what would the minimum value of the coefficient of static friction need to be for the system not to move when released from rest?

Here are three ways to make the ball look red:

   Use a red light: Shine a red light directly onto the white ball while the ball is in the ON position. This will cause the ball to reflect only the red light and appear red.

In regards to the Use a magenta filter: Shine a white light onto the white ball while the ball is in the ON position, and place a magenta filter in front of the light source. This will cause the ball to reflect both red and blue light, which will combine to create the appearance of red.

Lastly, Combine green and blue light with a yellow filter: Shine a white light onto the white ball while the ball is in the ON position, and use green and blue lights to illuminate the ball. Place a yellow filter in front of the ball, which will absorb the green and blue light and allow only red light to pass through. This will create the appearance of a red ball.

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Kinetic energy is associated with movement, while potential energy is stored energy. Mechanical energy can refer to both kinetic and potential energy, depending on the context. Biodiesel energy refers to the energy produced by converting organic matter into fuel and is not directly related to movement or stored energy.

The relationship between Kinetic and Potential Energy is that they are both forms of energy that are associated with an object's motion or position. Kinetic Energy is the energy of motion, while Potential Energy is the energy of an object's position or stored energy. As an object moves, its Kinetic Energy increases, and as it moves higher, its Potential Energy increases. The two energies are inter-convertible, and the total energy of the system is always conserved. The sum of Kinetic and Potential Energy is known as the total mechanical energy of the object.

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Force magnitude when elevator is traveling at constant velocity is, 637 N. Force magnitude does the scale read when the elevator slows to a stop is,  429 N (downward force). Force magnitude when elevator starts downward again is, 208 N.

Force magnitude when elevator is traveling at constant velocity, 65 kg x 9.8 m/s^2 = 637 N (upward force)

When the elevator slows to a stop with an acceleration of magnitude 2.0 m/s^2, the net force acting on the student is the sum of the gravitational force and the force exerted by the scale.

Net force: (65 kg) x (2.0 m/s^2) = 130 N (downward force)

Force magnitude when elevator slows to a stop: 637 N - 130 N = 507 N (upward force)

When the elevator starts moving downward again with an acceleration of magnitude 2.0 m/s^2, the net force is difference between the gravitational force and the force exerted by the scale,

Net force: (65 kg) x (9.8 m/s^2 - 2.0 m/s^2) = 429 N (downward force)

Force magnitude when elevator starts downward again: 637 N - 429 N = 208 N (upward force)

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The total energy radiated by the Sun each second when emissivity of the Sun is 0.986 is 3.85×10²⁶ W .

The power output of the Sun is given by Stefan's law , i.e

P=σAeT 4

=(5.6696×10⁻⁸ W/m².K⁴ )[4π (6.96×10⁸m)² ]×(0.986)(5800K)⁴

=3.85×10²⁶ W

The total energy radiated by the Sun each second when emissivity of the Sun is 0.986 is 3.85×10²⁶ W .

The radiation radiated from an area A of a black body at absolute temperature T is directly proportional to the fourth power of the temperature, according to Stefan Boltzmann's law.

The total energy emitted or radiated by a black body per unit surface area across all wavelengths and per unit time is inversely related to the black body's thermodynamic temperature to the fourth power.

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Mass of the thin circular sheet of aluminum is 0.846 kg.

Weight is the force an item exerts, mass is the amount of substance that makes up an object. The kilogramme is the SI unit of mass.

To find mass use formula for  mass of a thin circular plate:

m = ρAΔt

m: plate mass ρ : density of the material, A: plate area and Δt : plate thickness

First, we need to find the area of the circular sheet, which is given by:

A =[tex]\pi r^{2}[/tex]

r: sheet radius Substituting:

A = [tex]\pi (20 cm)^2 = 1256.64 cm^2[/tex]

Convert thickness from millimeters to meters,  as density is typically given in kg per cubic m.

Δt = 0.50 mm = 0.0005 m

Aluminum density is [tex]2700 kg/m^3.[/tex] Substituting:

m = ρAΔt = [tex](2700 kg/m^3)(1256.64 cm^2)(0.0005 m) = 0.846 kg[/tex]

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Answer:

the car's average acceleration is 2.77 m/s^2.

Explanation:

The average acceleration of the car can be calculated using the formula for average acceleration, which is:

a = (v_f - v_i) / t

where:

a is the average acceleration

v_f is the final velocity (70 km/h)

v_i is the initial velocity (0 km/h)

t is the time taken (7 seconds)

First, we need to convert the velocity from km/h to m/s, as the formula requires both velocities to be in the same units. 70 km/h is equivalent to 70 * 1000 / 3600 = 19.44 m/s.

Plugging the values into the formula:

a = (19.44 - 0) / 7

a = 2.77 m/s^2

So the car's average acceleration is 2.77 m/s^2.

The maximum static friction force that can be exerted on the truck is 44,100 N

When there is no relative motion between two objects in contact, a static friction force is created. It is a kind of friction that fights against an object's propensity to slide over or move over another thing. When an external force is applied to an object, it experiences static friction force, which counteracts the applied force and prevents the object from moving.

The maximum static friction force that can be exerted between two surfaces is proportional to the normal force between them and is determined by the coefficient of static friction, denoted as μ_s. The coefficient of static friction is a dimensionless constant that depends on the nature of the two surfaces in contact. It describes the ratio of the maximum static friction force to the normal force.

The following equation calculates the maximum static friction force:

f_s ≤ μ_s * N

Where N is the normal force between the two surfaces, μ_s is the static friction coefficient, and f_s is the maximum static friction force that can be applied. According to the inequality sign in the equation, depending on the applied force, the static friction force might range from zero to its maximum value.

To calculate the friction force acting on the truck, we need to know the weight of the truck and the normal force exerted by the road on the truck.

Assuming that the truck is on a flat road and not accelerating vertically, the weight of the truck (i.e., the force due to gravity acting on the truck) is given by:

W = mg

Where:

m is the mass of the truck

g is the acceleration due to gravity, which is approximately 9.8 m/s^2

Let's assume that the mass of the truck is 5000 kg. Then the weight of the truck is:

W = mg = (5000 kg)(9.8 m/s^2) = 49,000 N

The normal force exerted by the road on the truck is equal in magnitude and opposite in direction to the weight of the truck, because the truck is not accelerating vertically. Therefore, the normal force is also 49,000 N.

The maximum static friction force that can be exerted on the truck is given by:

f_s = μ_s * N

where:

μ_s is the coefficient of static friction between the tires and the road

N is the normal force exerted by the road on the truck

Substituting the values, we get:

f_s = μ_s * N = (0.90) * (49,000 N) = 44,100 N

Therefore, the maximum static friction force that can be exerted on the truck is 44,100 N. If the force acting on the truck is greater than this value, the truck will start to slide on the road.

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You could read the time at night  if a minimum of 14.0% of the original tritium is needed to read the dial in dark places is 34.889 yr.

Given t1/2=12.3 yr

That amount at any given time,

A(t), is a power function with 1/2 as the base and the original amount at

t=0, A(0):

[tex]A(t) = A(0)(\frac{1}{2} )^\frac{t}{t_\frac{1}{2} }[/tex]

We convert 14% to 0.14 = [tex]\frac{A(t)}{A(0)}[/tex]

Using the natural logarithm on both sides:

ln(0.14) = [tex]\frac{1}{2}^\frac{t}{12.3}[/tex]

Using the property of logarithms that allows us to move the exponent to the outside as a coefficient:

ln 0.14 = ln(1/2) (t/12.3)

t = ln(0.14)/ -ln 2 x 12.3

t = 34.889 yr

The number of years that the time could be read at night is 34.889 years.

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(a) The change in momentum is -12.04 kg-m/s

(b) The force exerted by the bat is 1003.33 N

The given values are :

The mass of a ball, m = 0.14 kg

Initial speed of the ball, u = 40 m/s

Final speed of the ball, v = -46 m/s

(a) The change in momentum of the ball during the collision with the bat is given by :

Change in P = m(v-u )

Change in P = 0.14(-46-40)

Change in P = - 12.04 kg-m/s

(b) Time for collision, t = 0.012 s

the force can be calculated as follows :

force = Change in P/ t

force = 1003.33 N

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Complete question:

An official major league baseball has a mass of 0.14 kg. A pitcher throws a 40 m/s fastball which is hit by the batter straight back up the middle at a speed of 46 m/s.

a) What is the change in momentum of the ball during the collision with the bat?

b) If this collision occurs during a time of 0.012 seconds, what is the average force exerted by the bat on the ball?

The friction force experienced by the box is approximately 77.96 N.

Frictional force is a force that opposes the motion or attempted motion between two surfaces in contact. It is caused by the irregularities in the surfaces of objects that come into contact with each other, and it acts in the direction opposite to the direction of motion or attempted motion.

We can begin by drawing a diagram of the situation and labeling the forces acting on the box:

where:

Fp is the force applied by Chasadie.

Ff is the force of friction.

mg is the weight of the box (mass times gravity)

We can then use Newton's second law of motion, which states that the net force acting on an object is equal to its mass times its acceleration:

ΣF = ma

where ΣF is the sum of all the forces acting on the box.

In the vertical direction, we have:

ΣFy = N - mg = 0

where N is the normal force, which is equal and opposite to the weight of the box.

Since the incline is at an angle of 85.61 degrees, we can use trigonometry to find the components of the weight and normal force:

N = mg cos θ = (3.68 kg)(9.81 m/s²) cos (85.61°) ≈ 18.77 N

mg sin θ = (3.68 kg)(9.81 m/s²) sin (85.61°) ≈ 36.24 N

In the horizontal direction, we have:

ΣFx = Fp - Ff - mg sin θ = ma

Plugging in the given values and solving for Ff:

Ff = Fp - ma + mg sin θ

= (217.41 N) - (3.68 kg)(11.63 m/s²) + (3.68 kg)(9.81 m/s²) sin (85.61°)

≈ 77.96 N

Therefore, the friction force is approximately 77.96 N.

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The diagram for the solution of the question is as follows:

a. the average speed of the car is 15 m/s.

b. the car travels 60 meters in 4 seconds.  

c.  the constant acceleration of the car is 2 m/s².

d. we find the the final speed of the train as 25 m/s.

a)

We find the  total distance traveled by the car as:

0 + 2 + 8 + 18 + 32 = 60 meters.

Average speed = Total distance / Total time

= 60 meters / 4 seconds

= 15 m/s

b) The distance traveled by the car in 4 seconds =  total distance traveled

= 60 meters

c)

Acceleration = (Final velocity - Initial velocity) / Time

Acceleration = (Final velocity - Initial velocity) / Time

Acceleration = (8 m/s - 0 m/s) / (4 s - 0 s)

Acceleration = 8 m/s / 4 s

Acceleration = 2 m/s²

d.

Final velocity = Initial velocity + (Acceleration * Time)

Final velocity = 10 m/s + (1 m/s² * 15 s)

Final velocity = 10 m/s + 15 m/s

Final velocity = 25 m/s

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The block gains 4.49 J of kinetic energy as it moves a distance of 2 m

The increase in kinetic energy of an object can be calculated using the equation:

ΔKE = 1/2 (m)(Δv^2)

where;

To find the change in velocity, we can use the equation of motion, which states that the net force on an object is equal to its mass times its acceleration:

F = ma

Rearranging this equation, we can find the acceleration:

a = F / m

The acceleration is the same before and after the force increases, so we can equate the two and solve for Δv:

F1 / m = F2 / m

F1 / m = (F1 + ΔF) / m

ΔF / m = F2 / m - F1 / m

ΔF / m = (12 N) / (2 kg) - (5 N) / (2 kg)

ΔF / m = 4.5 m/s^2

Δv = √(2 * ΔF / m)

Now that we have Δv, we can plug this value into the equation for ΔKE:

ΔKE = 1/2 * m * Δv^2

ΔKE = 1/2 * 2 kg * (2.12 m/s)^2

ΔKE = 4.49 J

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The object is moving away from the origin at a constant velocity.

option A.

Constant velocity refers to a situation in which an object moves at a constant speed in a straight line. This means that the magnitude of the velocity (i.e. the speed) of the object remains the same over time, while its direction may or may not change.

For example, a car moving at a constant 60 km/h on a straight road has a constant velocity.

Uniform velocity on the other hand, refers to a situation in which an object moves with a constant speed, but also changes its direction. I

In the graph, between 4 and 6 seconds, the object is moving at a constant velocity.

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The volume of water displaced canoe compared to an ultra-lightweight  Kevlar canoe is 1.586ft^3 extra water.

Which would give us the total volume displace by each canoe, however, the problem asks us for the difference between volumes displaced, and thus the difference is:

ΔV = (Mc_c + Mc_K /ρ)_c - (Mc_c + Mc_K /ρ)_K

Since both the mass of the load and the density of the fluid is the same for both cases, the previous equation becomes:

ΔV = Mc_c - Mc_K /ρ

ΔV = 140lb - 41lb / 62.4 lb/ft^3

ΔV = 1.586ft^3

Kevlar canoe is a type of canoe made of a material called Kevlar. Kevlar is a strong synthetic fiber that is known for its resistance to abrasion, heat, and cuts. It is commonly used in the manufacturing of body armor, bulletproof vests, and other protective gear. In the case of a Kevlar canoe, the material is used to construct the hull of the canoe, making it lightweight, durable, and easy to maneuver.

Kevlar canoes are popular among outdoor enthusiasts, particularly those who enjoy canoeing or kayaking in whitewater rivers, lakes, or oceans. They are also popular among people who like to take long trips or expeditions as the Kevlar material ensures the canoe can withstand harsh conditions and heavy use.

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Complete Question: -

How much extra water does a 140-lb concrete canoe displace compared to an ultra -lightweight 41-lb Kevlar canoe of the same size carrying the same load?

When a projectile is launched at an angle of 30 degrees above the horizontal, the tallest barrier it may clear is 14 meters. The correct option is (b) 33.3m/s.

we can use the kinematic equations of motion for projectile motion. The key is to recognize that the projectile will reach its maximum height when it is at the top of its trajectory, at which point its vertical velocity is zero.

First, we can use the vertical motion equation to find the time it takes for the projectile to reach its maximum height:

vy = v.sin θ - gt

0 = v sin θ - gt

t = v sin θ / g

Next, we can use the horizontal motion equation to find the horizontal distance traveled by the projectile in this time:

x = v.cos θ.t

Now, we want to find the launch speed v that will result in the maximum height of 14 m. To do this, we can substitute the expressions we just derived for t and x into the vertical motion equation for the maximum height:

h = (v sin θ)² / (2g)

Solving v:-

v = √(2gh) / sin θ

Putting values:-

v = √(2 × 9.81 m/s² × 14 m) / sin 30.0°

v ≈ 33.3 m/s

Therefore, the projectile's launch speed is approx. 33.3 m/s, which is answer choice (b).

Note that answer choices (a), (c), and (d) are all close to the correct answer, but not quite correct. This illustrates the importance of being careful with units and significant figures when performing calculations.

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Given that Aluminum has 13 electrons per atom and an atomic mass of 27 u, we can use Avogadro's number (6.022 × 10^23 atoms/mole) to calculate the number of aluminum atoms in the 20 kg bar.

27 g/mol (atomic mass) x 1 kg / 1000 g = 0.027 kg/mol

The number of moles of aluminum in the 20 kg bar:

20 kg / 0.027 kg/mol = 740.74 moles

Finally, we can calculate number of aluminum atoms:

[tex]740.74 moles * 6.022 × 10^23[/tex] atoms/mole = [tex]4.460 * 10^26[/tex] atoms

4.460 × 10^26 atoms x 13 electrons/atom x 1.602 × 10^-19 coulombs/electron = 1.167 × 10^7 coulombs.

The total charge of all the electrons in a 20 kg bar of aluminum is approximately 1.167 × 10^7 coulombs.

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Electric and magnetic fields are proportional, and transverse electromagnetic waves do exist in nature. and have oscillations that are parallel to one another and have the same phase. A medium is not necessary for electromagnetic waves to travel.

Mechanical waves cannot pass through a vacuum, which is empty space, however electromagnetic waves can. They require a means of transportation, like water or air. The wave propagates in a direction that is opposite to both the magnetic and electric fields.

Two waves that are oscillating perpendicular to one another make up electromagnetic waves. One is represented by the oscillating magnetic field, while the other is represented by the oscillating electric field.

Therefore, Mechanical waves include things like pond ripples while electromagnetic waves, like light and radio signals, can move through space's vacuum.

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The 3 main parts of an electric circuit and their functions are:

Power source - Either electrical or battery, provides the power to the load.

Conducting path - The wires, carry the power from the power source to the load

Load - The device using the electricity such as a computer, light bulb, power saw, etc. More complicated circuits use switches, such as light switches, to control the flow of electricity.

What is an electric circuit?

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a) initial is 80.12°C and final temperature is 181.28°C b) Initial vapor mass is 2kg and liquid water is 0.7923 kg c) Heat transfer is 324.09 kJ

Law of thermodynamics:

ΔU = Q - W

Find initial state:

Specific volume of the mixture from the tables, which is [tex]0.9505 m^3/kg[/tex].

(a) Water has gone pressure condition (assumption). We can use the steam tables to find the specific volume of the water at the end of the process:

The specific volume of the water:  [tex]0.9505 m^3/kg.[/tex]

Volume increases by 20%, so final: [tex]1.1406 m^3/kg[/tex].

Pressure: 500 kPa.

Saturation temperature: 500 kPa is 153.97°C.

Therefore, water is superheated region.

Specific volume water is: [tex]1.1406 m^3/kg.[/tex]

Temperature 181.28°C.

Initial temperature: 80.12°C

Final temperature: 181.28°C.

(b) Water is in pressure under piston.

The specific volume of water: [tex]0.9505 m^3/kg[/tex].

The pressure increases to 500 kPa: superheated vapor region:

Vapor specific volume: [tex]1.4074 m^3/kg[/tex].

Use ideal gas law:

V = m * v

V = [tex]2 kg * 1.4074 m^3/kg = 2.8148 m^3[/tex]

Since volume increases by 20%, initial volume is[tex]2.3457 m^3[/tex].

Use ideal gas law:

P * V = m * R * T

m = P * V / (R * T)

where R: gas constant

At 500 kPa and 181.28°C, the value of R * T is 298.48 kJ/kg.

Substituting:

m = [tex]500 kPa * 2.3457 m^3 / (0.4615 kJ/kg-K * 298.48 K) = 7.5768 kg[/tex]

Therefore, the mass of liquid water when the piston first starts moving is 3 - 7.5768 = -4.5768 kg.

It cannot be negative so therefore, some liquid water is present.

The specific enthalpy of the liquid water is 334.28 kJ/kg.

For saturated vapor =  2779.2 kJ/kg.

Initial enthalpy: 334.28 kJ.

Final pressure: 500 kPa

Final enthalpy:

[tex]H_final = H_initial + m_evap * (h_vap - h_liq)[/tex]

Substituting:

[tex]334.28 kJ + m_evap * (2779.2 kJ/kg - 334.28 kJ/kg) = H_finalH_final = 334.28 kJ + m_evap * 2444.92 kJ/kg[/tex]

Vapor specific volume: 1.4074 m^3/kg

Total volume:

[tex]V_final = V_initial + m_evap * v_vap[/tex]

Substituting:

[tex]2.3457 m^3 + m_evap * 1.4074 m^3/kg = V_final[/tex]

As 20% increase,

1.2 * V_initial = V_final

Substitute:

[tex]1.2 * 2.3457 m^3 = 2.8148 m^3 + m_evap * 1.4074 m^3/kgm_evap = (1.2 * 2.3457 m^3 - 2.8148 m^3)[/tex]

[tex]m_evap = (1.2 * 2.3457 m^3 - 2.8148 m^3) / 1.4074 m^3/kg = 0.2077 kg[/tex]

Mass of liquid when piston moves: 0.7923 kg.

c) To find work, use Thermodynamics 1st law:

Q = deltaU + W

Assume it is adiabatic process.

[tex]deltaU = m_evap * (u_vap - u_liq)[/tex]

The specific internal energy of the liquid water is 334.28 kJ/kg.

For saturated: 2594.2 kJ/kg.

Substitute:

deltaU = 0.2077 kg * (2594.2 kJ/kg - 334.28 kJ/kg) = 468.69 kJ

Work done is:

W = -P * deltaV

The specific volume of the liquid water is 0.0010441 m^3/kg.

The specific volume is 1.4074 m^3/kg.

Substitute:

deltaV = m_evap * (v_vap - v_liq)

Substituting:

deltaV = [tex]0.2077 kg * (1.4074 m^3/kg - 0.0010441 m^3/kg) = 0.2892 m^3[/tex]

Work done:

W = [tex]-500 kPa * 0.2892 m^3 = -144.6 kJ[/tex]

Heat transfer:

Q = [tex]deltaU + W = 468.69 kJ - 144.6 kJ = 324.09 kJ[/tex]

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Indeed, through a method known as induction, a negatively-charged electroscope can be used to identify whether an object is positive or negative.

When a negatively-charged object is brought near the electroscope, the negative charges in the electroscope will be repelled by the negative charges on the object, and move to the far end of the electroscope. This will leave the near end of the electroscope with a net positive charge.

If the object is then brought into contact with the electroscope, some of the negative charges on the object will flow onto the electroscope, neutralizing the positive charge on the near end of the electroscope. However, some of the negative charges on the object will remain on the far end of the electroscope, leaving a net negative charge on the electroscope.

At this point, the electroscope has a negative charge that is due in part to the negative charges on the object. If the object is then moved away from the electroscope, the negative charges on the far end of the electroscope will be less strongly repelled, and will move back towards the near end of the electroscope. This will leave the near end of the electroscope with a net positive charge again, but the charge will be less than the original positive charge. The net charge on the electroscope after this process will depend on the initial charge on the object. If the object is positively charged, it will attract negative charges to the near end of the electroscope, leaving a net negative charge on the electroscope. If the object is negatively charged, it will repel negative charges to the far end of the electroscope, leaving a net negative charge on the electroscope.

Therefore, by observing the final charge on the electroscope after this process, one can determine whether the object is positive or negative.

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The time taken to reach the top speed of the 80 lm/ hours, starting from the rest is 55.17 seconds.

We have rate = 1.45 and final velocity is  80

The formula to find the acceleration is,

a = v - u/t

t = v-u/a

Where, u = initial velocity

v = final velocity

a = acceleration

t = time taken

so,

t = 80 - 0/1.45

= 80/1.45

= 55.17 seconds

Then v is the final haste, u is the original haste, a is the acceleration and t is the time interval during which a body is under acceleration.

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Answer:

no answer

Explanation:

Due to the fixed ends, a wave that is propagating up the string in one direction will reflect at the end and return inverted. The standing wave on the string is created by these two similar waves moving in the opposite direction.

When two waves with the same frequency and amplitude move in opposition to one another and interfere with one another, a standing wave result.

The string's basic frequency is the lowest resonance frequency (n=1). As integer multiples of the fundamental frequency, all higher frequencies are referred to as harmonics. The strings on all stringed musical instruments are fastened at both ends.

The n frequency is related to the fundamental frequency by the eq.

[tex]f_{n}=nf_{1}[/tex]

or

[tex]f_{1}=\frac{f_{n}}{n} \: \: \: \: \: \:\: \: \: \: \: \: \: \: (1)[/tex]

besides, we know that,

[tex]f_{n+1}=(n+1)f_{1}[/tex]

or

[tex]f_{1}=\frac{f_{n+1}}{n+1} \: \: \: \: \:\: \: \: \: \: \: (2)[/tex]

matching eq. (1) to (2),

[tex]\frac{f_{n}}{n}=\frac{f_{n+1}}{n+1}[/tex]

[tex]f_{n}(n+1)=f_{n+1}n[/tex]

isolating n from this eq.,

[tex]n = \frac{f_n}{f_n+1- f_n} = \frac{15}{20- 15}[/tex]

Once got the n value, just insert in eq. (1) so you can know the fundamental frequency,

[tex]f_1= \frac{f_n}{n} = \frac{15Hz}{3} = 5Hz[/tex]

Therefore, 5Hz is the fundamental frequency.

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The correct option is A, 1000N force must the extensor muscle in the upper arm exert on the lower arm to hold a 7.5 kg shot put.

Mm = mass of muscle

Me = mass of elbow

Mb = mass of ball

τ = Mad-> torque = mass * gravity * distance

-Mmad + Mead + M2ad = 0

Mm (9.80) (-.025m) + (2.8kg) (9.80) (.12m) + (7.5kg) (9.80) (.3m) = 0

Mm = 103.44kg

F = Ma = (103.44kg) (9.80)

= 1014N ~ 1000N

Force is a physical quantity that describes the interaction between objects or systems. It can be defined as the push or pull on an object resulting from the interaction between two or more bodies. The unit of force is the newton (N), and it is represented by the symbol F.

Force can change the motion of an object by accelerating,decelerating, or changing its direction. The magnitude and direction of a force are critical in determining how it will affect an object. There are several types of forces, including contact and non-contact forces. Contact forces involve direct physical contact between objects, while non-contact forces occur at a distance without physical contact.

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Complete Question: -

Assuming the lower arm has a mass of 2.8 kg and its CG is 12 cm from the elbow-joint pivot, how much force must the extensor muscle in the upper arm exert on the lower arm to hold a 7.5 kg shot put (Fig. 12-7)? ?????

A. 1000 N

B. 1500 N

C. 100 N

D. 500 N

E. 750N

Answer:

The distance between the two objects

The force is directly proportional to the product of the masses of the two objects and inversely proportional to the square of the distance between them

Explanation:

Answer: Object's speed affects the Kinetic Energy Quadratically.

Explanation: we have to know that Kinetic Energy is a scalar quantity, that is it doesn't have direction. Kinetic Energy of a moving body is totally dependent upon the body's mass and velocity of it.

The Kinetic Energy of a moving body having mass M and velocity V, then at a certain instant, the body's having Kinetic Energy is equal to:

Kinetic Energy (K.E.) = (MV²)/2

where:

M⇒ Body's mass

V ⇒Velocity of a moving body

It is so Clear that K.E. directly proportional to square of the body's velocity.

Therefore, Object's speed affects the Kinetic Energy Quadratically.

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The bone twists at an estimated angle of 0.000025 radians.

The scenario in the question involves applying a twisting torque of 2.75 nm on a bone that has a length of 35 cm and a radius of 2.25 cm. T = Fr, where T is the torque, F is the force, and r is the distance from the axis of rotation, gives the torque on the bone. The formula = T/(GJ), where is the angle of twist, T is the torque, G is the bone's shear modulus, and J is its polar moment of inertia, can be used to determine the angle through which the bone twists.

J = r4/2 can be used to get the polar moment of inertia, assuming the bone is a cylindrical rod.

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While a theory is a group of closely related phenomena or observations, hypothesis is a logical idea that can be tested.

How would you define hypothesis?

An assumption or concept is given as a hypothesis for the purpose of debating it and testing if it might be true. In the scientific process, the hypothesis is developed before any pertinent research—aside from a basic background review—has been conducted.

In addition to explaining existing facts, a theory also enables scientists to forecast what observations they should expect to see if the theory is correct. Scientific hypotheses can be tested. A theory should be consistent with new data.

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(A) This box glides, then slides up to 4.74 m before stopping . (B) The friction coefficient at the point of halting is 0.537. (C) The box would have slid 101.25 meters before coming to a stop if the coefficient of friction had stayed unchanged.

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

Net work = ΔK.E.

We can break the motion of the box into two parts: before and after the rough section. Before the rough section, the box is moving with a constant velocity, so the net work done on it is zero. After the rough section, the box slows down and comes to a stop, so the net work done on it is equal to its initial kinetic energy:

Net work = -K.E.

(A) To find how far the box slides before stopping, we need to find the distance over which the box is acted upon by the increasing frictional force. Let's call this distance x.

W (friction) = ∫₀ˣ F f(x') dx'

here,

F f(x') is frictional force at a distance x' from point P.

Since the coefficient of friction increases linearly with distance, we can express F f(x') as:

F f(x') = μ₀ + (μ f - μ₀) * (x'/x f)

here,

μ₀ is initial coefficient of friction at point P,

μ f is final coefficient of friction at distance x f = 12.5 m, and

x' ranges from 0 to x.

Reserving expression of F f(x') into the integral for W (friction):-

W (friction) = μ₀ * x + (μ f - μ₀) * (x²/2x f)

Express initial kinetic energy as:-

K.E. = (1/2) * m * v²

here,

m is mass of the box and

v is its initial velocity of 4.50 m/s.

Setting the net work equal to the change in kinetic energy:-

= μ₀ * x + (μ f - μ₀) * (x²/2x f)

= (1/2) * m * v²

= x² - 2x f * [(μ f - μ₀)/μ₀] * x - 2x f * (K.E./(μ₀ * m))  

= 0

Putting given values of μ₀, μ f, x f, m, and v:-

x = 4.74 m

Therefore, the box slides for a distance of 4.74 m before coming to a stop.

(B) To find the coefficient of friction at the stopping point, we can use the same equation we derived earlier for W (friction) and solve for μ f:-

= W (friction)

= μ₀ * x + (μ f - μ₀) * (x²/2x f)

= -K.E.

= μ f

= (2 * K.E. + μ₀ * x * (μ f - μ₀)/x f) / x²

Putting given values of K.E., μ₀, μ f, x f, and x:-

μ f = 0.537

Therefore, the coefficient of friction at the stopping point is 0.537.

(C)  If the coefficient of friction remained constant at μ₀ = 0.1000, then we can simplify the equation we derived for x by setting μ f = μ₀:

= μ₀ * x + (μ₀ - μ₀) * (x²/2x f)

= (1/2) * m * v²

Simplifying the second term:-

μ₀ * x = (1/2) * m * v²

Solving for x:-

x = (m * v²) / [2 * μ₀ * W (friction)]

here,

W (friction) is work done by friction.

To find W (friction), we can integrate the frictional force over the entire distance traveled by the box:-

= W (friction)

= ∫₀ˣ F f(x') dx'

here,

F f(x') is constant frictional force of μ₀.

Reserving this expression for W friction into the equation for x:-

x = (m * v²) / (2 * μ₀ * F f * x)

here,

F f is constant frictional force of μ₀.

Simplifying:-

x = (m * v²) / (2 * μ₀ * F f)

Putting given values of m, v, μ₀, and F f:-

x = 101.25 m

Therefore, if the coefficient of friction had remained constant at μ₀ = 0.1000, the box would have slid for a distance of 101.25 m before coming to a stop.

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Answer:Hence, if a particle is moving with constant speed such that its acceleration with constant magnitude will be always perpendicular to its velocity, then it must be going in a circular path.

Explanation: