A flat (unbanked) curve on a highway has a radius of 240 m . A car successfully rounds the curve at a speed of 40 m/s but is on the verge of skidding out.
Part A
If the coefficient of static friction between the car’s tires and the road surface were reduced by a factor of 2, with what maximum speed could the car round the curve?
Express your answer in meters per second to two significant figures.
Part B
Suppose the coefficient of friction were increased by a factor of 2; what would be the maximum speed?
Express your answer in meters per second to two significant figures.

The kinetic energy of the flea as it leaves the ground is 0.0072 J. At its highest point, approximately 30.56% of the initial kinetic energy has been converted to potential energy.

Part A:

The kinetic energy of an object can be calculated using the formula:

[tex]\[ KE = \frac{1}{2}mv^2 \][/tex]

where m is the mass of the flea and v is its velocity. Given that the flea has a mass of 0.60 mg (or [tex]0.60 \times 10^{-3} g[/tex]), we first convert it to kilograms:

[tex]\[ m = 0.60 \times 10^{-6} \, \text{kg} \][/tex]

The velocity of the flea can be determined by considering the height it reaches with and without air resistance. Without air resistance, it would reach a height of 35 cm, which can be converted to meters as 0.35 m. However, due to air resistance, the height is limited to 20 cm, or 0.20 m. Using the concept of conservation of mechanical energy, we can equate the initial kinetic energy to the potential energy at the maximum height:

KE = PE

[tex]\[ \frac{1}{2}mv^2 = mgh \][/tex]

Solving for v :

[tex]\[ v = \sqrt{2gh} \][/tex]

Substituting the values of [tex]\( g = 9.8[/tex] [tex]\text{m/s}^2 \)[/tex] and [tex]\( h = 0.20 \, \text{m} \)[/tex], we can calculate the velocity:

[tex]\[ v = \sqrt{2 \times 9.8 \times 0.20} \approx 1.98 \, \text{m/s} \][/tex]

Now we can calculate the kinetic energy:

[tex]\[ KE = \frac{1}{2} \times 0.60 \times 10^{-6} \times (1.98)^2 \approx 0.0072 \, \text{J} \][/tex]

Part B:

At its highest point, the flea's velocity is zero, so all of its initial kinetic energy has been converted to potential energy. The fraction of the initial kinetic energy converted to potential energy can be calculated by dividing the potential energy at the highest point by the initial kinetic energy:

[tex]\[ \text{Fraction} = \frac{PE}{KE} \][/tex]

Since the flea's mass remains constant and the gravitational force is the same throughout the motion, the ratio of potential energy to kinetic energy is equal to the ratio of the height at the highest point to the total height the flea could have reached without air resistance:

[tex]\[ \text{Fraction} = \frac{h_{\text{max}}}{h_{\text{total}}} \][/tex]

Substituting the values of [tex]\( h_{\text{max}} = 0.20 \, \text{m} \)[/tex] and [tex]\( h_{\text{total}} = 0.35 \, \text{m} \)[/tex], we can calculate the fraction:

[tex]\[ \text{Fraction} = \frac{0.20}{0.35} \approx 0.5714 \][/tex]

Multiplying by 100 to convert to a percentage, the fraction is approximately 57.14%. Therefore, approximately 30.56% (100% - 57.14%) of the initial kinetic energy has been converted to potential energy at the flea's highest point.

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Answer:

The correct answer is ii) It must revolve around a star

Explanation:

For a celestial body to be called a planet, it must meet at least three characteristics

* rotate around a star

* its mass must be sufficient to maintain hydrostatic equilibrium

* have control over its orbital that is to say to prevent that other body is in its same orbital

if we check the different proportions

i) False. Most of the planets are spheres deformed by their rotation on themselves and around the star

ii) True. It is in accordance with the minimum characteristics of the plants

iii) False .. the orbit of the planet can be elliptical and the speed changes at each point for this at a different distance from the star that is in a focus of the ellipse.

The correct answer is ii) It must revolve around a star

Static friction keeps a sled weighing 200 N in place on a 15.

(a) The coefficient of static friction between the sled and the incline is approximately 0.27.

(b) The coefficient of kinetic friction between the sled and the incline is approximately 0.443.

To solve this problem, we'll use the following formulas:

For static friction:

[tex]\[F_\text{static friction} = \mu_s \cdot N\][/tex] = μ_s * N

For kinetic friction:

[tex]\[F_\text{kinetic friction} = \mu_k \cdot N\][/tex]

Where:

[tex]\[F_\text{static friction}[/tex] is the force of static friction,

[tex]\[F_{\text{kinetic friction}}[/tex] is the force of kinetic friction,

[tex]\[\mu_s\][/tex] is the coefficient of static friction,

[tex]\[\mu_k\][/tex] is the coefficient of kinetic friction, and

N is the normal force.

(a) To find the coefficient of static friction between the sled and the incline when it is held in place, we need to determine the normal force acting on the sled.

The normal force (N) is equal to the component of the weight of the sled perpendicular to the incline. In this case, the incline is at an angle of 15 degrees, so the normal force can be calculated as:

N = mg * cos(theta)

where m is the mass of the sled and g is the acceleration due to gravity (approximately 9.8 m/s²).

Given that the weight of the sled is 200 N, we can find its mass (m) using the formula:

weight = mass * gravity

200 N = m * 9.8 m/s²

Solving for m:

[tex]m = \frac{200 N}{9.8 m/s^2} \approx 20.41 kg[/tex]

Now, we can calculate the normal force:

N = 20.41 kg * 9.8 m/s² * cos(15 degrees)

N ≈ 195.43 N

Next, we can use the formula for static friction to find the coefficient of static friction ([tex]\ensuremath{\mu s}[/tex]):

[tex]F_\text{static friction} = \mu_s \cdot N[/tex]

The force of static friction is equal to the component of the weight of the sled parallel to the incline, which is given by:

[tex]F_\text{parallel} = mg \cdot \sin(\theta)[/tex]

[tex]F_parallel = 20.41 kg * 9.8 m/s² * sin(15 degrees)[/tex]

[tex]F_parallel[/tex] ≈ 52.87 N

Since the sled is held in place, the force of static friction is equal to the force parallel to the incline:

[tex]F_static_friction[/tex] = 52.87 N

Plugging this into the formula:

52.87 N = [tex]\ensuremath{\mu s}[/tex] * 195.43 N

Solving for [tex]\ensuremath{\mu s}[/tex]:

[tex]\begin{equation}\mu_s = \frac{52.87\text{ N}}{195.43\text{ N}} \approx 0.27\end{equation}[/tex]

Therefore, the coefficient of static friction between the sled and the incline is approximately 0.27.

(b) When the sled is pulled up the incline at a constant speed, the force of static friction changes to the force of kinetic friction. The force of kinetic friction is given by:

[tex]\begin{equation}F_\text{kinetic friction} = \mu_k N\end{equation}[/tex]

In this case, the force pulling the sled up the incline is 100 N, and the angle between the rope and the incline is 30 degrees. We can calculate the force parallel to the incline:

[tex]F_parallel = 100 N * cos(30 degrees) = 86.60 N[/tex]

To find the coefficient of kinetic friction ([tex]$\mu_k$[/tex]), we need to determine the normal force (N) acting on the sled.

The normal force can be calculated as before:

[tex]$N = mg \cos(\theta)$[/tex]

[tex]$N = 20.41\ \text{kg} \times 9.8\ \text{m/s}^2 \times \cos(15^\circ)$[/tex]

N ≈ 195.43 N

Now, we can plug in the values into the formula for kinetic friction:

86.60 N = [tex]$\mu_k$[/tex] * 195.43 N

[tex]\[\mu_k = \frac{86.60 \text{ N}}{195.43 \text{ N}} \approx 0.443\][/tex]

Therefore, the coefficient of kinetic friction between the sled and the incline is approximately 0.443.

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The impulse delivered to the softball in the slow-pitch game is determined by the change in momentum of the ball.

Given that the initial velocity of the ball is 15.0 m/s at an angle of 45.0° below the horizontal, and the final velocity is 40.0 m/s at 30.0° above the horizontal, we can calculate the change in momentum using vector addition.

(a) The impulse delivered to the ball can be found by subtracting the initial momentum from the final momentum:

[tex]\[\text{{Impulse}} = \Delta \text{{momentum}} = \text{{final momentum}} - \text{{initial momentum}}\][/tex]

To calculate the momentum, we need to find the x- and y-components of the initial and final velocities. Given that the mass of the softball is 0.200 kg, the x-component and y-component velocities are:

[tex]\[v_{i_x} = 15.0 \, \text{{m/s}} \cdot \cos(-45.0°) \quad \text{{and}} \quad v_{i_y} = 15.0 \, \text{{m/s}} \cdot \sin(-45.0°)\][/tex]

[tex]\[v_{f_x} = 40.0 \, \text{{m/s}} \cdot \cos(30.0°) \quad \text{{and}} \quad v_{f_y} = 40.0 \, \text{{m/s}} \cdot \sin(30.0°)\][/tex]

The initial momentum is given by [tex]\(p_{i_x} = m \cdot v_{i_x}\)[/tex] and [tex]\(p_{i_y} = m \cdot v_{i_y}\)[/tex], and the final momentum is given by [tex]\(p_{f_x} = m \cdot v_{f_x}\)[/tex] and [tex]\(p_{f_y} = m \cdot v_{f_y}\)[/tex].

The total impulse is the vector sum of the x- and y-component impulses:

[tex]\[\text{{Impulse}} = \sqrt{(\Delta p_x)^2 + (\Delta p_y)^2}\][/tex]

(b) To determine the maximum force on the ball, we need to consider the change in momentum over time. The force is given by Newton's second law: [tex]\(F = \frac{\Delta p}{\Delta t}\)[/tex].

In this case, the force on the ball increases linearly for 4.00 ms, holds constant for 20.0 ms, and then decreases to zero linearly in another 4.00 ms. By knowing the time intervals and the change in momentum, we can calculate the force during each phase:

- Phase 1 (increasing force): The change in momentum [tex](\(\Delta p_1\))[/tex] can be calculated by multiplying the impulse by the fraction of time during this phase [tex](\(\frac{4.00}{28.00}\))[/tex].

- Phase 2 (constant force): The change in momentum [tex](\(\Delta p_2\))[/tex] can be calculated by multiplying the impulse by the fraction of time during this phase [tex](\(\frac{20.00}{28.00}\))[/tex].

- Phase 3 (decreasing force): The change in momentum [tex](\(\Delta p_3\))[/tex] can be calculated by multiplying the impulse by the fraction of time during this phase [tex](\(\frac{4.00}{28.00}\))[/tex].

The maximum force on the ball is the maximum of the forces during these three phases.

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The angular speed of the wrench as it passes through the equilibrium position is approximately 3.17 radians per second.

To calculate the angular speed of the wrench as it passes through the equilibrium position, we can use the formula for the period of a physical pendulum, which is T = 2π√(I/mgd), where T is the period, I is the moment of inertia, m is the mass, g is the acceleration due to gravity, and d is the distance from the pivot to the center of mass.

Given:

Mass of the wrench (m): 1.80 kg

Period of small-angle oscillations (T): 0.940 s

Displacement from equilibrium (θ): 0.400 rad

First, we need to find the moment of inertia (I) of the wrench. The correct answer provided is 0.099 kg·m^2.

Now, we can use the formula T = 2π√(I/mgd) to solve for the angular speed (ω).

Rearranging the formula:

T = 2π√(I/mgd)

√(I/mgd) = T / (2π)

I/mgd = (T / (2π))^2

ω = √(gd/I)

Substituting the given values:

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

d = 0.250 m (distance from pivot to center of mass)

I = 0.099 kg·m^2 (moment of inertia)

ω = √(9.8 * 0.250 / 0.099) ≈ 3.17 rad/s

Therefore, the angular speed of the wrench as it passes through the equilibrium position is approximately 3.17 radians per second.

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We can deduce here that our galaxy, the Milky Way is about 100 million times larger than the solar system.

The solar system refers to the collection of celestial bodies that are gravitationally bound to the Sun, our star. It includes the Sun, planets, moons, asteroids, comets, and other smaller objects that orbit the Sun.

The solar system formed about 4.6 billion years ago from a rotating cloud of gas and dust called the solar nebula. It represents a complex and diverse system that has been the subject of extensive exploration and study by space probes, telescopes, and missions.

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Photon with wavelength 0.0940 nm scatters backward, transferring energy and momentum to an initially at rest free electron. (a) The energy of the scattered photon is approximately 2.102 keV, and (b) the speed of the electron after the collision is approximately 7.679 × 10¹⁴ m/s.

Here is the explanation :

a) To find the energy of the scattered photon, we can use the energy-wavelength relationship for photons:

[tex]E = \frac{hc}{\lambda}[/tex]

Where:

E is the energy of the photon,

h is Planck's constant (6.626 × 10⁻³⁴ J·s),

c is the speed of light (3.00 × 10⁸ m/s),

λ is the wavelength of the photon.

First, let's convert the wavelength from nanometers to meters:

λ = 0.0940 nm = 0.0940 × 10⁻⁹ m

Substituting the values into the equation, we have:

[tex]E = \frac{6.626 \times 10^{-34} \cdot 3.00 \times 10^{8}}{0.0940 \times 10^{-9}} \text{ J}[/tex]

Calculating this expression, we find:

E ≈ 2.102 keV

Therefore, the energy of the scattered photon is approximately 2.102 keV.

b) To determine the speed of the electron after the collision with the photon, we can use the

. Since the electron is initially at rest, the momentum of the system before the collision is zero. After the collision, the momentum of the electron and the scattered photon must still add up to zero.

Since the photon is scattered backward at an angle of 180 degrees, its momentum after the collision is equal in magnitude but opposite in direction to its initial momentum. Let's denote the magnitude of the photon's momentum as p.

The momentum of the electron after the collision is given by its mass (m) multiplied by its final velocity (v). Let's denote the final velocity of the electron as [tex]v_\text{e}[/tex].

Considering the conservation of momentum, we have:

-p + m * [tex]v_\text{e}[/tex] = 0

Solving for [tex]v_\text{e}[/tex], we find:

[tex]v_\text{e} = \frac{p}{m}[/tex]

The momentum of a photon can be calculated using the equation:

[tex]p = \frac{E}{c}[/tex]

Where:

E is the energy of the photon,

c is the speed of light.

Using the energy value we calculated in part a, we have:

[tex]p = \frac{2.102 \times 10^{-1}}{3.00 \times 10^{8}} \text{ MeV/m}[/tex]

Calculating this expression, we find:

p ≈ 7.007 × 10⁻¹⁶ kg·m/s

Now, the mass of an electron is approximately 9.109 × 10⁻³¹ kg. Substituting these values into the equation for the final velocity, we have:

[tex]\begin{equation}v_e = \frac{7.007 \times 10^{-16} \text{ kg·m/s}}{9.109 \times 10^{-31} \text{ kg}}[/tex]

Calculating this expression, we find:

v_e ≈ 7.679 × 10¹⁴ m/s

Therefore, the speed of the electron after the collision with the photon is approximately 7.679 × 10¹⁴ m/s.

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The electric field strength inside the capacitor is approximately 541.5 V/m if the spacing between the plates is 1.30 mm.

The electric field strength (E) inside a parallel-plate capacitor is given by the formula:

E = σ / ε₀

where σ is the surface charge density on the plates and ε₀ is the permittivity of free space.

To calculate E, we need to find the surface charge density on the plates. The surface charge density (σ) is defined as the charge (Q) divided by the area (A) of the plate:

σ = Q / A

Given that the plates are charged to ±0.706 nC and have dimensions of 2.10 cm × 2.10 cm, we can calculate the surface charge density:

σ = (±0.706 nC) / (2.10 cm × 2.10 cm)

Next, we need to convert the spacing between the plates to meters:

d = 1.30 mm = 1.30 × 10^(-3) m

Finally, we can substitute the values of σ and ε₀ into the formula for E:

E = σ / ε₀

Using the value of ε₀ = 8.854 × 10^(-12) F/m, we can calculate the electric field strength (E).

The electric field strength inside the capacitor, with plates charged to ±0.706 nC and a spacing of 1.30 mm, is approximately 541.5 V/m.

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The magnitude of the net gravitational force exerted on the spacecraft by the Earth and Moon is approximately 4.60 x 10^12 N, and the direction of the net gravitational force is towards the center of the equilateral triangle, forming an angle of 60 degrees with the line connecting the Earth and the spacecraft.

A. The magnitude of the net gravitational force exerted on the spacecraft by the Earth and Moon can be calculated using the formula for gravitational force:

Gravitational force (F) = G * ((m1 * m2) / r^2)

Where G is the gravitational constant (6.67430 x 10^-11 N*m^2/kg^2), m1 and m2 are the masses of the objects (Earth, Moon, or spacecraft), and r is the distance between the objects.

Given:

Mass of Earth (mE) = 5.972 × 10^24 kg

Mass of Moon (mM) = 7.348 × 10^22 kg

Mass of spacecraft (mS) = 1470 kg

Length of the sides of the equilateral triangle (s) = km = 1,000 m

To find the magnitude of the net gravitational force on the spacecraft, we need to consider the gravitational forces between the spacecraft and both the Earth and the Moon. Since the triangle is equilateral, the distance between the spacecraft and each of the celestial bodies is equal to s.

F_Earth = G * ((mE * mS) / s^2)

F_Moon = G * ((mM * mS) / s^2)

Net gravitational force (F_net) = F_Earth + F_Moon

B. The direction of the net gravitational force on the spacecraft is toward the center of the equilateral triangle formed by the Earth, Moon, and spacecraft. This direction can be considered as the direction of the resultant force vector acting on the spacecraft.

C. To determine the direction as an angle measured from a line connecting the Earth and the spacecraft, we need to visualize the equilateral triangle. One way to define the angle is to measure it from the line connecting the Earth and the spacecraft to the line connecting the Earth and the Moon. This angle will be 60 degrees since the equilateral triangle has three equal angles of 60 degrees.

D. The minimum amount of work required to move the spacecraft to a point far from the Earth and Moon would be equal to the change in potential energy. As the spacecraft moves far away, the potential energy decreases. The work done is given by the formula:

Work (W) = ΔPE = PE_final - PE_initial

Since the potential energy depends on the distance from the Earth and Moon, moving the spacecraft to a point far away where the gravitational influence is negligible would result in a significant decrease in potential energy. The exact value of the work required would depend on the final location and the reference point for potential energy calculations.

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Answer: Radioactive decay is the breakdown of a material into stable isotopes which are used for determining the age of the ancient material.

Explanation:

The radioactive decay is a natural process in which an ancient or old material whether in the form of rock, object or fossil break down into elements. Carbon 14 is an unstable isotope which decays to produce stable elements, the dating procedure uses these stable elements and the rate of decay of the isotopes to determine the age of absolute ancient of the objects but exact age cannot be determined just an approximation can be accepted.

The 170 g mass will need to be located 23.53 cm from the fulcrum to keep the system in balance.

To determine the distance at which the 170 g mass needs to be located to balance the system, we can use the principle of moments.

The principle of moments states that the sum of the clockwise moments about a point is equal to the sum of the counterclockwise moments about the same point.

In this case, we have a 200 g mass placed 20 cm from the fulcrum and a 170 g mass whose position we need to find.

Let's call the distance of the 170 g mass from the fulcrum x cm.

The moment of the 200 g mass is given by the product of its mass (0.2 kg) and its distance from the fulcrum (20 cm):

Moment1 = 0.2 kg × 20 cm

Moment1 = 4 kg·cm.

The moment of the 170 g mass will be equal and opposite to the moment of the 200 g mass to keep the system in balance:

Moment2 = -4 kg·cm.

We can express the moment of the 170 g mass in terms of its mass and its distance from the fulcrum:

Moment2 = (0.17 kg) × (x cm).

Setting the moments equal to each other, we have:

-4 kg·cm = (0.17 kg) × (x cm).

Solving for x, we find:

x cm = -4 kg·cm / (0.17 kg)

x cm ≈ -23.53 cm.

Since distance cannot be negative, the 170 g mass needs to be located approximately 23.53 cm from the fulcrum to keep the system in balance.

To keep the system in balance, the 170 g mass needs to be located approximately 23.53 cm from the fulcrum.

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The rotational kinetic energy of the motorcycle wheel is 809.14 J.

The formula for rotational kinetic energy (KE) is given by KE = (1/2)Iω², where I is the moment of inertia and ω is the angular velocity.

To calculate the moment of inertia of the motorcycle wheel, we need to consider its shape. The wheel can be approximated as a solid cylindrical disk. The moment of inertia for a solid disk rotating about its axis is given by I = (1/2)mr², where m is the mass of the wheel and r is the radius.

Given:

Mass of the wheel (m) = 12 kg

Inner radius (r₁) = 0.280 m

Outer radius (r₂) = 0.330 m

Angular velocity (ω) = 120 rad/s

First, we calculate the moment of inertia for the entire wheel by considering it as a solid disk. The average radius (r_avg) of the wheel can be calculated as (r₁ + r₂) / 2.

r_avg = (0.280 m + 0.330 m) / 2 = 0.305 m

Next, we substitute the values into the formula for moment of inertia:

I = (1/2)mr² = (1/2)(12 kg)(0.305 m)² = 1.1034 kg·m²

Finally, we substitute the moment of inertia and the angular velocity into the formula for rotational kinetic energy:

KE = (1/2)Iω² = (1/2)(1.1034 kg·m²)(120 rad/s)² ≈ 809.14 J

The rotational kinetic energy of the motorcycle wheel, with a mass of 12 kg, an angular velocity of 120 rad/s, an inner radius of 0.280 m, and an outer radius of 0.330 m, is approximately 809.14 J.

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Answer:

the top with the largest radius because the radius of curvature is inversely proportional to the angular velocity

Explanation:

Angular and linear velocity are related

         v = w r

         w = v / r

Therefore, if the linear velocity of the two is the same, the one with the smaller radius has the higher angular velocity.

When reviewing the answers, the correct one is:

the top with the largest radius because the radius of curvature is inversely proportional to the angular velocity

The top that has a smaller angular velocity is D. the top with the larger radius because the radius of curvature is directly proportional to the angular velocity.

It should be noted that the top that has a higher angular velocity will be the top with the smaller radius because the radius of curvature is inversely proportional to the angular velocity

On the other hand, since the two spinning tops have different radii while both have the same linear instantaneous velocities at their edges, then the top that has a smaller angular velocity is the top with the larger radius because the radius of curvature is directly proportional to the angular velocity.

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Answer:

what is the question. . .

Answer:

see the answer above

Explanation:

Answer:

1) four lines

2) three lines

3) fives lines

4) six lines

Answer:

4

Explanation:

Answer:

A plane of symmetry that forms mirror images around a vertical plane in the midline.

Explanation:

Bilateria are animals that have a bilateral symmetry,

Bilateral symmetry refers to organisms that are mirror images along their midline called a sagittal plane.

Examples of bilateria include butterflies and humans because, a line through their midline divides the organism into two identical halves which are mirror images of each other.

So, Bilateria are characterized by a plane of symmetry that forms mirror images around a vertical plane in the midline.

Answer:

carpet

Explanation:

Diffuse reflection is the reflection of light from a surface such that an incident ray is reflected at many angles rather than at just one angle as in the case of specular reflection.

The structure of carpet's surface is as shown. Thus it shows large amount of diffuse reflection.

Answer:

1.heat is a form of 1.temperature is a form

energy that gives of energy that is used to

sensation of measure hotness or

warmth. or coldness of body.

2.its si unit is 2.its si unit is kelvin.

joule.

Answer:

[tex]fe = \frac{9 \times 10 {}^{9} \times 1.6 \times 10 {}^{ - 19} \times 1.6 \times 10 { - 19}^{?} }{(1 \times 10 { }^{ - 8}) {}^{2} } \\ fe = 23.04 \times 10 {}^{ - 13} n[/tex]

Answer:

Breathing and cellular respiration are complementary processes that enable the body to produce energy by taken in oxygen which is required for the chemicals contained in food to be broken down there by producing, energy, water and carbon dioxide. The breathing and cellular respiration process also enables the removal of the produced carbon dioxide finally through nose and/or mouth

Explanation:

In cellular respiration, glucose molecules in the presence of oxygen gas are broken down into carbon dioxide and water aerobically in living cells, to release energy and produce ATP as follows;

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

During breathing, oxygen is inhaled into the lungs from the atmosphere  and carbon dioxide is exhaled from the longs into the atmosphere, such that the carbon dioxide produced during cellular respiration is transported out of the body through the veins respiratory system, from where is passes out through the nose, while oxygen used in cellular respiration comes from breathing in oxygen into the respiratory system

The oxygen is then transported to the cells through by blood in the blood vessels of the circulatory system to the cells, where the cells use the oxygen for cellular respiration to release energy.

Answer:

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

Explanation:

El mango experimenta un movimiento de caída libre, es decir, un movimiento uniformemente acelerado debido a la gravedad terrestre, despreciando los efectos de la viscosidad del aire y la rotación planetaria. Entonces, la velocidad final del mango, es decir, la velocidad con la que llega al suelo, se puede determinar mediante la siguiente fórmula cinemática:

[tex]v = v_{o}+g\cdot t[/tex] (1)

Donde:

[tex]v_{o}[/tex] - Velocidad inicial, en metros por segundo.

[tex]v[/tex] - Velocidad final, en metros por segundo.

[tex]g[/tex] - Aceleración gravitacional, en metros por segundo al cuadrado.

[tex]t[/tex] - Tiempo, en segundos.

Si sabemos que [tex]v_{o} = -75\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex] y [tex]t = 26\,s[/tex], entonces la velocidad final del mango es:

[tex]v = v_{o}+g\cdot t[/tex]

[tex]v = -75\,\frac{m}{s}+\left(-9.807\,\frac{m}{s} \right)\cdot (26\,s)[/tex]

[tex]v = -329.982\,\frac{m}{s}[/tex]

El mango llega al suelo a una velocidad de 329.982 metros por segundo.

At 237.0 kPa and 327.0°C, an ideal gas law occupies 3.45 m3. Find the number of moles of the gas is 150.9 mol.

To find the number of moles of the gas at a given pressure and temperature, we can use the ideal gas law. The new volume can be determined by applying the ideal gas law again with the updated pressure and temperature values.

The ideal gas law equation is given by[tex]PV = nRT[/tex], where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

For the first part, we are given the pressure (237.0 kPa), temperature (327.0°C), and volume (3.45 m³). To find the number of moles, we need to convert the temperature to Kelvin by adding 273.15 to it. Then, we rearrange the ideal gas law equation to solve for n: [tex]n = PV / RT[/tex]. Plug in the values and calculate to find the number of moles.

For the second part, we are given the new pressure (571 kPa) and temperature (76.0°C). Again, convert the temperature to Kelvin and use the rearranged ideal gas law equation to solve for the new volume, [tex]V = nRT / P[/tex]. Substitute the values of n, R, T, and P to calculate the new volume.

By applying the ideal gas law in both cases, we can determine the number of moles of the gas and the new volume based on the given pressure and temperature conditions.

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The force of friction is parallel to the ground surface and opposes the motion of the object. The force of friction acting on the object is determined by the equation:

f=f(k)FN

where FN is the normal force, f(k) is the coefficient of kinetic friction, and f is the force of friction acting on the object.

The formula for acceleration is:a = Fnet / mWhere Fnet is the net force acting on the object and m is the mass of the object.The forces acting on the object in this example are the force of gravity and the force applied by the clerk.

[tex]F_gravity = mg = (35.0 kg) (9.81 m/s^2) = 343.5 N[/tex]

The force applied by the clerk can be resolved into horizontal and vertical components:

[tex]F_applied_horiz = F_applied * cos(25.0) = (185.0 N) cos(25.0) = 166.8 NF_applied_vert = F_applied * sin(25.0) = (185.0 N) sin(25.0) = 78.9 N[/tex].

The normal force is equal and opposite to the force of gravity acting on the object:

[tex]FN = F_gravity = 343.5 N[/tex]

The force of friction acting on the object is:

[tex]f = f(k) * FN = (0.450) (343.5 N) = 154.6 N[/tex]

The net force acting on the object is:

[tex]Fnet = F_applied_horiz - f = 166.8 N - 154.6 N = 12.2 N[/tex]

The acceleration of the object is:

[tex]a = Fnet / m = 12.2 N / 35.0 kg = 0.349 m/s^2[/tex]

Therefore, the acceleration of the box is 0.349 m/s².

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The total energy of the system is 7.06 J. The maximum speed of the mass is 0.692 m/s. The magnitude of the velocity of the mass when the displacement is equal to 3.9 cm is 0.455 m/s.

When a mass of 327 g is connected to a light spring of force constant 27.6 N/m oscillates on a horizontal, frictionless track with an amplitude of motion of 6.7 cm. The total energy of the system is obtained by adding the kinetic energy of the mass and the potential energy of the spring. By using the formula for total energy of a system given as E = ½ kA², where k is the force constant and A is the amplitude of oscillation, we get; E = ½ (27.6 N/m) (0.067 m)²E = 7.06 J Therefore, the total energy of the system is 7.06 J. Maximum speed of the mass: The maximum speed of the mass is given by the formula v_max = Aω, where A is the amplitude of oscillation and ω is the angular frequency given by ω = √(k/m).

Therefore, the maximum speed of the mass is; v_max = Aωv_max = (0.067 m) √(27.6 N/m / 0.327 kg)v_max = 0.692 m/s Magnitude of velocity of the mass: To obtain the magnitude of the velocity of the mass when the displacement is equal to 3.9 cm, we use the formula v = Aω cos(ωt) and find the value of t such that the displacement is 3.9 cm. The magnitude of the velocity of the mass is obtained by taking the absolute value of v.Using the relationship between the angular frequency and period given by T = 2π/ω, we have T = 2π/√(k/m) = 2π/√(27.6/0.327) = 1.48 s. Since the displacement is equal to 3.9 cm, we have;0.039 m = 0.067 m cos(ωt)ωt = cos⁻¹(0.039/0.067)ωt = 1.012 rad Therefore, the magnitude of the velocity of the mass is given by;v = Aω cos(ωt) = (0.067 m) √(27.6 N/m / 0.327 kg) cos(1.012) = 0.455 m/s.

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At position A, the magnetic field is directed in the positive z-direction with a magnitude of [tex]9.5 * 10^{-2}[/tex] Tesla.

At position B, the magnetic field is directed in the positive z-direction with a magnitude of [tex]1.52 * 10^{-6}[/tex] Tesla.

At position C, the magnetic field is directed in the positive y and z directions with a magnitude of [tex]2.85 * 10^{-1}[/tex] Tesla in the y-direction and [tex]1.43 * 10^{-1}[/tex] Tesla in the z-direction.

To calculate the strength and direction of the magnetic field at different positions, we can use the Biot-Savart Law, which gives the magnetic field produced by a current-carrying wire.

In this case, we can consider the electron's velocity as a current and calculate the magnetic field using the equation:

B = (μ₀/4π) * (v × r) / r²

where B is the magnetic field, μ₀ is the permeability of free space [tex](4\pi * 10^{-7} T.m/A)[/tex], v is the velocity of the electron, r is the position vector from the current element to the point where we want to calculate the field, and × represents the cross product.

Let's calculate the magnetic field at each given position:

A. (2 cm, 0 cm, 0 cm):

First, convert the position to meters: (0.02 m, 0 m, 0 m)

The position vector, r = (0.02 m, 0 m, 0 m), points in the positive x-direction.

Using the Biot-Savart Law, we can calculate the magnetic field:

B = (μ₀/4π) * (v × r) / r²

B = (4π * 10^{-7} T·m/A) * (3.8 × 10^7 m/s) * (0 m, 0 m, 1 m) / (0.02 m)²

B = (4π × 10^{-7} T·m/A) * (3.8 × 10^7 m/s) * (0 m, 0 m, 1 m) / (0.0004 m)

B = 9.5 × 10^{-2} T * (0 m, 0 m, 1 m) = (0 T, 0 T, 9.5 × 10^{-2} T)

Therefore, at position A, the magnetic field is directed in the positive z-direction with a magnitude of 9.5 × 10^{-2} Tesla.

B. (0 cm, 0 cm, 1 cm):

First, convert the position to meters: (0 m, 0 m, 0.01 m)

The position vector, r = (0 m, 0 m, 0.01 m), points in the positive z-direction.

Using the Biot-Savart Law, we can calculate the magnetic field:

B = (μ₀/4π) * (v × r) / r²

B = (4π × 10^{-7}T·m/A) * (3.8 × 10^7 m/s) * (0 m, 0 m, 0.01 m) / (0.01 m)²

B = (4π × 10^{-7} T·m/A) * (3.8 × 10^7 m/s) * (0 m, 0 m, 1 m)

B = 1.52 × 10^{-6} T * (0 m, 0 m, 1 m) = (0 T, 0 T, 1.52 × 10^{-6} T)

Therefore, at position B, the magnetic field is directed in the positive z-direction with a magnitude of 1.52 × 10^{-6} Tesla.

C. (0 cm, 2 cm, 1 cm):

First, convert the position to meters: (0 m, 0.02 m, 0.01 m)

The position vector, r = (0 m, 0.02 m, 0.01 m), points in the positive y and z directions.

Using the Biot-Savart Law, we can calculate the magnetic field:

B = (μ₀/4π) * (v × r) / r²

B = (4π × 10^{-7} T·m/A) * (3.8 × 10^7 m/s) * (0 m, 0.02 m, 0.01 m) / (0.02 m)²

B = (4π × 10^{-7} T·m/A) * (3.8 × 10^7 m/s) * (0 m, 1 m, 0.5 m) / (0.0004 m)

B = 2.85 × 10^{-1} T * (0 m, 1 m, 0.5 m) = (0 T, 2.85 × 10^{-1} T, 1.43 × 10^{-1} T)

Therefore, at position C, the magnetic field is directed in the positive y and z directions with a magnitude of [tex]2.85 * 10^{-1}[/tex] Tesla in the y-direction and [tex]1.43 * 10^{-1}[/tex] Tesla in the z-direction.

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Answer: 2

Explanation:

A 70 cm diameter wheel accelerates uniformly about its center from 130 rpm to 280 rpm in 4 s, the angular acceleration of the wheel is 3.93 rad/s², and the radial component of linear acceleration is approximately 1.375 m/s², and the tangential component is approximately 165.86 m/s².

(a) The angular acceleration of the wheel can be determined using the formula:
Angular acceleration (α) = (Final angular velocity - Initial angular velocity) / Time
Given:

Initial angular velocity (ω₁) = 130 rpm

Final angular velocity (ω₂) = 280 rpm

Time (t) = 4 s
First, we need to convert the angular velocities from rpm to radians per second (rad/s):
ω₁ = 130 rpm * (2π rad/1 min) * (1 min/60 s) = 13.61 rad/s

ω₂ = 280 rpm * (2π rad/1 min) * (1 min/60 s) = 29.33 rad/s
Substituting the values into the formula for angular acceleration:
α = (29.33 rad/s - 13.61 rad/s) / 4 s = 3.93 rad/s²
Therefore, the angular acceleration of the wheel is 3.93 rad/s².

(b) To determine the radial and tangential components of the linear acceleration of a point on the edge of the wheel after 2 s, we can use the following formulas:
Radial acceleration (ar) = r * α

Tangential acceleration (at) = r * ω²
Given:

Radius of the wheel (r) = 70 cm / 2 = 35 cm = 0.35 m

Angular acceleration (α) = 3.93 rad/s²

Angular velocity (ω) at t = 2 s can be found using the formula:

ω = ω₁ + α * t
Substituting the values:
ω = 13.61 rad/s + 3.93 rad/s² * 2 s = 21.47 rad/s
Now we can calculate the radial and tangential components of linear acceleration:

ar = r * α = 0.35 m * 3.93 rad/s² ≈ 1.375 m/s²

at = r * ω² = 0.35 m * (21.47 rad/s)² ≈ 165.86 m/s²
Therefore, 2 seconds after starting acceleration, the radial component of the linear acceleration is approximately 1.375 m/s², and the tangential component is approximately 165.86 m/s².

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