replacement of the resistor by a resistor will theoretically have what effect on the circuit waveform at the output, ?

Answer:

I (shell) = 2/3 M R^2

I(sphere) = 2/5 M R^2

I(shell) / I(sphere) = (2/3) / (2/5) = 5/3

R(shell) / R(sphere) = (5/3)^1/2 = (15)^1/2 / 3

The required speed of the man when mass and speed of the stone are specified is calculated to be 0.00326 m/s.

Mass of the man is given as M = 92 kg.

Mass of the stone is given as m = 75 g = 0.075 kg.

Speed of the stone is given as u = 4 m/s.

Speed of the man is to be found out, v = ?

Using the conservation of momentum, we have,

The initial velocities of the man and the stone is zero(V).

So, mathematically,

(M+m)V = M v + m u

(M+m) × 0 = 92 × v + 0.075 × 4

92 × v + 0.3 = 0

92 v = - 0.3

v = - 0.00326 m/s

Thus, the speed of the man is calculated to be 0.00326 m/s.

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The star's (u-b) color can be calculated by taking the logarithm base 10 of the brightness ratio between the u and b filters, which yields a value of 0.63 magnitudes. Therefore, the star has a blue color.

The newly discovered star in this instance is revealed to be 2.33 times brighter when measured with the u filter than with the b filter. This ratio's logarithm in base 10 gives us log(2.33) = 0.37. The (u-b) colour index is 0.63 magnitudes since we are interested in the magnitude difference between the u and b filters, thus we must multiply this number by a factor of 1.7. This number being positive leads us to the conclusion that the star is blue. We cannot, however, draw any firm conclusions about the size, age, or makeup of the star without knowing more about its absolute brightness or other features.

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In the other motion,  your weight also be greater than your normal weight if the elevator moves downward while slowing in speed, then the weight of the person would also be greater than the normal weight.

The normаl force is the perpendiculаr force thаt opposes the weight of аn object in contаct with а surfаce. The normаl force equаls the object's weight only in situаtions where the object is directly on а horizontаl surfаce or the incline аngle is 0 degrees.

The force exerted by аn object perpendiculаr to а surfаce thаt prevents the object from sinking into the surfаce is referred to аs the normаl force. When аn object is plаced on а surfаce, the surfаce responds by exerting а force thаt is perpendiculаr to the object's weight. Аnother motion thаt would аlso result in your weight being greаter thаn your normаl weight is when the elevаtor moves downwаrd while slowing in speed. This occurs due to the аccelerаtion of the person within the elevаtor.

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The frequency of a wave is the number of cycles it completes in a given period of time, and it can be calculated using the following equation: frequency = velocity/wavelength. In the case of this wave, the frequency is 0.1 Hz (or 10 cycles/second).

frequency = 20 m/s/200 m = 0.1 Hz
The frequency of a wave that travels at 20 m/s with a wavelength of 200 meters is 0.1 Hz.What is frequency?Frequency is the number of occurrences of a periodic event per unit of time. It is commonly used to determine the number of occurrences of a specific event in a given period of time.

The frequency equation is:f = v/λwhere:f is the frequency of the  v is the velocity of the wave (m/s)λ is the wavelength of the wave (m)Using the formula given above:f = v/λwherev = 20 m/sλ = 200 metersf = 20/200f = 0.1 HzTherefore, the frequency of the wave is 0.1 Hz.

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A softball weighs 0.18427 kg. Its average rebound height on concrete, grass, and wood is 80.8 cm, 75.3 cm, or 87.6 cm, respectively. Kinetic energy 81 multiplied by a speed of 30 m/s and mass equals 2430.

The amount of matter inside an item is referred to as mass in mathematics. The most common way to determine mass is to weigh something. Anything will weigh more the more matter it contains. For instance, a mouse will have a higher mass than an ant since it contains more stuff.

The quantity of substance contained within an item is expressed in terms of mass. Typically, mass is expressed in kilogrammes (kg) or grammes (g) (kg). No matter where in the cosmos it is or how much gravitational force is exerted on it, mass is a measure of how much matter there is.

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When a particle is located a distance 2 meters from the origin, a force of 4 newtons acts on it.

The work done in moving the particle from A to B is 16J

The work done in moving the particle from B to C is -8J

The work done in moving the particle from C to D is 8J

Force F = 4N

Displacement dx = 2m

Total displacement of the particle from point A to B is dAB = 4m.

Work done is given by: W = F.dx (cosθ)  

Where,θ is the angle between the force and displacement.

1. Work done in moving the particle from A to B:

Let the particle is located at point A (x = 0). The force F acts in the positive direction of the x-axis. Therefore, the angle between the force and displacement is 0°.The work done in moving the particle from point A to B is

WAB = F(dx) cosθ

= (4 N)(4m) cos 0°

= (4 N)(4m)

= 16 J

2. Work done in moving the particle from B to C:

The displacement of the particle from B to C is dBC = 2m.

Therefore, the total displacement of the particle from point A to C is

dAC = dAB + dBC

=4m + 2m = 6m.

The force F acts in the negative direction of the x-axis. Therefore, the angle between the force and displacement is 180°.The work done in moving the particle from point B to C is

WBC = F(dx) cosθ

= (4 N)(2m) cos 180°

= (4 N)(-2m)

= -8 J

Note: Here, cos 180° = -1.

3. Work done in moving the particle from C to D:

Let the particle is located at point D (x = 6m).

The force F acts in the positive direction of the x-axis. Therefore, the angle between the force and displacement is 0°.

The work done in moving the particle from point C to D is WCD = F(dx) cosθ

= (4 N)(2m) cos0°

= (4 N)(2m)

=8J

Therefore, the work done in moving the particle from A to B is 16 J, the work done in moving the particle from B to C is -8 J, and the work done in moving the particle from C to D is 8 J.

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Given Angular velocity ω = 13.0 rad/s, magnitude v of the velocity of the particle in m/s = 13 m/s.

Explanation:

The formula used to calculate the linear velocity of a particle that is rotating at a distance from a fixed point at an angular velocity of ω radians per second is given as, v = r * ω.

Where, v = Linear velocity of a particle that is rotating at a distance r from a fixed point, r = Distance from the fixed point, ω = Angular velocity of the particle. Substituting the given values in the formula, v = r * ω= 1 * 13= 13 m/s. Therefore, the magnitude v of the velocity of the particle in m/s is 13 m/s.

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The initial velocity of the ball is approximately 73.65 m/s. So, the correct option is A. 73.6 m/s.

A particle travelling in a straight line has its acceleration plotted against time on an acceleration-time graph.

To solve this problem, we need to use the formula for the displacement of an object under constant acceleration:

Δy = v₀t + 1/2at²

where:

Δy = 0 (because the ball returns to its starting position)

v₀ = initial velocity (what we're trying to find)

t = 15.0 s (the time it takes for the ball to return to its starting position)

a = acceleration due to gravity (approximately -9.81 m/s², assuming the ball is thrown on Earth)

Plugging in these values, we get:

0 = v₀(15.0 s) + 1/2(-9.81 m/s²)(15.0 s)²

Simplifying:

0 = 15.0v₀ - 1104.75

Solving for v₀:

v₀ = 1104.75/15.0

v₀ ≈ 73.65 m/s

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The person-ladder system's centre of gravity is located 0.3 metres horizontally from the point where the ladder hits the ground.

The average position of an object's weight is known as its centre of gravity. Any object's travel through space may be entirely explained in terms of how its centre of gravity moves from one location to another.

According to the principle of moments, the total of the clockwise and anticlockwise moments is equal. In this instance, we could type:

[tex]W1 * d1 = (W1 + W2) * x[/tex]

We know that the total of the vertical forces acting on the ladder and the person is zero since they are both in equilibrium. Hence, we may write:

W1 + W2 = F

where F is the system's weight multiplied by the vertical force exerted on the ladder-person arrangement.

The two equations together give us:

W1 * d1 = F * x

Solving for x, we get:

x = (W1 * d1) / F

W1 = 200 N

W2 = 600 N

F = W1 + W2 = 800 N

we can see that d1 = 1.2 m and d3 = 0.8 m. Therefore:

x = (W1 * d1) / F = (200 N * 1.2 m) / 800 N = 0.3 m

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The center of gravity for the person-ladder system is 0.3 meters away from where the ladder touches the ground.

The sum of the clockwise and anticlockwise moments is equal, as per the moments' principle.

W1 × d1 = (W1 + W2)

Since the ladder and the person are both in equilibrium, we know that the sum of the vertical forces acting on them is zero. So, we may say:

W1 + W2 = F

Where F is the system's weight multiplied by the vertical force exerted on the ladder-person arrangement.

The two equations together give us:

W1 × d1 = F × x

Solving for x, we get:

x = (W1 × d1) / F

W1 = 200 N

W2 = 600 N

F = W1 + W2 = 800 N

d1 = 1.2 m and d3 = 0.8 m.

Therefore:

x = (W1 × d1) / F

= (200 N × 1.2 m) / 800 N

Distance = 0.3 m

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Use a 10mH inductor to design a low-pass passive filter with a cutoff frequency of 1600rad/s.

a) To specify the cutoff frequency in hertz, you can use the formula:
f = ω / 2π
where f is the frequency in hertz, and ω is the frequency in radians per second.

Given the cutoff frequency of 1600 rad/s, you can calculate the frequency in hertz as follows:
f = 1600 / (2 * π) ≈ 254.65 Hz

b) To specify the value of the filter resistor, use the formula:
R = 1 / (ω * L)
where R is the resistor value, ω is the cutoff frequency in rad/s, and L is the inductor value.

Given the cutoff frequency of 1600 rad/s and an inductor value of 10mH (0.01 H), the resistor value can be calculated as follows:
R = 1 / (1600 * 0.01) ≈ 0.0625 Ω

c) To find the smallest value of load resistance that can be connected across the output terminals of the filter without decreasing the cutoff frequency by more than 10%, you can use the following formula:

R_ load_ min = R / ((1 - 0.9) * (1 + 0.9))
Given the resistor value calculated in (b) is 0.0625 Ω:
R_ load_ min = 0.0625 / ((1 - 0.9) * (1 + 0.9)) ≈ 3.125 Ω

d) If the resistor found in (c) is connected across the output terminals, the magnitude of H(jω) when ω=0 can be calculated using the formula:
H(jω) = R_ load / (R + R_ load)

Given the resistor values calculated in (b) and (c):
H(jω) = 3.125 / (0.0625 + 3.125) ≈ 0.9804

So, the magnitude of H(jω) when ω=0 is approximately 0.9804.

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Answer:

The equation that relates the wave period and wavelength is:

wave speed = wavelength / wave period

Explanation:

The equation that relates the wave period (T) and wavelength (λ) is v = λ / T, Where v represents the velocity of the wave.

The wave period (T) is the time it takes for one complete cycle of a wave to pass a given point. It is measured in seconds (s) and represents the time taken for a wave to repeat its pattern.

The wavelength (λ) is the distance between two corresponding points on a wave, usually measured from crest to crest or trough to trough. It is denoted in units of length, such as meters (m) or centimeters (cm). The wavelength determines the spatial extent of a wave's repeating pattern.

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a) The correct statements are: Electron can be diffracted by matter. This confirms their wave nature.

The wavelength of the electron is given by the de Broglie equation.

The incorrect statements are:

The wave associated with a moving electron is an electromagnetic wave.

The kinetic energy of the electron is given by the equation E = hf.

b) The de Broglie wavelength of a particle is given by the equation:

λ = h / p

where λ is the wavelength, h is Planck’s constant, and p is the momentum of the particle. The momentum of the carbon atom is given by:

p = mv

where m is the mass of the carbon atom and v is its velocity. Substituting the given values, we get:

p = (2.0 x 10⁻²⁶kg) v

λ = h / p

λ = h / (2.0 x 10⁻²⁶ kg) v

Substituting the given value of λ, we get:

6.8 x 10⁻²⁶ m = (6.626 x 10⁻³⁴ J s) / (2.0 x 10⁻²⁶kg) v

Solving for v, we get:

v = (6.626 x 10⁻³⁴ J s) / (2.0 x 10⁻²⁶ kg) (6.8 x 10⁻²⁶ m)

v = 1.62 x 10³ m/s

Therefore, the speed of the carbon atom is 1.62 x 10³ m/s.

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In order to ensure that everyone on the team is aware of the goals, rules, and steps involved in the science experiment, communication is crucial. It permits the transparent exchange of information and thoughts.

The chance to get input from stakeholders, experts, and other people with a professional or academic interest in the topic is provided by the presentation of your research findings.

Science gains support, understanding of its broader relevance to society is promoted, and it encourages more informed decision-making at all levels, from government to communities to individuals, when scientists are able to effectively communicate outside of their peer group.

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Tension required to make the 5.0 gram piano wire vibrate at the D4 note is approximately 268,679.21 N.

Let's discuss it further below.

To find the tension required to make a 5.0 gram piano wire with a length of 44.0 cm vibrate at the D4 note (f = 293.7 Hz), follow these steps:

1. Convert the given mass and length to SI units.
Mass (m) = 5.0 g = 0.005 kg
Length (L) = 44.0 cm = 0.44 m

2. Use the formula for the fundamental frequency of a vibrating string:
f = (1/2L) * sqrt(T/μ), where T is the tension and μ is the linear mass density.

3. Calculate the linear mass density (μ) using the given mass and length:
μ = m / L = 0.005 kg / 0.44 m = 0.01136 kg/m

4. Rearrange the formula for the fundamental frequency to solve for tension (T):
T = (2L * f)² * μ

5. Plug in the values and calculate the tension:
T = (2 * 0.44 m * 293.7 Hz)² * 0.01136 kg/m
T = (517.968)² * 0.01136 kg/m
T = 268679.21 N

Therefore, the tension required to make the 5.0 gram piano wire vibrate at the D4 note is approximately 268,679.21 N.

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The collision between the tennis ball and the ground is an inelastic collision, as some energy was lost during the collision, indicating that it was not perfectly elastic.

The collision between the tennis ball and the ground is an example of an inelastic collision. During the collision, some energy is lost due to the deformation of the ball and the ground. This loss of energy is evidenced by the fact that the ball does not rebound to the same height from which it was dropped. In an elastic collision, the kinetic energy of the system is conserved, but in an inelastic collision, it is not. Inelastic collisions are characterized by permanent deformation of the objects involved, as energy is transformed into other forms such as heat and sound.

Therefore, based on the information given, we can conclude that the tennis ball and the ground experienced an inelastic collision when the ball was dropped from a height of 1.0 m, bounced off the ground, and rose to a height of 0.85 m.

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Answer: Geological records show that there have been a number of large variations in the Earth's climate. These have been caused by many natural factors, including changes in the sun, emissions from volcanoes, variations in Earth's orbit and levels of carbon dioxide (CO2).

For a shorter answer: changes in the sun, emissions from volcanoes, variations in Earth's orbit and levels of carbon dioxide (CO2).

The braking force needed to bring the car to a halt in 2.0 seconds, given that the car has amass of 1100 Kg and was moving at 24 m/s is -13200 N

We'll begin our calculation by obtaining the deceleration of the car. This is shown below:

a = (v - u) / t

a = (0 - 24) / 2

a = -24 / 2

a = -12 m/s²

Haven obtained the deceleration, we shall determine the breaking force needed to halt the car. Details below:

Force = mass × deceleration

Breaking force = 1100 × -12

Breaking force = -13200 N

Thus, the breaking force needed to stop the car is -13200 N

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Explanation:

We can start by using the fact that the linear velocity of the belt is equal to the tangential velocity of the flywheel at the point where the belt contacts it. We can use this to find the tangential velocity of the outer edge of the flywheel using the ratio of the radii.

Let's call the tangential velocity of the outer edge of the flywheel "v". Then we have:

v / 45.0 m/s = R / r

where R is the radius of the flywheel and r is the radius of the axle. We can rearrange this to solve for v:

v = (45.0 m/s) * (R / r)

Substituting in the given values for R and r, we get:

v = (45.0 m/s) * (27.4 cm / 3.25 cm)

Converting the radius to meters:

v = (45.0 m/s) * (0.274 m / 0.0325 m)

Simplifying:

v = 379.6 m/s

Therefore, the tangential velocity of the outer edge of the flywheel is approximately 379.6 m/s

The terminal velocity depends directly on the drag force and the relationship between them is nonlinear.

The terminal velocity is the maximum velocity that a falling object can reach when the drag force of the surrounding fluid is equal to the gravitational force acting on the object. The drag force is dependent on the velocity of the object, and as the velocity increases, the drag force also increases.

However, the relationship between the drag force and velocity is nonlinear because the drag force is proportional to the square of the velocity. This means that as the velocity of the object increases, the drag force increases more rapidly. Therefore, the terminal velocity, which is the point at which the drag force balances the gravitational force, is reached when the nonlinear relationship between the drag force and velocity is balanced by the gravitational force.

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A coal-burning power station that burned 250 tonnes of coal and also was 40% efficient could produce 7.3E5 kWh of energy. 0.89 kWh/pound for coal. 0.14 kWh/cubic foot for natural gas.

A power over 1 kW in use for 1 hour is equal to 1 kWh, as are powers of 05 kW used for two h, 2 kW used for 05 hours, etc. 1 k W h is equal to 1 kilowatt multiplied by 1 hour, 1000 watts, 3600 seconds, or 3,600,000 watt-seconds, or joules.

The calorific value for hard coal, which varies depending on the type, is somewhere between 29.3 MJ/kg (fuel coal) and 33.5 MJ/kg. One kilogramme of coal is equal to 7,000 kilocalories (7,000 kwh 29.3 MJ 8.141 kWh) (anthracite).

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The coal-burning power plant could generate 667 kWh of energy.

What is Energy?

Energy is the ability to do work, and it comes in many different forms. It can be in the form of mechanical energy, thermal energy, electrical energy, electromagnetic radiation, or nuclear energy, among others. Energy can be converted from one form to another, but it cannot be created or destroyed, only transferred or transformed.

To calculate the energy generated by the coal-burning power plant, we need to use the following formula:

Energy Generated = Efficiency x Energy Content x Amount of Coal Burned

Efficiency is given as 40%, which can be converted to a decimal by dividing by 100:

Efficiency = 40% = 0.40

The energy content of coal varies depending on the type of coal, but a reasonable estimate is around 24 megajoules per kilogram (MJ/kg). To convert this to kilowatt-hours (kWh), we need to divide by 3.6 million (the number of joules in a kWh):

Energy Content = 24 MJ/kg / 3.6 million = 0.00667 kWh/kg

The amount of coal burned is given as 250 tons, which can be converted to kilograms by multiplying by 1000:

Amount of Coal Burned = 250 tons x 1000 kg/ton = 250,000 kg

Now we can substitute these values into the formula:

Energy Generated = 0.40 x 0.00667 kWh/kg x 250,000 kg

Energy Generated = 667 kWh

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The work required to pump all the water out of the tank is [tex](1000/3)\pi g[/tex] J.

The volume of a cone can be calculated using the formula:

[tex]V = (1/3)\pi r^2h[/tex]

where r is the base radius, h is the height, and π is the mathematical constant pi.

Assuming the base radius of the cone is 1 meter and the height is 1 m.

In this case, the tank is full of water, so its volume is:

[tex]V = (1/3)\pi (1^2)(1) = (1/3)\pi[/tex]

The mass of the water in the tank is its volume times its density:

[tex]m = V\rho = (1/3)\pi (1000) = (1000/3)\pi[/tex]

To pump out the water, we need to lift it to the top of the tank, which has a height of 2 meters. So the work required is:

[tex]W = mgh = (1000/3)\pi g\ J[/tex]

Where g is the acceleration due to gravity.

Therefore, it requires approximately [tex](1000/3)\pi g[/tex] Joules of work to pump all the water out of the tank.

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The resultant force of an object being subjected to two forces in the positive y-direction of 20N each, a force in the positive x-y direction of 65N at an angle of 60 degrees with respect to the positive x-axis, and a force in the positive x-direction of 15N is 103.85N.

This can be calculated by the use of vector addition. Vector addition involves combining the vectors mathematically so that the total effect of all the vectors can be seen.

Step 1: Represent each of the forces as vectors. For the two forces in the positive y-direction, these can be represented as Fy1 = 20N and Fy2 = 20N. The force in the positive x-y direction can be represented as Fxy = 65N at an angle of 60 degrees. Lastly, the force in the positive x-direction can be represented as Fx = 15N.

Step 2: Calculate the components of the force in the positive x-y direction. This can be done using trigonometry. The horizontal component is found by multiplying the force by the cosine of the angle. This gives the horizontal component as Fxy,h = 65N x cos60 = 65N x 0.5 = 32.5N. Similarly, the vertical component is found by multiplying the force by the sine of the angle. This gives the vertical component as Fxy,v = 65N x sin60 = 65N x 0.866 = 56.59N.

Step 3: Calculate the resultant force. This can be done by summing the components of each of the forces. For the x-direction, this is simply Fx = 15N + 32.5N = 47.5N. For the y-direction, this is Fy = 20N + 20N + 56.59N = 96.59N.

Step 4: Calculate the magnitude and direction of the resultant force. This can be done using the Pythagorean theorem, since the magnitude is the hypotenuse of a right triangle. The magnitude of the resultant force is thus Fres = √(47.52 + 96.592) = 103.85N. The direction is found by using the inverse tangent of the x and y components. This gives the direction as θ = tan-1(96.59/47.5) = 77.94°.

Therefore, the resultant force of the object being subjected to two forces in the positive y-direction of 20N each, a force in the positive x-y direction of 65N at an angle of 60 degrees with respect to the positive x-axis, and a force in the positive x-direction of 15N is 103.85N at an angle of 77.94° with respect to the positive x-axis.

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The man moves to the left with a speed of 0.084 m/s (or about 8.4 cm/s) after he throws the shoe to the right.

To solve this problem, we need to apply the law of conservation of momentum, which states that the total momentum of a system is conserved if no external forces act on it. In this case, the man and the shoe form a closed system, and their initial momentum is zero because they are at rest. After the man throws the shoe to the right, the system's momentum remains zero, but the man will move to the left to conserve the momentum.

We can use the formula for momentum, which is given by:

[tex]p = m .v[/tex]

Where p is momentum, m is mass, and v is velocity.

Before the man throws the shoe, the total mass of the system is:

[tex]m_{total} = m_{man} + m_{shoe}\\\\m_{total} = \frac{720 N}{9,81 m/s^2} + \frac{4,0 N}{9,81 m/s^2} \\\\m_{total} = 73.4 kg[/tex]

The initial momentum of the system is:

[tex]p_{initial} = m_{total} . 0\\p_{initial} = 0 kg m/s[/tex]

After the man throws the shoe, the shoe's momentum is:

[tex]p_{shoe} = m_{shoe} . v_{shoe}\\\\p_{shoe} = \frac{4,0 N}{9,81 m/s^2}. 15 m/s \\\\p_{shoe} = 6.10 kg m/s[/tex]

To conserve the momentum, the man's momentum must be equal and opposite:

[tex]p_{man} = -p_{shoe}\\p_{man} = -6.10 kg m/s[/tex]

Finally, we can solve for the man's velocity using the formula for momentum:

[tex]p_{man} = m_{man} . v_{man}\\\\v_{man} = \frac{p_{man}}{m_{man}} \\\\v_{man} = \frac{(-6).10 kg m/s}{\frac{720 N}{9,81 m/s^2} } \\\\v_{man} = -0.084 m/s[/tex]

Therefore, the man moves to the left with a speed of 0.084 m/s (or about 8.4 cm/s) after he throws the shoe to the right.

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The sample size required to be 90% confident that the estimated mean has an error less than 0.02 is 5534.

The formula used to calculate the required sample size is given by:

n = ((zα/2 × σ) / E)²

where:

n = sample size

zα/2 = z-value for the level of confidence (α/2

)σ = population standard deviation

E = maximum error

Population standard deviation, σ = 6.8

Maximum error, E = 0.02

Confidence level, α = 0.9

Therefore, α/2 = 0.45 (since the confidence interval is symmetric)

The z-value for 0.45 level of confidence is 1.645.

Thus:

n = ((1.645 × 6.8) / 0.02)²

n = (11.066 / 0.02)²

n = 5533.0256

Rounding up, the required sample size is n = 5534.

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In the given collision scenario, the law of conservation of momentum is not satisfied, indicating the idealized nature of perfectly elastic collisions.

To determine whether the law of conservation of momentum is satisfied in this collision, we need to calculate the total momentum of the system before and after the collision and see if they are equal.

If we assume that the two balls are moving in opposite directions with the same speed, then their momenta before the collision are:

p₁ = m₁v₁ = (0.5 kg)(v) and p₂ = m₂v₂ = -(0.5 kg)(v)

where v is the speed of the balls, and the negative sign for p₂ indicates that it is in the opposite direction.

The total momentum before the collision is:

p_before = p₁ + p₂ = (0.5 kg)(v) - (0.5 kg)(v) = 0

This means that the total momentum of the system before the collision is zero.

After the collision, the two balls will stick together and move with a common speed. Let's assume that their final speed is v_f.

The total momentum after the collision is:

p_after = (m₁ + m₂)*v_f = (0.5 kg + 0.5 kg)v_f = 1 kgv_f

Since the two balls stick together and move with a common speed, the momentum is conserved and the total momentum after the collision is equal to the total momentum before the collision:

p_before = p_after = 0 = 1 kg*v_f

This is a contradiction, as there is no value of v_f that satisfies this equation. Therefore, the law of conservation of momentum is not satisfied in this collision.

In reality, the collision between the two balls would not be perfectly elastic, and some energy would be lost to friction and other factors. This would result in a partial loss of momentum, and the law of conservation of momentum would be approximately satisfied, but not exactly.

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1) To apply maximum pushing force, use the largest possible number of segments and move through a large range of motion.
2) Additionally, move the segments in an ordered sequence, one after the other, to ensure maximum pushing force.
3) For example, the sequence could be 1, 3, 2, 4.
To apply maximum pushing force, one should move through a large range of motion.

Maximum pushing force refers to the amount of force that a person can exert on an object when pushing it. The amount of maximum pushing force that a person can apply is dependent on various factors such as the strength of their muscles, the weight of the object being pushed, and the range of motion.

To apply maximum pushing force, one should move through a large range of motion. Moving through a large range of motion helps to recruit a larger number of muscle fibers which in turn helps to generate more force. Therefore, option 3 is the correct answer. Option 1 is incorrect because using the largest possible number of segments does not necessarily translate to more force. Option 2 is also incorrect because using the smallest segments may not result in more force. Option 4 is incorrect because moving the segments in an ordered sequence, one after the other does not always translate to more force.

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The new voltage between the plates of the capacitor in the terminals of a 6.33-V battery is 6.33-V.

The voltage between the plates of the capacitor will remain the same after the spacing between the plates is doubled. This is because the voltage of a capacitor is determined solely by the amount of charge stored in the capacitor. Increasing the spacing between the plates does not change the charge stored on the capacitor, so the voltage between the plates stays the same.

In this case, the new voltage between the plates of the capacitor would remain at 6.33-V.

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The block sliding downwards on an inclined rough plane accelerates with a magnitude of g sinθ − µg cosθ, where µ is the coefficient of kinetic friction. The maximum angle that the plane can make with the horizontal is tan−1 µ.There are no circumstances in which tilting over is excluded.

When a block of mass m slides down on an inclined rough plane, the force acting on it is its weight, which is in a downward direction. This can be resolved into two components: one that is parallel to the plane and the other that is perpendicular to it. The former tends to move the block down the plane, while the latter counteracts the normal force acting on the block. The acceleration of the block can be calculated as a result of the net force acting on it.

µ is the coefficient of kinetic friction.

The angle of the plane with the horizontal is denoted by θ.

The acceleration of the block is given by:

Under certain circumstances, tilting over is avoided. A block can be prevented from tilting over on an inclined plane by ensuring that the center of gravity of the block lies within the base of the plane.

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Work = 222 cm x 2. The joule 444(J), sometimes known as the newton metre (N m), is the Si derived unit for work. The work required to move an item 2 metres with 1 N much force is measured in joules.

The work performed by a force from one newton acting via one metre is equivalent to one joule, a unit of work of energy in the Internacional System of Units (SI). Its name honours English physicist William Prescott Joule and its equivalent in ergs is 107, or around 0.7377 foot-pounds.

The work (or heat expended) by the a force with one newton (N) operating more than a distance of 1 m is equivalent to one joule (m). A force of one newton causes a mass of one kilogramme (kg) to accelerate by one m per second (s) every second. Thus, one joule is equivalent to one newton metre.

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