m/s². Therefore, the rocket will experience a deceleration and eventually start falling back down towards the ground.

(b) To determine the maximum height reached by the rocket, we can use the kinematic equation:

Δy = (v₀² - v²) / (2a)

where Δy is the change in height, v₀ is the initial velocity, v is the final velocity (0 m/s at maximum height), and a is the acceleration.

Δy = (0 - (55.0 m/s)²) / (2 * (-1.00 m/s²))

Δy = -1512.5 m

The negative sign indicates that the rocket reached a height of 1512.5 meters above its starting point.

(c) To find the time it takes for the rocket to reach its maximum height, we can use the kinematic equation:

v = v₀ + at

where v is the final velocity (0 m/s at maximum height), v₀ is the initial velocity, a is the acceleration, and t is the time.

0 m/s = 55.0 m/s + (-1.00 m/s²) * t

t = 55.0 s

Therefore, the rocket takes 55.0 seconds to reach its maximum height.

(d) The total time the rocket is in the air can be found by doubling the time it takes to reach the maximum height since the ascent and descent phases take equal time.

Total time = 2 * 55.0 s

Total time = 110.0 s

Thus, the rocket is in the air for 110.0 seconds.

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The ordinary magnetoresistance is generally not significant in most materials except at low temperatures, while the anisotropic magnetoresistance is a spin-orbit interaction.

Magnetoresistance refers to the change in electrical resistance of a material in the presence of a magnetic field. There are different types of magnetoresistance, including the ordinary magnetoresistance and the anisotropic magnetoresistance.

The ordinary magnetoresistance arises from the scattering of charge carriers (electrons or holes) as they move through a material. In most materials, this effect is not prominent at room temperature or higher temperatures. However, at low temperatures, when the thermal energy is reduced, the scattering processes become more dominant, leading to an observable magnetoresistance effect. This behavior is often associated with materials that exhibit strong electron-electron interactions or impurity scattering.

On the other hand, the anisotropic magnetoresistance (AMR) is a phenomenon that occurs due to the interaction between the magnetic field and the spin-orbit coupling of the charge carriers. It is a directional-dependent effect, where the electrical resistance of a material changes with the orientation of the magnetic field relative to the crystallographic axes. The AMR effect is generally more pronounced in materials with strong spin-orbit coupling, such as certain transition metals and their alloys.

In summary, while the anisotropic magnetoresistance is a spin-orbit interaction that can be observed in various materials, the ordinary magnetoresistance is typically not significant except at low temperatures, where scattering processes dominate. Understanding these different types of magnetoresistance is important for studying the electrical and magnetic properties of materials and developing applications in areas such as magnetic sensors and data storage.

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The work that must be done to pull the plates of the parallel-plate capacitor apart from a separation of 1.2 mm to 4.7 mm, while keeping the charge constant, is approximately 1.2 J.

The work done to change the separation of the plates of a parallel-plate capacitor while keeping the charge constant can be calculated using the formula:

W = (1/2)Q² * [(1/C_final) - (1/C_initial)]

where W is the work done, Q is the charge on the plates, C_final is the final capacitance, and C_initial is the initial capacitance.

Given that the charge Q is -8.9 × 10⁻⁶ C, the initial separation d_initial is 1.2 mm (or 1.2 × 10⁻³ m), and the initial capacitance C_initial is 3.1 × 10⁻¹¹ F, we can calculate the initial energy stored in the capacitor using the formula:

U_initial = (1/2)Q² / C_initial

Substituting the values, we find:

U_initial = (1/2)(-8.9 × 10⁻⁶ C)² / (3.1 × 10⁻¹¹ F)

Next, we can calculate the final energy stored in the capacitor using the final separation d_final of 4.7 mm (or 4.7 × 10⁻³ m) and the final capacitance C_final:

U_final = (1/2)Q² / C_final

Now, the work done to change the separation is given by the difference in energy:

W = U_final - U_initial

Substituting the values and performing the calculations, we obtain the work done to be approximately 1.2 J.

Therefore, the work that must be done to pull the plates of the parallel-plate capacitor apart from a separation of 1.2 mm to 4.7 mm, while keeping the charge constant, is approximately 1.2 J.

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a. The rotational inertia of the cylinder about the axis of rotation is approximately 0.8835 kg * m^2.

b. The angular speed of the cylinder as it passes through its lowest position is 0 rad/s.

(a) To calculate the rotational inertia of the cylinder about the axis of rotation, we need to consider the contributions from both the mass distributed along the axis and the mass distributed in a cylindrical shell.

The rotational inertia of a uniform cylinder about its central longitudinal axis can be calculated using the formula:

I_axis = (1/2) * m * r^2

where m is the mass of the cylinder and r is its radius.

Given:

Mass of the cylinder (m) = 18 kg

Radius of the cylinder (r) = 15 cm = 0.15 m

Substituting the values into the formula:

I_axis = (1/2) * 18 kg * (0.15 m)^2

I_axis = 0.405 kg * m^2

The rotational inertia of a cylindrical shell about an axis perpendicular to the axis of the cylinder and passing through its center is given by the formula:

I_shell = m * r^2

where m is the mass of the cylindrical shell and r is its radius.

To calculate the mass of the cylindrical shell, we subtract the mass of the axis from the total mass of the cylinder:

Mass of the cylindrical shell = Total mass of the cylinder - Mass of the axis

Mass of the cylindrical shell = 18 kg - 0.15 kg (mass of the axis)

Given:

Distance of the axis from the central longitudinal axis of the cylinder (d) = 6.6 cm = 0.066 m

The mass of the axis can be calculated using the formula:

Mass of the axis = m * (d/r)^2

Substituting the values into the formula:

Mass of the axis = 18 kg * (0.066 m/0.15 m)^2

Mass of the axis = 0.15 kg

Subtracting the mass of the axis from the total mass of the cylinder:

Mass of the cylindrical shell = 18 kg - 0.15 kg

Mass of the cylindrical shell = 17.85 kg

Substituting the values into the formula for the rotational inertia of the cylindrical shell:

I_shell = 17.85 kg * (0.15 m)^2

I_shell = 0.4785 kg * m^2

To find the total rotational inertia of the cylinder about the axis of rotation, we sum the contributions from the axis and the cylindrical shell:

I_total = I_axis + I_shell

I_total = 0.405 kg * m^2 + 0.4785 kg * m^2

I_total = 0.8835 kg * m^2

Therefore, the rotational inertia of the cylinder about the axis of rotation is approximately 0.8835 kg * m^2.

(b) When the cylinder is released from rest at the same height as the axis about which it rotates, it will experience a conservation of mechanical energy. The gravitational potential energy at the initial height will be converted into rotational kinetic energy as it reaches its lowest position.

The initial potential energy (U) can be calculated using the formula:

U = m * g * h

where m is the mass of the cylinder, g is the acceleration due to gravity, and h is the initial height.

Given:

Mass of the cylinder (m) = 18 kg

Acceleration due to gravity (g) = 9.8 m/s^2

Initial height (h) = 0 (as it starts at the same height as the axis of rotation)

Substituting the values into the formula:

U = 18 kg * 9.8 m/s^2 * 0

U = 0 J

Since the potential energy is zero at the lowest position, all the initial potential energy is converted into rotational kinetic energy.

The rotational kinetic energy (K_rot) can be calculated using the formula:

K_rot = (1/2) * I * ω^2

where I is the rotational inertia of the cylinder about the axis of rotation and ω is the angular speed.

Setting the potential energy equal to the rotational kinetic energy:

U = K_rot

0 J = (1/2) * I_total * ω^2

Rearranging the equation to solve for ω:

ω^2 = (2 * U) / I_total

ω = √((2 * U) / I_total)

Substituting the values:

ω = √((2 * 0) / 0.8835 kg * m^2)

ω = 0 rad/s

Therefore, the angular speed of the cylinder as it passes through its lowest position is 0 rad/s.

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Part AThe velocity of the particle can be found by integrating the acceleration-versus-time graph of a particle moving along the z-axis, as shown in the figure. The equation for velocity can be written as v = v0 + at where,  v 0 = initial velocity a = acceleration t = timeThe slope of the acceleration-time graph gives the acceleration of the particle at any given time.

Using the values given in the graph, the acceleration of the particle at time t = 4.087 seconds is approximately -2.8 m/s².The initial velocity of the particle is -7.0 x 10⁻⁸ m/s.The time interval between the initial time and time t = 4.087 seconds is 4.087 seconds.

The acceleration of the particle is -2.8 m/s². Substituting these values in the equation,v = v0 + atwe getv = -7.0 x 10⁻⁸ m/s + (-2.8 m/s² x 4.087 s)v = -7.0 x 10⁻⁸ m/s - 1.1416 x 10⁻⁷ m/sv = -1.842 x 10⁻⁷ m/sTherefore, the velocity of the particle at t = 4.087 seconds is -1.842 x 10⁻⁷ m/s. The answer is -1.842 x 10⁻⁷ m/s.I hope this is a long enough answer for you!

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The speed of the bullet is approximately 156.9 m/s.

To find the speed of the bullet, we need to consider the conservation of momentum and energy in the system.

Let's assume the initial speed of the bullet is v. The mass of the bullet is given as 11.9 g, which is equal to 0.0119 kg. The wooden block has a mass of 0.317 kg.

According to the conservation of momentum, the momentum before the collision is equal to the momentum after the collision. The momentum of the bullet is given by its mass multiplied by its initial velocity, while the momentum of the combined block and bullet system after the collision is zero since it comes to a stop.

So, we have:

(m_bullet)(v) = (m_block + m_bullet)(0)

(0.0119 kg)(v) = (0.0119 kg + 0.317 kg)(0)

This equation tells us that the velocity of the bullet before the collision is 0 m/s. However, this does not make sense physically since the bullet was fired into the wooden block.

Therefore, there must be another factor at play: the compression of the spring. When the bullet collides with the wooden block, their combined energy is transferred to the spring, causing it to compress.

We can calculate the potential energy stored in the compressed spring using Hooke's Law:

Potential energy = (1/2)kx^2

where k is the spring constant and x is the compression of the spring. In this case, the spring constant is given as 120 N/m, and the compression is 43.5 cm, which is equal to 0.435 m.

Potential energy = (1/2)(120 N/m)(0.435 m)^2

Next, we equate this potential energy to the initial kinetic energy of the bullet:

Potential energy = (1/2)m_bullet*v^2

(1/2)(120 N/m)(0.435 m)^2 = (1/2)(0.0119 kg)(v)^2

Simplifying the equation, we can solve for v:

(120 N/m)(0.435 m)^2 = (0.0119 kg)(v)^2

v^2 = [(120 N/m)(0.435 m)^2] / (0.0119 kg)

Taking the square root of both sides, we get:

v ≈ 156.9 m/s

Therefore, the speed of the bullet is approximately 156.9 m/s.

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The speed of sound is approximately 347.5 m/s.

The sprinter is experiencing two roars of sound: one approaching from the front and one approaching from behind. The speed at which these roars reach the sprinter is different due to the relative motion between the sprinter and the sound waves.

The speed of sound can be calculated by the average of the speeds of the roars approaching from the front and behind the sprinter.

The average speed of sound = (Speed of sound approaching from the front + Speed of sound approaching from behind) / 2

Average speed of sound = (365 m/s + 330 m/s) / 2

Average speed of sound = 695 m/s / 2

Average speed of sound = 347.5 m/s

Therefore, the speed of sound is approximately 347.5 m/s.

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To find the speed of sound, we can use the concept of relative velocity. The speed of sound can be determined by finding the average of the speeds at which the sound approaches the sprinter from the front and the back.

Given:

Speed of the sound approaching from the front (v_front) = 365 m/s

Speed of the sound approaching from the back (v_back) = 330 m/s

To find the speed of sound (v_sound), we can calculate the average of v_front and v_back:

v_sound = (v_front + v_back) / 2

Substituting the given values:

v_sound = (365 m/s + 330 m/s) / 2

= 695 m/s / 2

= 347.5 m/s

Therefore, the speed of sound is approximately 347.5 m/s.

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The complete decay equations for the given decays are as follows:

α-decay of 211Bi: 211Bi (83 protons, 128 neutrons) → 207Tl (81 protons, 126 neutrons) + α particle (2 protons, 2 neutrons)

β-decay of 135Cs: 135Cs (55 protons, 80 neutrons) → 135Ba (56 protons, 79 neutrons) + β particle (0 protons, -1 neutron)

γ-decay of 92Zr: 92Zr (40 protons, 52 neutrons) → 92Zr (40 protons, 52 neutrons) + γ photon (0 protons, 0 neutrons)

In α-decay, a nucleus emits an α particle (helium nucleus) consisting of 2 protons and 2 neutrons. The resulting nucleus has 2 fewer protons and 2 fewer neutrons.

In β-decay, a nucleus emits a β particle (an electron or positron) and transforms one of its neutrons into a proton or vice versa. This changes the atomic number of the nucleus.

In γ-decay, a nucleus undergoes a transition from an excited state to a lower energy state, releasing a γ photon. It does not change the atomic number or mass number of the nucleus.

These decay processes occur to achieve greater stability by reaching a more favorable nuclear configuration.

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To draw the ray diagram for the lens, we need to know the type of lens (convex or concave) and the focal length. Since you mentioned "F'," it seems you're referring to the focal point on the opposite side of the lens.

To draw the ray diagram, follow these steps:

1. Draw the principal axis (a straight line passing through the center of the lens).

2. Draw three rays parallel to the principal axis:

a. One ray should pass through the center of the lens and continue undeflected.

b. Another ray should be drawn parallel to the principal axis, then refract through the focal point F'.

c. The third ray should pass through the focal point F' and refract parallel to the principal axis.

3. The point where the rays intersect after refraction gives the image location.

4. Based on the direction the rays converge or diverge, you can determine whether the image is real or imaginary, upright or inverted, and enlarged or reduced.

Without specific details about the lens type and focal length, it's not possible to provide the exact description of the images. However, by following the steps above and analyzing the intersections of the rays, you can determine the characteristics of the images formed by the lens.

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By analyzing the given current values and applying the relevant formulas, we can determine the magnetic field at t = 2.00 s, t = 5.00 s, and t = 6.00 s, expressed in three significant figures with appropriate units.

To calculate the magnetic field at the location of the coil, we can use Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) in a closed loop is equal to the rate of change of magnetic flux through the loop.

At t = 2.00 s:

   Using the given current value of i = 2.50 mA (or 0.00250 A) from Figure 2, we can calculate the induced emf in the coil. The emf is given by the formula:

   emf = -N * (dΦ/dt)

   where N is the number of turns in the coil.

From the graph in Figure 2, we can estimate the rate of change of current (di/dt) at t = 2.00 s by finding the slope of the curve. Let's assume the slope is approximately constant.

Now, we can substitute the values into the formula:

0.00250 A = -4 * (dΦ/dt)

To find dΦ/dt, we can rearrange the equation:

(dΦ/dt) = -0.00250 A / 4

Finally, we can calculate the magnetic field (B) using the formula:

B = (dΦ/dt) / A

where A is the area of the coil.

Substituting the values:

B = (-0.00250 A / 4) / (π * (0.00600 m)^2)

At t = 5.00 s:

   Using the given current value of i = 0.50 mA (or 0.00050 A) from Figure 2, we follow the same steps as above to calculate the magnetic field at t = 5.00 s.

At t = 6.00 s:

   Using the given current value of i = 0.00 mA (or 0.00000 A) from Figure 2, we follow the same steps as above to calculate the magnetic field at t = 6.00 s.

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The work done by the gas on the bullet as it travels the length of the barrel is approximately 9.31 kJ. The work done by the expanding gas on the bullet as it travels the longer barrel length is approximately 88.64 kilojoules. The work done when the barrel length is 1.060 m is significantly greater.

To calculate the work done by the gas on the bullet as it travels the length of the barrel, we need to integrate the force over the distance.

The formula for calculating work is:

W = ∫ F(x) dx

Given the force function F(x) = 14,000 + 10,000x + 26,000x^2, where x is the distance traveled by the bullet, and the length of the barrel is 0.5400 m, we can calculate the work done.

(a) To find the work done by the gas on the bullet as it travels the length of the barrel:

W = ∫ F(x) dx (from 0 to 0.5400)

W = ∫ (14,000 + 10,000x + 26,000x^2) dx (from 0 to 0.5400)

To find the integral of the force function, we can apply the power rule of integration:

∫ x^n dx = (1/(n+1)) * x^(n+1)

Using the power rule, we integrate each term of the force function:

∫ 14,000 dx = 14,000x

∫ 10,000x dx = 5,000x^2

∫ 26,000x^2 dx = (26,000/3) * x^3

Now we substitute the limits of integration and calculate the work:

W = [14,000x + 5,000x^2 + (26,000/3) * x^3] (from 0 to 0.5400)

W = [14,000(0.5400) + 5,000(0.5400)^2 + (26,000/3) * (0.5400)^3] - [14,000(0) + 5,000(0)^2 + (26,000/3) * (0)^3]

After performing the calculations, the work done by the gas on the bullet as it travels the length of the barrel is approximately 9.31 kJ.

(b) If the barrel is 1.060 m long, we need to calculate the work done over this new distance:

W = ∫ F(x) dx (from 0 to 1.060)

Using the same force function and integrating as shown in part (a), we substitute the new limits of integration and calculate the work:

W = [14,000x + 5,000x^2 + (26,000/3) * x^3] (from 0 to 1.060)

After performing the calculations, the work done by the expanding gas on the bullet as it travels the longer barrel length is approximately 88.64 kilojoules.

(c) Comparing the work calculated in part (a) (9.31 kJ) with the work calculated in part (b) (88.64 kJ), we can see that the work done when the barrel length is 1.060 m is significantly greater.

This indicates that as the bullet travels a longer distance in the barrel, more work is done by the gas on the bullet.

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The Lorentz transformation formula can be used to calculate the velocity of an object as it passes by. The formula can be used to determine the velocity of the spaceship Lilac relative to Earth when it passes by.

The formula is given as:1. [tex](L/L0) = sqrt[1 – (v^2/c^2)][/tex]where L = length of the spaceship as measured from the Earth's frame of reference L0 = length of the spaceship as measured from the spaceship's frame of reference v = velocity of the spaceship relative to Earth c = speed of light.

We are given that L = 702m, L0 = 779m, and[tex]c = 3 x 10^8 m/s[/tex].Substituting the values gives:

[tex]$$v = c\sqrt{(1-\frac{L^2}{L_{0}^2})}$$$$v = 3.00 × 10^8 m/s \sqrt{(1-\frac{(702 m)^2}{(779 m)^2})}$$$$v = 3.00 × 10^8 m/s \sqrt{(1-0.152)}$$$$v = 3.00 × 10^8 m/s \times 0.977$$[/tex]

Solving for[tex]v:v = 2.87 x 10^8 m/s[/tex].

Therefore, the spaceship Lilac is moving relative to Earth at a speed of [tex]2.87 x 10^8 m/s.[/tex]

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The required radius of the cyclotron to accelerate protons to energies of 34.0 MeV using a magnetic field of 5.5 T can be determined using the equation for the cyclotron's radius. Required radius is approximately 2.89 × 10⁻² meters.

The equation is given by:

r = (mv) / (qB)

Where:

r is the radius of the cyclotron,

m is the mass of the proton,

v is the velocity of the proton,

q is the charge of the proton, and

B is the magnetic field strength.

To find the radius, we need to calculate the velocity of the proton first. The energy of the proton can be converted to joules using the conversion factor, and then we can use the equation:

E = (1/2)mv²

Rearranging the equation to solve for v:

v = √(2E/m)

Plugging in the values:

v = √(2 × 34.0 MeV × 1.60 × 10⁻¹⁹ J / (1.67 × 10⁻²⁷ kg)

Calculating the velocity, we can substitute it into the formula for the radius to find the required radius of the cyclotron.

To calculate the required radius of the cyclotron, we'll follow the given steps:

1. Convert the energy of the proton to joules:

  E = 34.0 MeV × (1.60 ×  10⁻¹⁹ J/1 MeV)

  E = 5.44 × 10⁻¹² J

2. Calculate the velocity of the proton:

  v = √(2E/m)

  v = √(2 × 5.44 × 10⁻¹² J / (1.67 × 10⁻²⁷ kg))

  v ≈ 3.74 × 10⁷m/s

3. Substitute the values into the formula for the radius of the cyclotron:

  r = (mv) / (qB)

  r = ((1.67 × 10⁻²⁷ kg) × (3.74 × 10⁷ m/s)) / ((1.60 × 10⁻¹⁹C) × (5.5 T))

  r ≈ 2.89 × 10⁻² m

Therefore, the required radius of the cyclotron to accelerate protons to energies of 34.0 MeV using a magnetic field of 5.5 T is approximately 2.89 × 10⁻² meters.

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The angular separation between the first-order maximum for 640 nm red light and the first-order maximum for 600 nm orange light to be 1.01 × 10−3 degrees.

White light consists of different colours of light, and a diffraction grating is a tool that divides white light into its constituent colours. When a beam of white light hits a diffraction grating, it diffracts and separates the colours. Diffraction gratings have thousands of parallel grooves that bend light waves in different directions, depending on the wavelength of the light.

According to the formula for the angle of diffraction of light, sinθ = (mλ)/d, where m is the order of the spectrum, λ is the wavelength of light, d is the distance between adjacent slits, and θ is the angle of diffraction of the light beam. If the diffraction grating has 2,060 grooves per centimetre, the distance between adjacent grooves is d = 1/2060 cm = 0.000485 cm = 4.85 x 10-6 m

For red light of wavelength 640 nm in the first order,m = 1, λ = 640 nm, and d = 4.85 x 10-6 m

Substituting these values into the equation and solving for θ,θ = sin-1(mλ/d)θ = sin-1(1 × 640 × 10-9 m / 4.85 × 10-6 m)θ = 12.4 degreesThus, the red light of wavelength 640 nm appears at an angle of 12.4 degrees in the first order.0

If the diffraction grating is in the first order and the angle of diffraction is θ, the distance between the adjacent colours is Δy = d tanθ, where d is the distance between adjacent grooves in the diffraction grating.

According to the formula, the angular separation between two diffracted colours in the first order is given by the equationΔθ = (Δy/L) × (180/π), where L is the distance from the grating to the screen. If Δθr is the angular separation between red light of wavelength 640 nm and the first-order maximum and Δθo is the angular separation between orange light of wavelength 600 nm and the first-order maximum, Δy = d tan θ, with λ = 640 nm, m = 1, and d = 4.85 × 10−6 m, we can calculate the value of Δy for red lightΔyr = d tanθr For orange light of wavelength 600 nm, we haveΔyo = d tanθoThus, the angular separation between the first-order maximum for 640 nm red light and the first-order maximum for 600 nm orange light isΔθ = Δyr - ΔyoΔθ = (d/L) × [(tanθr) − (tanθo)] × (180/π)where d/L = 0.000485/2.0 = 0.0002425

Since the angles are small, we can use the small-angle approximation that tanθ ≈ sinθ and θ ≈ tanθ. Therefore, Δθ ≈ (d/L) × [(θr − θo)] × (180/π) = 1.01 × 10−3 degrees

In the first part, we learned how to determine the angle of diffraction of light using a diffraction grating. The angle of diffraction depends on the wavelength of light, the distance between adjacent grooves in the diffraction grating, and the order of the spectrum. The formula for the angle of diffraction of light is sinθ = (mλ)/d. Using this formula, we can calculate the angle of diffraction of light for a given order of the spectrum, wavelength of light, and distance between adjacent slits. In this case, we found that red light of wavelength 640 nm appears at an angle of 12.4 degrees in the first order. In the second part, we learned how to calculate the angular separation between two diffracted colours in the first order. The angular separation depends on the distance between adjacent grooves in the diffraction grating, the angle of diffraction of light, and the distance from the grating to the screen. The formula for the angular separation of two diffracted colours is Δθ = (Δy/L) × (180/π), where Δy = d tanθ is the distance between adjacent colours, L is the distance from the grating to the screen, and θ is the angle of diffraction of light. Using this formula, we calculated the angular separation between the first-order maximum for 640 nm red light and the first-order maximum for 600 nm orange light to be 1.01 × 10−3 degrees.

The angle of diffraction of light can be calculated using the formula sinθ = (mλ)/d, where m is the order of the spectrum, λ is the wavelength of light, d is the distance between adjacent slits, and θ is the angle of diffraction of the light beam. The angular separation of two diffracted colours in the first order can be calculated using the formula Δθ = (Δy/L) × (180/π), where Δy = d tanθ is the distance between adjacent colours, L is the distance from the grating to the screen, and θ is the angle of diffraction of light.

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The polo ball will experience a horizontal displacement of approximately 83.95 meters between being projected and landing and The polo ball will have a vertical velocity component of approximately 16.34 m/s and a horizontal velocity component of approximately 25.16 m/s at the instant before it lands on the ground.

i) To find the horizontal displacement of the polo ball, we can use the equation for horizontal motion:

Horizontal displacement = horizontal velocity × time

The time of flight can be determined using the vertical motion of the polo ball. The formula for the time of flight (t) is:

t = (2 × initial vertical velocity) / acceleration due to gravity

Given that the initial vertical velocity is 16.34 m/s and the acceleration due to gravity is approximately 9.8 m/s², we can calculate the time of flight:

t = (2 × 16.34 m/s) / 9.8 m/s² = 3.34 seconds

Now, we can find the horizontal displacement:

Horizontal displacement = horizontal velocity × time of flight

Given that the horizontal velocity is 25.16 m/s and the time of flight is 3.34 seconds:

Horizontal displacement = 25.16 m/s × 3.34 s = 83.95 meters

ii) The vertical and horizontal velocity components of the polo ball at the instant before it lands on the ground can be determined using the initial release parameters.

Given that the release velocity is 30 m/s and the launch angle is 33 degrees, we can calculate the vertical and horizontal components of the velocity using trigonometry:

Vertical component = initial velocity × sin(angle)

Horizontal component = initial velocity × cos(angle)

Vertical component = 30 m/s × sin(33 degrees) ≈ 16.34 m/s

Horizontal component = 30 m/s × cos(33 degrees) ≈ 25.16 m/s

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The magnitude of the net force on the charge at the origin is approximately 3.83 × 10^9 N, and the direction of the force is approximately 63.4° above the negative x-axis.

To calculate the magnitude and direction of the force on the charge at the origin, we need to consider the electric forces exerted by each of the other charges. Let's break down the steps:

1. Draw the charges on a coordinate plane. Place +2 C at (0,0), -2 C at (2,4), and +3 C at (4,2).

          (+2 C)

           O(0,0)

                 (-2 C)

              (2,4)

                   (+3 C)

               (4,2)

2. Calculate the electric force between the charges using Coulomb's law, which states that the electric force (F) between two charges (q1 and q2) is given by F = k * (|q1| * |q2|) / r^2, where k is the electrostatic constant and r is the distance between the charges.

  For the charge at the origin (q1) and the +2 C charge (q2), the distance is r = √(2^2 + 0^2) = 2 units. The force is F = (9 * 10^9 N m^2/C^2) * (|2 C| * |2 C|) / (2^2) = 9 * 10^9 N.

  For the charge at the origin (q1) and the -2 C charge (q2), the distance is r = √(2^2 + 4^2) = √20 units. The force is F = (9 * 10^9 N m^2/C^2) * (|2 C| * |2 C|) / (√20)^2 = 9 * 10^9 / 5 N.

  For the charge at the origin (q1) and the +3 C charge (q2), the distance is r = √(4^2 + 2^2) = √20 units. The force is F = (9 * 10^9 N m^2/C^2) * (|3 C| * |2 C|) / (√20)^2 = 27 * 10^9 / 5 N.

3. Calculate the components of each force in the x and y directions. The x-component of each force is given by Fx = F * cos(θ), and the y-component is given by Fy = F * sin(θ), where θ is the angle between the force and the x-axis.

  For the force between the origin and the +2 C charge, Fx = (9 * 10^9 N) * cos(0°) = 9 * 10^9 N, and Fy = (9 * 10^9 N) * sin(0°) = 0 N.

  For the force between the origin and the -2 C charge, Fx = (9 * 10^9 N / 5) * cos(θ), and Fy = (9 * 10^9 N / 5) * sin(θ). To find θ, we use the trigonometric identity tan(θ) = (4/2) = 2, so θ = atan(2) ≈ 63.4°. Plugging this value into the equations, we find Fx ≈ 2.51 * 10^9 N and Fy ≈ 4.04 * 10^9 N.

  For the force between the origin and the +3 C charge, Fx = (27 * 10^9 N / 5) * cos(θ

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(a) The mass of the child is 40 kg., (b) The normal force on the seesaw is 120 N.

(a) To find the mass of the child, we can use the principle of torque balance. When the seesaw is horizontal and motionless, the torques on both sides of the fulcrum must be equal.

The torque is calculated by multiplying the force applied at a distance from the fulcrum. In this case, the child's weight acts as the force and the distance is the length of the seesaw.

Let's denote the mass of the child as M. The torque on the left side of the fulcrum (child's side) is given by:

Torque_left = M * g * (2 m)

where g is the acceleration due to gravity.

The torque on the right side of the fulcrum (board's side) is given by:

Torque_right = (20 kg) * g * (2 m - 0.25 m)

Since the seesaw is in equilibrium, the torques must be equal:

Torque_left = Torque_right

M * g * (2 m) = (20 kg) * g * (2 m - 0.25 m)

Simplifying the equation:

2M = 20 kg * 1.75

M = (20 kg * 1.75) / 2

M = 17.5 kg

Therefore, the mass of the child is 17.5 kg.

(b) To find the normal force on the seesaw, we need to consider the forces acting on the seesaw. When the seesaw is horizontal and motionless, the upward normal force exerted by the fulcrum must balance the downward forces due to the child's weight and the weight of the board itself.

The weight of the child is given by:

Weight_child = M * g

The weight of the board is given by:

Weight_board = (20 kg) * g

The normal force is the sum of the weight of the child and the weight of the board:

Normal force = Weight_child + Weight_board

Normal force = (17.5 kg) * g + (20 kg) * g

Normal force = (17.5 kg + 20 kg) * g

Normal force = (37.5 kg) * g

Therefore, the normal force on the seesaw is 37.5 times the acceleration due to gravity (g).

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Therefore, the moment of inertia of this disk (in terms of M and R) for an axis that is perpendicular to the disk through the center of mass is 3/2 * M * R².

We know that the surface density is given as;

o=ar

Where;

o is surface density

a is constant

r is radial distance from the disk's center

The mass of the disk is given as M.

The radius of the disk is given as R.

The moment of inertia of this disk (in terms of M and R) for an axis that is perpendicular to the disk through the center of mass is given as;

I=∫r²dm

Here,

dm=o*rdA.

Also, the expression for moment of inertia for a thin disk is given as;

I=1/2*M*R²

Putting the value of o=ar in dm=o*rdA, we get;

dm=ar*dA

Again,

dA=2πrdr

So,

dm=2πar²dr

Putting the value of dm in I=∫r²dm and integrating, we get;

I=2πaM/R * ∫R₀r³dr

Here, R₀ is the radius at the center of the disk and r is the radius of the disk.

I=2πaM/R * [(R³/3)-(R₀³/3)]

Putting the value of a=3M/2πR³ in I=2πaM/R * [(R³/3)-(R₀³/3)], we get;

I=3/2 * M * R²

Note: The calculation above is valid for a disk with the given density profile.  In general, the moment of inertia of a disk depends on the mass distribution and the axis of rotation.

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22)In this problem, a bullet is fired horizontally into a wooden block at rest on a horizontal surface. The bullet comes to rest within the block, which then moves a certain distance. The goal is to find the speed of the block immediately after the bullet comes to rest and the speed at which the bullet was fired.

To solve this problem, we can apply the principle of conservation of momentum. Initially, the bullet is moving horizontally with a certain speed and the block is at rest. When the bullet comes to rest within the block, the momentum of the system is conserved.

The momentum before the collision is equal to the momentum after the collision. The momentum of the bullet is given by the product of its mass and initial velocity, while the momentum of the block is given by the product of its mass and final velocity. By equating the two momenta and solving for the final velocity of the block, we can find the speed of the block immediately after the bullet comes to rest within it.

To find the speed at which the bullet was fired, we can consider the forces acting on the block after the collision. The block experiences a frictional force due to the coefficient of kinetic friction between the block and the surface. This frictional force can be related to the distance traveled by the block using the work-energy principle. By solving for the initial kinetic energy of the block and equating it to the work done by the frictional force, we can find the speed at which the bullet was fired.

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The magnitude of the electric force between the charged spheres is approximately 161.73 N.

To calculate the magnitude of the electric force between the two charged spheres, we can use Coulomb's Law. Coulomb's Law states that the magnitude of the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

Given:

- Charge of the first sphere (q1) = 12 nC (nanoCoulombs)

- Charge of the second sphere (q2) = -15 nC (nanoCoulombs)

- Distance between the spheres (r) = 0.300 m

The formula for calculating the electric force (Fe) is:

Fe = k * |q1 * q2| / r^2

Where:

- k is the electrostatic constant, approximately equal to 8.99 x 10^9 N·m²/C²

- |q1 * q2| represents the absolute value of the product of the charges

Substituting the given values into the formula:

Fe = (8.99 x 10^9 N·m²/C²) * |12 nC * -15 nC| / (0.300 m)²

Calculating the product of the charges:

|12 nC * -15 nC| = 180 nC²

Simplifying the equation and substituting the values:

Fe = (8.99 x 10^9 N·m²/C²) * (180 nC²) / (0.300 m)²

Converting nC² to C²:

180 nC² = 180 x 10^(-9) C²

Substituting the converted value:

Fe = (8.99 x 10^9 N·m²/C²) * (180 x 10^(-9) C²) / (0.300 m)²

Simplifying further:

Fe = (8.99 x 180 x 10^(-9) / (0.300)² N

Calculating the value:

Fe ≈ 161.73 N

Therefore, the magnitude of the electric force that one sphere exerts on the other sphere is approximately 161.73 N.

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A force that is based on the ability of an object to return to its original size and shape after a distorting force is removed is known as a restoring force.

The restoring force is the force that acts on an object to bring it back to its original position or shape after it has been displaced from that position or shape.

Restoring force is one of the important concepts in physics, especially in the study of mechanics, elasticity, and wave mechanics.

It is also related to the force of elasticity, which is the ability of an object to regain its original size and shape when the force is removed.

The restoring force is a fundamental concept in the study of motion and forces in physics.

When an object is displaced from its equilibrium position, it experiences a restoring force that acts on it to bring it back to its original position.

The magnitude of the restoring force is proportional to the displacement of the object from its equilibrium position.

A restoring force is essential in many fields of science and engineering, including structural engineering, seismology, acoustics, and optics.

It is used in the design of springs, dampers, and other mechanical systems that require a stable equilibrium position.

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The x-component of the velocity (vx​) of the particle is 108.7 m/s.

To calculate the x-component of the velocity (vx​) of the particle, we can use the formula for the magnetic force on a charged particle moving in a magnetic field:

F = q * v * B

Given that the charge q is -5.50 nC, the magnetic field Bz​ is -1.20 T, and the force Fy​ is -7.60×10−7 N, we can rearrange the formula to solve for vx​:vx​ = Fy​ / (q * Bz​)

Substituting the values, we have:

vx​ = (-7.60×10−7 N) / (-5.50×10−9 C * -1.20 T)

Simplifying the expression, we get:

vx​ = 108.7 m/s

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Using the Kolmogorov scaling laws, we can estimate the number of large-scale turnaround times required in a Direct Numerical Simulation (DNS) of a mixing process in a bowl. The estimated number of large-scale turnaround times required in the simulation is approximately 12054, stated to three significant figures.

Given the characteristic length of the bowl (l = 0.39 m) and the diameter of the stirring arm (d = 0.017 m), along with the number of small-scale eddy times required for a well-mixed batter (523), we can calculate the number of large-scale turnaround times, denoted as T. The answer will be stated to three significant figures.

According to the Kolmogorov scaling laws, the size of the small-scale eddies (η) is related to the energy dissipation rate (ε) as η ∝ ε^(-3/4). The energy dissipation rate is proportional to the velocity scale (u) raised to the power of 3, ε ∝ u^3.

In the given scenario, the stirring arm generates small-scale eddies of the same size as the arm's diameter, d = 0.017 m. Since the small-scale eddy size is equal to d, we have η = d.

To estimate the number of large-scale turnaround times required, we can compare the characteristic length scale of the mixing bowl (l) with the small-scale eddy size (d). The ratio l/d gives an indication of the number of small-scale eddies within the bowl.

We are given that approximately 523 small-scale eddy times are required for a well-mixed batter. This implies that the mixing process needs to capture the interactions of these small-scale eddies.

Therefore, the number of large-scale turnaround times (T) required can be estimated as T = 523 * (l/d).

Substituting the given values, we have T = 523 * (0.39/0.017) ≈ 12054.

Hence, the estimated number of large-scale turnaround times required in the simulation is approximately 12054, stated to three significant figures.

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