Shannon and Chris push on blocks with identical force. SHannon's block is twice as massive as Chris'. After pushing for 5 seconds, who did more work?

Answer:

a) What is the flow speed in the second section? Assume the pressure remains constant.

Given:

Section 1 area = 10.0 cm^2  

Section 1 flow speed = 293 cm/s

Section 2 area = 3.00 cm^2

Pressure = 1.20 x 105 Pa (constant)

Since pressure remains constant, by Bernoulli's equation:    

P1/ρ + 1/2*v12 = P2/ρ+ 1/2*v22

Since P1 = P2 and ρ (density) is constant:

v22 = 2*(v12 - v22)    

v22 = 2*(293^2 - v22)

Solving for v2:    

v2 = √(2*293^2) = 846 cm/s

b) What is the kinetic energy loss per second between the two sections?

Kinetic energy = 1/2 * mass * velocity^2

Mass flow rate = density * area * velocity

= 1.65 g/cm^3 * 10.0 cm^2 * 293 cm/s = 485.5 g/s

Kinetic energy loss = 1/2 * (485.5 g/s) * (293^2 - 846^2) cm^2/s^2        

            = 1/2 * (485.5)*(86124 - 716256)

            = 20617 J/s

So the kinetic energy loss per second between the two pipe sections is 20617 J/s.

Explanation:

The elastic potential energy stored in the spring is 0.375 J.

The formula for elastic potential energy is:

E = 1/2 * k * x^2

where:

* E is the elastic potential energy in Joules

* k is the spring constant in N/m

* x is the distance the spring is stretched or compressed from its equilibrium position in meters

In this problem, we have:

* k = 3 N/m

* x = 0.5 m (50 cm)

Substituting these values into the formula, we get:

E = 1/2 * 3 * 0.5^2 = 0.375 J

Therefore, the elastic potential energy stored in the spring is 0.375 J.

Answer:

E: More than one option is correct  

Explanation:

Tension, weight, and friction produce torques on the hoop, while the normal force does not.

The net work done (in J) required to accelerate a 1500 kg car from 55 m/s to 65 m/s is 3127500 J

First, we shall obtain the initial kinetic energy. Details below:

KE₁ = ½mu²

= ½ × 1500 × 55²

= 41250 J

Next, we shall final kinetic energy. Details below:

KE₂ = ½mv²

= ½ × 1500 × 65²

= 3168750 J

Finally, we shall determine the net work done. Details below:

W = KE₂ - KE₁

= 3168750 - 41250

= 3127500 J

Thus, the net work done is 3127500 J

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A steam turbine receives steam with a velocity of 28 m/s, specific enthalpy 3000kJ/kg at a rate of 3500 kg per hour. The steam leaves the turbine with a specific enthalpy of 2200kJ/kg at 180 m/s then turbine output is 777.76kW.

To get the turbine output, we must first compute the change in specific enthalpy (h) and mass flow rate ().

Assume that the inlet steam velocity (v1) is 28 m/s.

Specific enthalpy at the inlet (h1) = 3000 kJ/kg

()=3500kg/h mass flow rate

2200 kJ/kg outlet specific enthalpy (h2)

v2 (outlet steam velocity) = 180 m/s

To begin, convert the mass flow rate from kg/h to kg/s as follows: =

[tex]3500 kg/h (1 h/3600 s) = 0.9722 kg/s[/tex]

The change in specific enthalpy (h) can then be calculated:

3000kJ/kg-2200kJ/kg=800kJ/kgh=h1-h2

The following formula can be used to compute the turbine output (P):

[tex]P = ṁ * Δh[/tex]

Substituting  P=0.9722kg/s*800kJ/kg=777.76kJ/sork W

As a result, ignoring losses, the turbine output is roughly 777.76kW

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The Force ≈ [tex]1.13 * 10^6 N[/tex]

The Weight ≈ [tex]1.66 * 10^5 N[/tex]

Pressure ≈ 6.32 × 10⁴Pa

(a)

Force = Pressure × Area

Force = (9.00 × 10⁴ Pa) × (π × (2.00 m)²)

Force ≈ [tex]1.13 * 10^6 N[/tex]

(b)

Weight = Density × Volume × g

Weight = (415 kg/m³) × (π × (2.00 m)² × 10.0 m) × (6.32 m/s²)

Weight ≈ 1.66 × 10⁵ N

(c)

Pressure = Pressure at the surface + Density × g × depth

Pressure = [tex](9.00 * 10^4 Pa) + (415 kg/m^3)* (6.32 m/^2)* (10.0 m)[/tex]

Pressure ≈ 6.32 × 10⁴Pa

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Explanation:

Rate  X  Time  =   Distance

Distance  / Time  = Rate

225 km / 3.35 hr = 67.2 km/hr

Answer:

Liberal arts education is important for ensuring continued innovation. It helps students think creatively, solve problems, collaborate effectively, and consider ethical factors. By exploring various subjects and adapting to new situations, liberal arts education equips individuals with the skills needed to generate new ideas and drive progress in different fields.

The velocity of the roller coaster at location 4 is 14.14 m/s (approx) using the given values of v₁ = 10 m/s, h₁ = 30 m, and h₂ = 15 m.

Roller coasters are fascinating machines that deliver an exhilarating experience by defying gravity and physics. Roller coaster physics is a significant concept to comprehend before riding a roller coaster or designing one. The laws of physics govern the motion of a roller coaster, including its velocity, acceleration, and potential energy.

A roller coaster moves over a crest at location 1 with a speed of 10 m/s. The question is how fast it's going at location 4, considering the neglect of friction and air resistance. To solve this, we'll need to consider the conservation of energy law.

The total energy of the roller coaster remains constant throughout the ride, and we can convert between potential and kinetic energy.Using the conservation of energy formula, which is: E1 = E2Where E1 is the total energy of the roller coaster at the crest and E2 is the total energy of the roller coaster at location 4.

Both E1 and E2 comprise kinetic energy (KE) and potential energy (PE). So,E1 = KE1 + PE1E2 = KE2 + PE2Since the roller coaster has no friction and air resistance, we can assume that PE1 = PE2 because the height of the roller coaster doesn't change. The energy is converted from potential energy at the crest to kinetic energy at location 4.

We can now use the formula for kinetic energy:KE = (1/2) mv²Where m is the mass of the roller coaster and v is its velocity. Both E1 and E2 can be written in terms of KE, so: E1 = (1/2) mv₁²E2 = (1/2) mv₂².

Substitute the values into the conservation of energy formula: E1 = E2(1/2) mv₁² = (1/2) mv₂²

Simplifying the equation gives:v₂² = v₁²×(h₁ / h₂)

where h₁ is the height of the crest and h₂ is the height of location 4.

To calculate the velocity, we need to take the square root of both sides:v₂ = v₁×√(h₁ / h₂)

Therefore, the velocity of the roller coaster at location 4 is 14.14 m/s (approx) using the given values of v₁ = 10 m/s, h₁ = 30 m, and h₂ = 15 m.

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(a) The force diagram is the image attached.

(b) The net force of the workers is F(net) = (T₁ + T₂) - (W₁ + W₂ + W₃).

(c) The net torque equation of the system is W₁(x/2) - W₂(x/2) = 0

(d) The cable that exerts the most tension is the tension on the left.

(e) The tension in wire 1 is 490 N and the tension in wire 2 is 392 N.

The net force of the workers is calculated by applying the following equations;

F(net) = (T₁ + T₂) - (W₁ + W₂ + W₃)

The net torque equation of the system is determined as follows;

τ (net) = W₁(x/2) - W₂(x/2) = 0

W₁(x/2) - W₂(x/2) = 0

where;

The cable that exerts the most tension is the tension on the left directly above the 50 kg worker because this work has the greatest downward force.

The tension in wire 1 is calculated as;

T₁ = 50 kg x 9.8 m/s²

T₁ = 490 N

The tension in wire 2 is calculated as;

T₂ = 40 kg x 9.8 m/s²

T₂ = 392 N

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The change in momentum for the baseball, which is hit back in the opposite direction by the batter, is -10.36 kg·m/s. This change in momentum is obtained by subtracting the initial momentum of 4.2 kg·m/s from the final momentum of -6.16 kg·m/s. The negative sign indicates the opposite direction of the momentum.

To find the change in momentum for the baseball, we can use the formula:

Change in momentum = Final momentum - Initial momentum

Momentum is defined as the product of mass and velocity.

Given data:

Mass of the baseball (m) = 0.140 kg

Initial velocity of the baseball ([tex]v_i_n_i_t_i_a_l)[/tex] = 30.0 m/s

Final velocity of the baseball ([tex]v_f_i_n_a_l_[/tex]) = -44.0 m/s (negative sign indicates opposite direction)

To calculate the initial momentum, we multiply the mass by the initial velocity:

Initial momentum = m * [tex]v_i_n_i_t_i_a_l[/tex] = 0.140 kg * 30.0 m/s = 4.2 kg·m/s

To calculate the final momentum, we multiply the mass by the final velocity:

Final momentum = m * [tex]v_f_i_n_a_l_[/tex] = 0.140 kg * (-44.0 m/s) = -6.16 kg·m/s

Now we can find the change in momentum:

Change in momentum = Final momentum - Initial momentum

Change in momentum = (-6.16 kg·m/s) - (4.2 kg·m/s)

Change in momentum = -10.36 kg·m/s

Therefore, the change in momentum for the baseball is -10.36 kg·m/s. The negative sign indicates a change in direction, as the ball is hit back in the opposite direction.

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4.44×[tex]10^{9}[/tex] kg/s is the rate at which the sun mass is decreasing.

The Sun radiates energy through a process called nuclear fusion, where hydrogen atoms combine to form helium, releasing a tremendous amount of energy in the process. According to Einstein's mass-energy equivalence principle (E=mc²), this energy release corresponds to a decrease in mass.

To calculate the rate at which the Sun's mass is decreasing, we can use the formula ΔE = Δmc², where ΔE is the change in energy, Δm is the change in mass, and c is the speed of light.

Given that the Sun radiates energy at a rate of 4×10^26 W, we can substitute this value into the equation as ΔE and solve for Δm.

ΔE = 4×10^26 W

c = 3×10^8 m/s (speed of light)

Using the equation ΔE = Δmc² and rearranging it, we get Δm = ΔE / c².

Substituting the values, we have:

Δm = (4×10^26 W) / (3×10^8 m/s)²

Evaluating this expression, we find that the rate at which the Sun's mass is decreasing is approximately 4.44×10^9 kg/s.

This calculation demonstrates that the Sun's mass is gradually decreasing as it continuously radiates energy into space, primarily through the process of nuclear fusion in its core.

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The minimum height (h) of the bag needed to infuse glucose into the vein is approximately 0.1235 meters.

The hydrostatic pressure is given by:

P = ρgh

Where:

P is the pressure,

ρ is the density of the fluid,

g is the acceleration due to gravity, and

h is the height of the fluid column.

Then we will have

h = P / (ρg)

h =[tex](1.21 * 10^3 Pa) / (1.01* 10^3 kg/m^3 * 9.8 m/s^2)[/tex]

h = 0.1235 meters

Therefore, the minimum height (h) of the bag needed to infuse glucose into the vein is approximately 0.1235 meters.

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The height (in meters) of the pipe, given that the pressure in the pipe is 320 KPa, and the velocity of water is 0.24 m/s, is 9.78 m

The following data were obtained from the question:

The height of the pipe can be obtained by using the Bernoulli's equation as illustrated below:

P₁ + 1/2ρv₁² + ρgh₁ = P₂ + 1/2ρv₂² + ρgh₂

415000 + (0.5 × 1000 × 1.4²) + (1000 × 9.81 × 0) = 320000 + (0.5 × 1000 × 0.24²) + (1000 × 9.81 × h₂)

415000 + 980 = 320000 + 28.8 + 9810h₂

Collect like terms

415000 + 980 - 320000 - 28.8 = 9810h₂

95951.2 = 9810h₂

Divide both sides by 9810

h₂ = 95951.2 / 9810

= 9.78 m

Thus, we can conclude that the height of the pipe is 9.78 m

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It takes 0.47 seconds for water to travel from the nozzle to the ground when the water pistol is fired horizontally.

We will denote the height of the water pistol above the ground as h and the initial velocity of water exiting the nozzle as v2. Assuming negligible air resistance, we will analyze the vertical motion of the water droplets.

The vertical displacement of the water droplets is calculated using equation: h = (1/2) * g * t^2.

Rearranging equation, we solve for time:

t = sqrt(2h / g).

Given data:

t = sqrt(2 * 1.10 / 9.8)

t = 0.47380354147

t = 0.47 seconds.

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Answer:

Therefore, the mass of the aluminum chunk is 15 g or 0.015 kg.

Explanation:

We have a container with 120 g of steel and 150 g of water at 25°C.

Two metal objects are dropped into the container:

A 244 g cube of copper initially at 90°C

An unknown mass of aluminum initially at 5.0°C

The surprise result is that the final water temperature is still 25°C.

Since the final water temperature is the same, the heat lost by the hot copper must equal the heat gained by the cold aluminum and the container.

The heat lost by the copper can be calculated as:

Qcopper = copper * copper * ΔTcopper

= 0.244 kg * 385 J/(kg•K) * (90°C - 25°C) = 29.5 kJ

The heat gained by the aluminum and container can be calculated as:

Qtotal = maluminum * cpaluminum * ΔTaluminum + mcontainer * cpcontainer * ΔTcontainer

Setting this equal to 29.5 kJ and plugging in the given values:

29.5 kJ = maluminum * 900 J/(kg•K) * (25°C - 5°C) + 0.120 kg * 450 J/(kg•K) * (25°C - 25°C)

Solving for aluminum :

aluminum = 0.015 kg

aluminum = 15 g

Answer:Certainly, let's solve the problem.

We can use the principle of conservation of energy to solve this problem. The heat lost by the hot object will be equal to the heat gained by the cold object and the water.

The heat lost by the copper cube can be calculated as:

Q1 = m1 * c1 * (T1 - Tfinal)

where

m1 = 244 g = 0.244 kg (mass of copper cube)

c1 = 0.385 J/g°C (specific heat of copper)

T1 = 90°C (initial temperature of copper cube)

Tfinal = 25°C (final temperature of the system)

Substituting the values, we get:

Q1 = 0.244 * 0.385 * (90 - 25) = 21.38 J

Similarly, the heat gained by the aluminum chunk can be calculated as:

Q2 = m2 * c2 * (Tfinal - T2)

where

m2 = ? (mass of aluminum chunk)

c2 = 0.902 J/g°C (specific heat of aluminum)

T2 = 5.0°C (initial temperature of aluminum chunk)

Substituting the values, we get:

Q2 = m2 * 0.902 * (25 - 5.0) = 18.144 m2 J

Now, since the water temperature doesn't change, we can assume that the heat gained by the water is equal to the heat lost by the hot objects. Therefore, we can write:

Q1 + Q2 = mwater * cwater * (Tfinal - Tinitial)

where

mwater = 150 g = 0.15 kg (mass of water)

cwater = 4.184 J/g°C (specific heat of water)

Tinitial = 25°C (initial temperature of water)

Substituting the values, we get:

21.38 J + 18.144 m2 J = 0.15 * 4.184 * (25 - 25)

Simplifying the equation, we get:

m2 = 0.266 kg

Therefore, the mass of the aluminum chunk is 0.266 kg or 266 grams.

Explanation:

In the diagram tha shows snapshots of an oscillator at different times, the frequency of the oscillation is 0.555 Hz.

The period of the oscillation is the time taken for the for the object to return to its original position. (ie. Displacement = 0). From the above snapshot,

Period of oscillation = 1.80s.

From here, finding the frequency is simple as, Frequency = 1/Period. Hence,

Frequency = 1/1.80

= 0.555 Hz (3 sf).

The frequency of the oscillation is indeed 0.555 Hz. The frequency represents the number of oscillations or cycles per second. In this case, the object completes approximately 0.555 oscillations per second.

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Answer:

a-greater than the original (solid disk)

Explanation:

The hollow drum's lower rotational inertia allows it to rotate faster as the wire unwinds. This absorbs more potential energy, leaving less to translate into the speed of the falling mass. Therefore, the final speed when using a hollow disk will be:

a-greater than the original (solid disk)

Explanation:

You want  N/m

 N = 66 * 9.81

 m = 2.3 x 10^-2 m

66* 9.81 / 2.3 x 10^-2  = 28150 =  28 000 N/m    to two S D

The weight of the man is approximately 888.6 Newtons.

The velocity of the train is approximately 10 m/s.

The resultant velocity of the man is 11 m/s.

The train has traveled a distance of 6000 meters in ten (10) minutes.

1.5.1 The weight of the man can be calculated using the formula:

Weight = mass * acceleration due to gravity

Given:

Mass of the man (m) = 90.6 kg

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

Weight of the man = 90.6 kg * 9.8 m/s² ≈ 888.6 N

Therefore, the weight of the man is approximately 888.6 Newtons.

1.5.2 The velocity of the train in m/s can be converted from its given velocity in km/h. Since 1 km/h is equal to 1000/3600 m/s, we can calculate:

Velocity of the train (v) = 36 km/h * (1000/3600) m/s ≈ 10 m/s

Therefore, the velocity of the train is approximately 10 m/s.

1.5.3 By taking into account the relative motion of the man and the train, it is possible to determine the man's final velocity. The resultant velocity will equal the vector sum of the man's walking velocity and the train's velocity because the man is traveling toward the rear of the train, which is moving eastward.

Given:

Velocity of the man relative to the train (v_man) = 1 m/s (backwards)

Resultant velocity of the man (v_resultant) = v_man + v_train

v_resultant = 1 m/s + 10 m/s = 11 m/s

Therefore, the resultant velocity of the man is 11 m/s.

1.5.4 The distance the train has traveled in ten (10) minutes can be calculated using the formula:

Distance = velocity * time

Given:

Time (t) = 10 minutes = 10 * 60 seconds = 600 seconds

Velocity of the train (v) = 10 m/s

Distance = 10 m/s * 600 seconds = 6000 meters

Therefore, the train has traveled a distance of 6000 meters in ten (10) minutes.

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Answer:

the density of the cube is approximately 2016.07 kg/m^3.

Explanation:

The buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

Let's first calculate the weight of the crate:

mass of crate = density * volume = density * (side length)^3 = density * 0.25^3 = 0.015625 * density

weight of crate = mass of crate * gravity = 0.015625 * density * 9.81 = 0.1530875 * density

where gravity is the acceleration due to gravity, which is approximately 9.81 m/s^2.

Since the crate is submerged in water, the buoyant force acting on it is:

buoyant force = weight of water displaced = density of water * volume of water displaced * gravity

The volume of water displaced is equal to the volume of the cube, which is 0.25^3 = 0.015625 m^3. Therefore, the buoyant force is:

buoyant force = 1000 kg/m^3 * 0.015625 m^3 * 9.81 m/s^2 = 1.534453125 N

According to the problem, it takes 310 N to lift the crate while it is still submerged. This means that the net force acting on the crate is:

net force = lifting force - buoyant force = 310 N - 1.534453125 N = 308.465546875 N

This net force is equal to the weight of the crate:

net force = weight of crate = 0.1530875 * density

Therefore, we can solve for the density of the crate:

density = net force / 0.1530875 = 308.465546875 / 0.1530875 = 2016.06666667 kg/m^3

Rounding to the nearest hundredth, we get:

density ≈ 2016.07 kg/m^3

Therefore, the density of the cube is approximately 2016.07 kg/m^3

Answer:

1 km

Explanation:

displacement =velocity ×time

displacement =40m/s ×25s

displacement =1000m equivalent to 1km

The maximum velocity is 8.66 m/s²

As we know, simple harmonic motion refers to a to-and-fro motion in a periodic manner and spring constant refers to the force required to stretch or compress a spring.

The spring constant for a simple harmonic oscillator is given as 280 N/m, the total energy is 150 J and the mass is 4 kgs. We have to find the maximum velocity of the given simple harmonic motion.

We know that Energy = force x perpendicular distance

In an SHM, energy is in the form of Kinetic Energy. Hence, we use the formula for kinetic energy.

To find the maximum velocity, we will apply the formula for kinetic energy.

Since Kinetic Energy = 1/2 mass x velocity²

Therefore, 150 = 1/2 mass x velocity² ; velocity = 8.66 m/s²

Hence, the maximum velocity for the given system of SHM is 8.66 m/s²

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The magnitude of the angular acceleration of the roller is approximately 108.8 rad/s².

To find the magnitude of the angular acceleration of the roller, we can use the rotational analog of Newton's second law: τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

First, let's calculate the moment of inertia of the roller. The moment of inertia of a solid cylinder rotating about its central axis is given by the formula: I = (1/2)mr², where m is the mass and r is the radius.

Given:

Mass of the roller (m) = 2.4 kg

Radius of the roller (r) = 3.8 cm = 0.038 m

Moment of inertia (I) = (1/2) * 2.4 kg * (0.038 m)² = 0.0021744 kg·m²

Next, we need to calculate the torque (τ) applied to the roller. Torque is given by the formula: τ = rFsin(θ), where r is the distance from the axis of rotation to the point of application of the force, F is the magnitude of the force, and θ is the angle between the force and the line connecting the axis of rotation and the point of application.

Given:

Force applied (F) = 16 N

Angle (θ) = 35°

Distance from the axis of rotation to the point of application (r) is equal to the radius of the roller, so r = 0.038 m.

Torque (τ) = (0.038 m) * (16 N) * sin(35°) = 0.2366 N·m

Now, we can use the equation τ = Iα and solve for the angular acceleration (α):

0.2366 N·m = (0.0021744 kg·m²) * α

α = 0.2366 N·m / 0.0021744 kg·m² ≈ 108.8 rad/s²

Therefore, the magnitude of the angular acceleration of the roller is approximately 108.8 rad/s².

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The amount of heat required to heat the ice from -20 °C to water at 30°C is 238,612 J.

The amount of heat required to heat the ice from -20 °C to water at 30°C is calculated as follows;

Q = Q₁ + Q₂ + Q₃

where;

The amount of heat required to heat the ice from -20 °C to water at 30°C is calculated as;

Q = (0.44 x 4186 x 20) + (3.33 x 10⁵ x 0.44) + (0.44 x 4186 x 30)

Q = 238,612 J

Thus, the total quantity of heat required to raise the temperature of the ice to the liquid is 238,612 J.

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The circled vectors represents acceleration.

The last option is correct.

We see in the  first motion diagram the length of velocity vector is increasing this shows that the velocity is increasing in the magnitude with time so this is an accelerated motion in which a uniform acceleration must be in the same direction of velocity must be there.

We also notice in the second motion diagram the length of velocity vector is decreasing with time which shows the velocity is decreasing me magnitude which shows a constant deceleration and the direction of acceleration must be opposite to that of velocity.

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The magnitude of the angular acceleration of the roller is approximately 104.2 rad/s^2.

The magnitude of the angular acceleration of the roller can be determined using the torque equation and Newton's second law for rotational motion.
Step 1: Calculate the moment of inertia of the roller.
The moment of inertia (I) of a solid cylinder is given by the formula I = (1/2) * m * r^2, where m is the mass of the object and r is the radius.
In this case, the mass of the roller is 2.4 kg and the radius is 0.038 m.
So, I = (1/2) * 2.4 kg * (0.038 m)^2.
Step 2: Calculate the torque applied to the roller.
Torque (τ) is equal to the force (F) applied multiplied by the perpendicular distance (r) from the axis of rotation.
In this case, the force applied by Joe is 16 N and the distance is equal to the radius of the roller, 0.038 m.
So, τ = F * r.
Step 3: Use the torque equation.
The torque applied to the roller causes an angular acceleration (α) according to the equation τ = I * α.
Rearranging the equation, we get α = τ / I.
Step 4: Substitute the values into the equation.
Using the values we calculated earlier, we can substitute them into the equation α = τ / I.
α = (16 N * 0.038 m) / [(1/2) * 2.4 kg * (0.038 m)^2].
Step 5: Calculate the magnitude of the angular acceleration.
Evaluating the expression, we find that the magnitude of the angular acceleration of the roller is approximately 104.2 rad/s^2.
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The mass of ice that was initially placed in the vessel of neglected heat capacity was found to be 1.135 kg.

Given that,

Mass of water = 14 kg

Mass of ice = m kg

P = Power of the burner = dQ/dt

Rate of the heat given by the burner is constant.

In the first 45 min,

dQ/dt = mL/t1 = m x (80 x 4.2 x 10³)/45 min.               (1)

From 45 to 60 min,

dQ/dt = (m+14) x δ(Η₂Ο) x Δθ / t2 =(m+14) x (4.2 x 10³) x 2/ 15 min.            (2)

From (1) and (2)

m x (80 x 4.2 x 10³)/45 min = (m+14) x (4.2 x 10³) x 2/ 15 min

80m/3 = (m+14) x 2

80m = 6m + 84

m = 1.135 kg

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Answer:

1. autonomy - C-believing you are capable of fixing something yourself

2. drive - E-the need to quench thirst or fill a hungry stomach

3. extrinsic - D-competing for the prize of first place

4. intrinsic - A-reading for pleasure

5. motive - B-wanting to appear smart