WHAT IS ACCURACY, PRECISION, AND REPRODUCIBILITY? AND WHY ARE THEY SO NECESSARY IN CONDUCTING/DESIGNING EXPERIMENTS? 30 POINTS AND WILL MARK BRAINLIEST

Explanation:

Joule, unit of work or energy in the International System of Units (SI); it is equal to the work done by a force of one newton acting through one metre. ... In electrical terms, the joule equals one watt-second—i.e., the energy released in one second by a current of one ampere through a resistance of one ohm.

Temperature is most closely related to molecular Kinetic energy. Hence, option (A) is correct.

The kinetic theory of gases is a theory that can be used to deduce many of the gas's macroscopic features from a simplified molecular or particle description of the gas.

The most basic kinetic model is predicated on the following premises:

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The force acting on the system of two blocks connected by a rope passing over a pulley is -13.26 N.

The system of two blocks connected by a rope passing over a pulley are M1 and M2, where M1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M1 and the surface is mu subscript 1. M2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle.

The coefficient of friction between box M2 and the surface is mu subscript 2. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle.M1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. M2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.The free-body diagram of M1 shows that the weight of M1 acts straight downwards (vertically) and the normal force acts perpendicular to the slope.

The force of friction opposes the motion and acts opposite to the direction of motion.M1 = 2.25 kgTheta subscript 1 = 42.5 degreesMu subscript 1 = 0.205g = 9.81 m/s²In the free-body diagram of M2, the normal force acts perpendicular to the incline of the slope, the weight of the object acts vertically downwards and parallel to the incline, and the force of friction opposes the motion and acts opposite to the direction of motion.M2 = 5.55 kgTheta subscript 2 = 33.5 degreesMu subscript 2 = 0.105g = 9.81 m/s²The tension in the string is the same throughout the rope. Since the masses are being pulled by the same rope, the acceleration of the objects is the same as the acceleration of the rope.

The tension in the string is directly proportional to the acceleration of the objects and the rope.A system of two blocks connected by a rope passing over a pulley has a total mass of M. The acceleration of the system is given by the formula below:a = [(m1-m2)gsin(θ1) - μ1(m1+m2)gcos(θ1)] / (m1 + m2)Where, μ1 = 0.205 is the coefficient of friction of block M1θ1 = 42.5 degrees is the angle of the incline of block M1M1 = 2.25 kg is the mass of block M1M2 = 5.55 kg is the mass of block M2g = 9.81 m/s² is the acceleration due to gravitysinθ1 = sin 42.5 = 0.67cosθ1 = cos 42.5 = 0.75The acceleration of the system is:a = [(2.25-5.55)(9.81)(0.67) - (0.205)(2.25+5.55)(9.81)(0.75)] / (2.25 + 5.55)a = -1.7 m/s² (the negative sign indicates that the system is accelerating in the opposite direction).

The force acting on the system is given by:F = MaWhere M is the total mass of the system and a is the acceleration of the system. The total mass of the system is:M = m1 + m2M = 2.25 + 5.55M = 7.8 kgThe force acting on the system is:F = 7.8(-1.7)F = -13.26 N (the negative sign indicates that the force is acting in the opposite direction).

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Answer: low = 1.73 m and high = 0.129 m

Explanation: All you have to do is use the formula (Wavelength = Velocity/Frequency. Which would be 340m/s divided by 196 Hz and then 340m/s divided by 2637Hz.

The wavelength of a wave is its speed divided by frequency. The wavelength corresponds to the lower frequency of 196 Hz is 1.73 m and that for 2637 Hz is 0.12 m. Hence the wavelength range is 1.73 m to 0.12 m.

Frequency of a wave is the number of wave cycles per unit time. It is the inverse of the time period. Thus, its unit is s⁻¹ which is equal to Hz. Frequency of a wave is inversely proportional to the wavelength.

The relation between speed and frequency with wavelength of the wave is given by,

c = νλ

Given frequency  ν1 = 196 Hz.

speed of sound wave c = 340 m/s

then, wavelength at this frequency λ1 = 340 m/s / 196 Hz = 1.73 m.

For a frequency ν2 = 2637 Hz.

λ2 = 340 m/s/ 2637 Hz = 0.12 m.

Therefore, the range of wavelength of the notes from the violin will be in between 1.73 m to 0.12 m.

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At 30°C, the rms speed of a sample of oxygen with a molar mass of 16 g/mol is approximately 482.34 m/s.

The root mean square (rms) speed of a gas molecule is a measure of the average speed of the gas particles in a sample. It can be calculated using the formula:

vrms = √(3kT/m)

Where:

vrms is the rms speed

k is the Boltzmann constant (1.38 x 10^-23 J/K)

T is the temperature in Kelvin

m is the molar mass of the gas in kilograms

To calculate the rms speed of oxygen at 30°C (303 Kelvin) with a molar mass of 16 g/mol, we need to convert the molar mass to kilograms by dividing it by 1000:

m = 16 g/mol = 0.016 kg/mol

Substituting the values into the formula, we have:

vrms = √((3 * 1.38 x 10^-23 J/K * 303 K) / (0.016 kg/mol))

Calculating this expression yields the rms speed of the oxygen sample:

vrms ≈ 482.34 m/s

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In this case, the demand (D1) moves to the left (D2), this also happens with supply (S1) leading to (S2), moreover, the intersections between these lines represent E1, E2, and E3.

Due to an increase in salary, it is expected the demand for dinners increase, which means this line would move to the left. This occurs as a higher wage for everyone implies people are more willing to pay for dinner than before.

This change would also mean restaurants are likely to provide more quantity, which increases the supply, and therefore in this process the equilibrium changes.

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Answer:

b is the answer of that question answer sapai

The potential energy of the roller coaster is 176,400 J (joules).

The potential energy of an object is given by the formula PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical position of the object.

In this case, the roller coaster is at a peak of 20m and has a mass of 900kg. The acceleration due to gravity, g, is approximately 9.8 \(m/s^2\).

Using the formula, we can calculate the potential energy:

PE = mgh

= (900 kg)(9.8 \(m/s^2\))(20 m)

= 176,400 J

Therefore, the potential energy of the roller coaster is 176,400 J (joules).

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Answer:

The time will be: 25.02 seconds

Explanation:

Given

A bicycle travels at a velocity of 2.33 m/s.

i.e.

and

As we know the formula to find the time

Time = Displacement/Velocity

        = d/v

        = 58.3/2.33

        = 25.02 seconds

Therefore, the time will be: 25.02 seconds

a. Upthrust on the solid:

\(Upthrust = volume of solid * density of fluid * g = 1000 kg/m^3 * volume of solid * 10 m/s^2\)

b. Volume of the solid:

\(volume = mass/density = 1.3 kg / (1000 kg/m^3) = 1.3 x 10^-3 m^3\)

c. Density of the solid:

So,\(density = mass/volume = 1.3 kg / (1.3 x 10^-3 m^3) = 1000 kg/m^3\)

Upthrust is the upward force exerted on an object immersed in a fluid. It is equal to the weight of the fluid displaced by the object and acts in the opposite direction to gravity. Upthrust helps to counteract the weight of the object and keep it afloat.

a. The upthrust on an object immersed in a fluid is equal to the weight of fluid displaced by the object. The weight of fluid displaced can be calculated using the formula:

Weight of fluid = volume of fluid * density of fluid * g

Since the solid is completely immersed in water, the volume of the fluid displaced is equal to the volume of the solid. The density of water is given as 1000 kg/m^3, and the acceleration due to gravity is given as\(10 m/s^2.\)

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For work to be considered done by a physicist, it must involve the application of physics principles and concepts to solve a problem or answer a research question. This can involve theoretical work, such as mathematical modeling, or experimental work, such as designing and conducting experiments to test hypotheses.

An example of work done by a physicist would be the design and analysis of experiments to study the properties of a new material. This could involve developing a theoretical model to predict the behavior of the material, designing and building experimental apparatus, conducting experiments, and analyzing the resulting data using statistical and mathematical techniques. The physicist would use their knowledge of physics principles and concepts to interpret the data and draw conclusions about the properties of the material.

A piece of copper at 100°C is transferred to a copper calorimeter with a liquid at 30°C. The final temperature is 50°C. By applying the principle of conservation of energy, the specific heat capacity of the liquid is calculated to be approximately 2100 J/kg/°C.

To calculate the specific heat capacity of the liquid, we can apply the principle of conservation of energy. The heat lost by the copper piece will be equal to the heat gained by the liquid and calorimeter.

The heat lost by the copper piece can be calculated using the formula:

Heat lost = Mass of copper × Specific heat capacity of copper × Temperature change

Given:

Mass of copper = 400 g

Specific heat capacity of copper = 390 J/kg/°C (assuming it remains constant)

Temperature change of copper = 100°C - 50°C = 50°C

Heat lost = 400 g × 390 J/kg/°C × 50°C

Heat lost = 7,800,000 J

The heat gained by the liquid and calorimeter can be calculated using the formula:

Heat gained = (Mass of liquid + Mass of calorimeter) × Specific heat capacity of liquid × Temperature change

Given:

Mass of liquid = 100 g

Mass of calorimeter = 10 g

Temperature change of liquid = 50°C - 30°C = 20°C

Heat gained = (100 g + 10 g) × Specific heat capacity of liquid × 20°C

Now, by equating the heat lost and heat gained:

7,800,000 J = (110 g) × Specific heat capacity of liquid × 20°C

Specific heat capacity of liquid = 7,800,000 J / (110 g × 20°C)

Specific heat capacity of liquid ≈ 3545.45 J/kg/°C

Therefore, the specific heat capacity of the liquid is approximately 3545.45 J/kg/°C.

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Answer:

Electromagnetic waves

Electromagnetic waves bring energy into a system by virtue of their electric and magnetic fields. These fields can exert forces and move charges in the system and, thus, do work on them. However, there is energy in an electromagnetic wave itself, whether it is absorbed or not.

So the answer is B electrical energy

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Answer:

to answer it just get the distance traveled (m) and divide it by the time it took

Explanation:

so for the first one just do 100 divided by 47 which would equal 2.12 m/s

A parallel circuit can be modelled by connecting the components parallel.

A parallel circuit is a form of electrical circuit in which the components are linked in parallel to one another, each having a separate channel for current flow and being directly connected to the power source. The circuit must first be schematically represented, with the power supply and each component linked in parallel. Next, the total resistance the total current in the circuit by using Ohm's Law must be calculated.

Additionally, it is necessary to confirm that the total current entering the circuit equals the total current leaving the circuit. In a parallel circuit, the total current passing through all of the components equals the current entering the circuit. The voltage drop between each component must then be once more computed using Ohm's Law.

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Answer:

S=d/t.

distance is= speed multiplied by the time

D= 115 multiplied by 20

the answer is 2300km

Driver is approximately 0.63888 kilometers away from where he received the text message after 20 seconds.

To determine the distance the driver travels 20 seconds after receiving the text message, we need to calculate the distance covered during that time interval.

First, we convert the driver's speed from km/h to m/s for consistent units. The conversion factor is 1 km/h = 0.27778 m/s.

Driver's speed = 115 km/h × 0.27778 m/s = 31.944 m/s

Next, we use the formula for distance traveled:

Distance = Speed × Time

Distance = 31.944 m/s × 20 s = 638.88 meters

Therefore, the driver will be 638.88 meters away from the point where he received the text message after 20 seconds.

To convert this distance to kilometers, we divide by 1000:

Distance = 638.88 m ÷ 1000 = 0.63888 km

So, the driver is approximately 0.63888 kilometers away from where he received the text message after 20 seconds.

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Answer: the average acceleration of the car is 3.33 mph/s.

Explanation:

To find the average acceleration of the car, we can use the formula:

average acceleration = (final velocity - initial velocity) / time

In this case, the initial velocity of the car is 45 mph, the final velocity is 65 mph, and the time taken to reach the final velocity is 6 seconds.

Substituting these values in the formula, we get:

average acceleration = (65 mph - 45 mph) / 6 s

average acceleration = 20 mph / 6 s

average acceleration = 3.33 mph/s (rounded to two decimal places)

Therefore, the average acceleration of the car is 3.33 mph/s.

Answer:

110

Explanation:

45+65=110

1,080,000 cal is the number of calories to change 1.5 kg of 0∘c ice water to 1.5 kg of 100∘c boiling water.

The outmoded caloric theory of heat gave rise to the calorie, a unit of energy. Two primary meanings of "calorie" are frequently used due to historical factors.

The amount of heat required to increase the temperature with one water kilogram with one degree Celsius is known as a big calorie, food caloric intake, dietary calorie, and kilogram calorie (or one kelvin).

The latent heat of fusion of water=80 cal/g.

The specific heat of water=1 cal/g-C.

The latent heat of vaporization of water= 540 cal/g.

Therefore, if we have 1.5 kg = 1500 g, the total heat requirement is:

1500 g[(80 cal/g) + (1 cal/g-C)(100 - 0)C + (540 cal/g)] = 1500 g(720 cal/g) = 1,080,000 cal

Therefore,  1,080,000 cal is the number of calories to change 1.5 kg of 0∘c ice water to 1.5 kg of 100∘c boiling water.

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The resultant force while pushing the car would be 20 N.

Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.

As given in the problem Mary and Carl's car broke down so they are pushing the car. Mary is pushing the car at 9N and Carl is pushing the car at 11N.

As both Mary and Carl apply the force in the same direction while pushing the broke car

Resultant force= 9 N + 11 N

                          = 20 N

Thus, the resultant force comes out to be 20N.

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The question seems incomplete, the complete question is,

Mary and Carl's car broke down so they are pushing the car. Mary is pushing the car at 9N and Carl is pushing the car at 11N. Find out the resultant force while pushing the car

Answer:

Explanation:

according to the graph at B the potential energy of the particle is 2J

therefore we can use the kinetic energy equation to calculate the particle's velocity or speed.

\(E_{k} =1/2mv^{2}\)

2J= 1/2*1/2kg*v^2

8=v^2

v= 2√2 ms-1

Answer:

A political ideology largely concerns itself with how to allocate power and to what ends it should be used. Some political parties follow a certain ideology very closely while others may take broad inspiration from a group of related ideologies without specifically embracing any one of them.

(A)The water will shoot out of the hole at a speed of 4.77 m/s, and the pressure of the water at the hole will be 9.91 × 10^4 Pa, and (B) The water will shoot out of the larger hole at a speed of 0.529 m/s, and the pressure of the water at the hole will be 1.012 × 10^5 Pa.

We can use Bernoulli's equation to solve this problem, which relates the pressure, velocity, and height of a fluid. The equation states that:

P + (1/2)ρv^2 + ρgh = constant

where P is the pressure, ρ is the density of the fluid, v is the velocity of the fluid, g is the acceleration due to gravity, and h is the height of the fluid.

(A) The diameter of the hole is 4.51 cm, which corresponds to a radius of 2.255 cm = 0.02255 m. The area of the hole is A = πr^2 = 1.587 × 10^-4 m^2. The volume flow rate of water is Q = 0.757 m^3/s.

We can calculate the velocity of the water as it exits the hole using the equation:

Q = Av

where A is the area of the hole and v is the velocity of the water. Solving for v, we get:

v = Q/A = 4.77 m/s

Now, we can use Bernoulli's equation to find the pressure of the water at the hole. Assuming that the height of the fountain is negligible compared to the height of the atmosphere, we can set the height term to zero. Also, we can assume that the pressure at the surface of the fountain is atmospheric pressure, which we can take as P = 1.013 × 10^5 Pa. Then, the equation becomes:

P + (1/2)ρv^2 = constant

Solving for P, we get:

P = constant - (1/2)ρv^2

At the hole, the velocity of the water is v = 4.77 m/s, and the density of water is ρ = 1000 kg/m^3. Substituting these values, we get:

P = 1.013 × 10^5 Pa - (1/2) × 1000 kg/m^3 × (4.77 m/s)^2 = 9.91 × 10^4 Pa

So, the water will shoot out of the hole at a speed of 4.77 m/s, and the pressure of the water at the hole will be 9.91 × 10^4 Pa.

(B) If the diameter of the hole is three times as large, then the area of the hole will be nine times as large. Therefore, the volume flow rate of water will be distributed over a larger area, resulting in a lower velocity. The new area of the hole is A = 9 × 1.587 × 10^-4 m^2 = 1.43 × 10^-3 m^2. The volume flow rate of water is still Q = 0.757 m^3/s.

Using the equation Q = Av, we can find the new velocity of the water:

v = Q/A = 0.529 m/s

Using Bernoulli's equation, we can find the pressure of the water at the larger hole:

P = 1.013 × 10^5 Pa - (1/2) × 1000 kg/m^3 × (0.529 m/s)^2 = 1.012 × 10^5 Pa

So, the water will shoot out of the larger hole at a speed of 0.529 m/s, and the pressure of the water at the hole will be 1.012 × 10^5 Pa.

Hence, Water will flow out of the smaller hole at a speed of 0.529 m/s and a pressure of 1.012 × 10^5 Pa, and the water will shoot out of the hole at a speed of 4.77 m/s and a pressure of 9.91 × 10^4 Pa.

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Answer:  If you are traveling at a speed of 60mph, you will go 220 feet.

Explanation: 60mph is a mile a minute. 5280 feet in a mile, 60 seconds in a minute. Divide to find that is 88 feet per second. Multiply by the number of seconds.

Answer:

the proton speed of the proton was 1.6 × 10⁶ m/s

Explanation:

Given the data in the question;

Radius r = 6.42 cm = 0.0642 m

magnetic field B = 0.260 T

we know that; charge of proton q = 1.602 × 10⁻¹⁹ C

And mass of proton m = 1.672 × 10⁻²⁷ kg

we know that; Magnetic Force F = qvBsinθ

where q is the charge of proton, v is velocity, B is the magnetic field and θ is  angle ( 90° )

Also the Centripetal force experienced by the particle is;

F = mv² / r

where r is radius, m is mass of proton and v is velocity

hence;

qvBsinθ = mv² / r

we solve for v

rqvBsinθ = mv²

divide both sides by mv

rqvBsinθ / mv = mv² / mv

rqBsinθ / m = v

so we substitute

v = [ 0.0642 m × (1.602 × 10⁻¹⁹ C) × 0.260 T × sin(90°) ] /  1.67 × 10⁻²⁷ kg

v = 2.6740584 × 10⁻²¹ / 1.672 × 10⁻²⁷

v = 1.6 × 10⁶ m/s

Therefore, the proton speed of the proton was 1.6 × 10⁶ m/s

Answer:

a

 \(a = 9.8 \ m/s\)

   Increasing  

b

 \(s = 6.22 \ m\)

Explanation:

From the question we are told that

    The  initial speed is \(u = 9.7 \ m/s\)

      The  time  taken is  \(t = 0.51 \ s\)

Generally given that the stone is moving downward the acceleration is  equivalent to the generally value of acceleration due to gravity and it would be increasing as the stone approaches the ground(toward the center of the earth )

Thus the acceleration is  \(a = 9.8 \ m/s\)

   Generally from the equation of motion we have that

       \(s = ut + \frac{1}{2} at^2\)

=>    \(s = 9.7 * 0.51 + \frac{1}{2} *9.8 * 0.51^2\)

=>     \(s = 6.22 \ m\)

Answer:

Newton's theory because he suggested that the greater the distance between objects, the weaker the pull of gravity.

Answer: The answer is Newton’s theory because he suggested that the greater the distance between objects, the weaker the pull of gravity

Explanation: Just took the test i hope this help you enjoy :D

1) The force acting on the object during the time interval 0-3 seconds is 1/3 N.

2) The force acting on the object during the time interval 6-10 seconds is -0.5 N.

3) There is no time interval in which no force acts on the object.

(i) Force acting on the object in time interval 0-3 seconds. Force acting on the object is equal to the product of its mass and acceleration, i.e.,F = ma.

In the given velocity-time graph, the acceleration of the object can be determined by determining the slope of the velocity-time graph from 0 to 3 seconds.

Slope = (change in velocity) / (change in time)= (20-0) / (3-0) = 20/3 m/s^2

Acceleration, a = slope= 20/3 m/s^2

Mass of the object, m = 50 g = 0.05 kg

∴ Force acting on the object, F = ma= 0.05 × 20/3= 1/3 N.

Therefore, the force acting on the object during the time interval 0-3 seconds is 1/3 N.

(ii) Force acting on the object in time interval 6-10 seconds. Similar to the first question, the force acting on the object in time interval 6-10 seconds can be determined by determining the acceleration of the object during this time interval.

The slope of the velocity-time graph from 6 seconds to 10 seconds can be determined as follows:

Slope = (change in velocity) / (change in time)= (-20-20) / (10-6) = -40/4= -10 m/s^2 (negative sign indicates that the object is decelerating)

Mass of the object, m = 50 g = 0.05 kg

∴ Force acting on the object, F = ma= 0.05 × (-10)= -0.5 N.

Therefore, the force acting on the object during the time interval 6-10 seconds is -0.5 N.

(iii) Time interval in which no force acts on the object. There is no time interval in which no force acts on the object. This is because, as per Newton's first law of motion, an object will continue to remain in a state of rest or uniform motion along a straight line unless acted upon by an external unbalanced force.In other words, if the object is moving with a constant velocity, there must be a force acting on the object to maintain its motion.

Therefore, there is no time interval in which no force acts on the object.

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Answer: To solve this problem, we'll analyze the forces acting on the piano and calculate the work done by different forces.

Given:

Mass of the piano (m) = 380 kg

Distance traveled along the incline (d) = 2.9 m

Angle of the incline (θ) = 25°

First, let's calculate the force of gravity acting on the piano:

Force of gravity (F_gravity) = m * g

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

F_gravity = 380 kg * 9.8 m/s²

F_gravity ≈ 3724 N

Next, we can calculate the force exerted by the man that keeps the piano from accelerating. This force is equal in magnitude but opposite in direction to the force of gravity acting down the incline.

Force exerted by the man (F_man) = F_gravity

F_man ≈ 3724 N

To calculate the work done by the man, we use the equation:

Work = Force × Distance × Cosine(θ)

Work done by the man (W_man) = F_man × d × Cos(θ)

W_man = 3724 N × 2.9 m × Cos(25°)

Now, let's calculate the work done by the force of gravity. Since the force of gravity is acting parallel to the incline, the work done by gravity is equal to the component of gravity force parallel to the incline multiplied by the distance traveled along the incline.

Work done by gravity (W_gravity) = F_gravity × d × Cos(θ)

W_gravity = 3724 N × 2.9 m × Cos(25°)

Finally, to find the net work done on the piano, we subtract the work done by gravity from the work done by the man:

Net work done on the piano (W_net) = W_man - W_gravity

Substituting the calculated values, we can find the net work done on the piano.

The force exerted by the man is 3724 N.

The work done on the piano by the man is 9591.55 J

The work done on the piano by the force of gravity is 9591.55 J and

The net work done on the piano is 19183.10 J.

To determine the various quantities in this scenario, we can apply the principles of work and energy.

Given:

Mass of the piano, m = 380 kg

Distance traveled down the incline, d = 2.9 m

Incline angle, θ = 25°

Friction is ignored.

First, let's calculate the force exerted by the man. Since the piano is kept from accelerating, the force exerted by the man must balance the force of gravity acting down the incline.

1)Force exerted by the man:

The force of gravity acting on the piano can be calculated as:

F_gravity = m * g

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

F_gravity = 380 kg * 9.8 m/s^2

F_gravity = 3724 N

Since the piano is kept from accelerating, the force exerted by the man must be equal in magnitude and opposite in direction to the force of gravity.

Therefore, the force exerted by the man is also 3724 N.

Work done on the piano by the man:

The work done is given by the formula:

Work = Force * Distance * cos(θ)

where θ is the angle between the force and the direction of motion (the incline).

Work = Force * Distance * cos(θ)

Work = 3724 N * 2.9 m * cos(25°)

Calculate the value using a calculator:

Work = 9591.55 J (rounded to two decimal places)

Work done on the piano by the force of gravity:

The work done by the force of gravity can be calculated using the same formula:

Work = Force * Distance * cos(θ)

The force of gravity acting down the incline is F_gravity = 3724 N.

Work = F_gravity * Distance * cos(θ)

Work = 3724 N * 2.9 m * cos(25°)

Calculate the value using a calculator:

Work = 9591.55 J (rounded to two decimal places)

Net work done on the piano:

The net work done on an object is the sum of the work done by all forces acting on it.

Net work = Work by the man + Work by gravity

Net work = 9591.55 J + 9591.55 J

Net work = 19183.10 J (rounded to two decimal places)

Therefore, the force exerted by the man is 3724 N, the work done on the piano by the man is 9591.55 J, the work done on the piano by the force of gravity is 9591.55 J, and the net work done on the piano is 19183.10 J.

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