A small ball is attached to one end of a spring that has an un- strained length of 0.200 m. The spring is held by the other end, and the ball is whirled around in a horizontal circle at a speed of 3.00 m/s. The spring remains nearly parallel to the ground during the motion and is observed to stretch by 0.010 m. By how much would the spring stretch if it were attached to the ceiling and the ball allowed to hang straight down, motionless?

Answer:Upward

Explanation:

To hang the object in the halfway we need to direct the Electric field toward upward direction

Suppose object has mass m and charge q

If electric field is Pointing towards upward direction then the force experienced by charged particle is in the upward direction which will be balanced by weight of particle(pointing downward) such that

qE=mg

where q=charge of particle

E=Electric field

m=mass of particle

g=acceleration due to gravity

Answer:

A) pbin = 1.535 Kgm/s (+)

B) pbf = 1.696 Kgm/s (-)

C) Δp = 3.3925 Kgm/s

D) Δvr = 10.249 m/s

Explanation:

Given

Mass of the ball: m = 57.5 g = 0.0575 Kg

Initial speed of the ball: vbi = 26.7 m/s

Mass of the racket: M = 331 g = 0.331 Kg

Final speed of the ball: vbf = 29.5 m/s

A) We use the formula

pbin = m*vbi = 0.0575 Kg*26.7 m/s = 1.535 Kgm/s (+)

B) pbf = m*vbf = 0.0575 Kg*29.5 m/s = 1.696 Kgm/s (-)

C) We use the equation

Δp = pbf - pbin = 1.696 Kgm/s - (-1.535 Kgm/s) = 3.3925 Kgm/s

D) Knowing that

Δp = 3.3925 Kgm/s

we can say that

Δp = M*Δvr

⇒  Δvr = Δp / M

⇒  Δvr = 3.3925 Kgm/s / 0.331 Kg

⇒  Δvr = 10.249 m/s

Technician B is correct. A voltmeter should be connected in parallel with the component or section of a circuit that is being tested, while an ammeter should be connected in series.

Technician B is correct when it comes to assessing the method of connecting a voltmeter. A voltmeter is designed to measure the voltage across elements of a circuit and is placed in parallel with the component or section of the circuit that is being tested. In this configuration, the voltmeter receives the full voltage.

On the other hand, Technician A is confusing a voltmeter with an ammeter. An ammeter, which measures the current flowing through a given branch of an electric circuit, should be connected in series in order to get a measure of the full current passing through that branch. The ammeter has a small resistance to limit its impact on the circuit.

By contrast, a voltmeter has to have a large resistance as it is connected in parallel and thus should have minimal impact on the circuit being tested.

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Answer:

Kinetic

Explanation:

Final answer:

The weight of the closed system of propane is 173.04 N. The mass-based specific volume is 25 m³/kmol and 0.567 m³/kg.

Explanation:

To answer the student's question about the weight and the mass-based specific volume of a closed system of propane:

Propane has a molar mass of approximately 44.1 kg/kmol. Since the system has 0.4 kmol of propane, the mass (m) is:

m = 0.4 kmol × 44.1 kg/kmol = 17.64 kg

Now, the weight (W) in Newtons can be calculated using the gravitational acceleration (g = 9.81 m/s2):

W = m × g = 17.64 kg × 9.81 m/s2 = 173.04 N

The specific volume (vsp) in m3/kmol is simply the volume divided by the number of kmols:

vsp = V / n = 10 m3 / 0.4 kmol = 25 m3/kmol

For m3/kg, we divide the specific volume by the molar mass of propane:

vsp = 25 m3/kmol ÷ 44.1 kg/kmol = 0.567 m3/kg

Answer:

Extension is 12.14m

Explanation:

mass of ornament=42g,weight of ornament=mg=42*9.8=411.6

Force constant(k)= 33.9N/m

F=ke

e=F/k

F is force, e is extension

F=weight of ornament

e=411.6/33.9

e=12.14m

Answer:

3894531 coulombs

Explanation:

1 hour = 3600 seconds

Let g = 10m/s2

The distance that the car travel at 25 m/s over an hour is

s = 25 * 3600 = 90000 m

The total mechanical energy of the car is the sum of its kinetic energy to reach 25 m/s, its potential energy to climb up 200m high hill and it work to travel a distance of s = 90000m with F = 500 N force:

[tex] \sum E = E_k + E_p + E_W[/tex]

[tex]\sum E = mv^2/2 + mgh + Fs[/tex]

[tex]\sum E = 750*25^2/2 + 750*10*200 + 500*90000 = 46734375 J[/tex]

This energy is drawn from the battery over an hour (3600 seconds), so its power must be

[tex]P = E / t = 46734375/3600 = 12982 W[/tex]

The system is 12V so its current is

[tex]I = P/U = 12982 / 12 = 1081.8 A[/tex] or 1081.8 Coulombs/s

The the total charge it needs for 1 hour (3600 s) is

C = 1081.8 * 3600 = 3894531 coulombs

The quantity of charge the batteries must be able to move is equal to 3.9 μC.

Given the following data:

Scientific data:

To find the quantity of charge the batteries must be able to move:

In this scenario, we would calculate the distance traveled and the total energy that is possessed by this battery-operated car.

Mathematically, the distance covered by an object is given by this formula:

[tex]Distance = speed \times time\\ \\ Distance = 25 \times 3600[/tex]

Distance = 90,000 meters.

[tex]E = mgh + \frac{1}{2} mv^2 + Fd\\ \\ E=[750\times 9.8 \times 200] + \frac{1}{2} \times 750 \times 25^2 + [500 \times 90000]\\ \\ E=1470000+234375+45000000[/tex]

Total energy = 46,704,375 Joules.

Next, we would calculate the power consumed by this battery-operated car:

[tex]Power = \frac{Energy}{time}\\ \\ Power = \frac{46,704,375}{3600} [/tex]

Power = 12,973.44 Watts.

Also, we would calculate the current:

[tex]Current = \frac{power}{voltage} \\ \\ Current = \frac{12,973.44}{12}[/tex]

Current = 1,081.12 Amperes.

Now, we can calculate the quantity of charge the batteries must be able to move:

[tex]Q = current \times time\\ \\ Q = 1081.12 \times 3600[/tex]

Q = 3,892,032 Coulombs.

Note: 1 μC = [tex]1 \times 10^6 \;C[/tex]

Q = 3.9 μC

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Answer:

47628 kg/hr

Explanation:

Total area = 18*20 =360 ft^(2)

no .of snowflakes per minute = 1050*360 =378000

mass of snowflakes per minute = 378000*2.1*10^(-3) =793.8 kg/min

mass accumulated per hour = 793.8kg/min * 60min/hr =47628 kg/hr

The water was moving out from the bottom at a velocity of 29 m/s

Applying Bernoulli's equation:

P + ρgh + (1/2)ρv² = constant

P is pressure, g is acceleration due to gravity, h is height, v is velocity, ρ is density.

At top:

P = 3.20 atm = 3.20 * 101300 Pa, ρ = 1025 kg/m³, v = 0, g = 9.81 m/s, h = 10.7 m, hence:

P + ρgh + (1/2)ρv² = ( 3.20 * 101300) + (1025 * 9.81 * 10.7) + 0

At bottom:

h = 0, P = 0

P + ρgh + (1/2)ρv² = 0 + 0 + (1/2* 1025)v²

Equating top and bottom:

( 3.20 * 101300) + (1025 * 9.81 * 10.7) = (1/2* 1025)v²

v = 29 m/s

The water was moving out from the bottom at a velocity of 29 m/s

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To calculate the speed of the water flowing out of the tank, we can use Bernoulli's equation. Plugging in the given values and solving for the final velocity, we can determine the speed of the water.

To determine how fast the water is moving, we can use Bernoulli's equation which relates the pressure, velocity, and height of a flowing fluid.

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

Where:

Plugging in the values, we can solve for v2:

3.20 atm + 0 + (1000 kg/m3)(9.8 m/s2)(10.7 m) = 1 atm + 1/2(1000 kg/m3)v22 + (1000 kg/m3)(9.8 m/s2)(0 m)

Simplifying the equation, we can solve for v2:

v22 = 2[(3.20 atm - 1 atm) / (1000 kg/m3)]

v2 = √[2(3.20 atm - 1 atm) / (1000 kg/m3)]

Using the equation, we can calculate v2 to find the speed at which the water is flowing.

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Answer:

R=19.5m

[tex]\theta[/tex] = 4.65° S of W

Explanation:

Refer the attached fig.

displacement  of the x and y components

x-component displacement is ([tex]R_{x}[/tex]) = [tex]A_{x}+B_{x}[/tex]

= A [tex]\sin[/tex](20°) + B [tex]\cos[/tex](40°)

= -12.0[tex]\sin[/tex](20°) + 20.0[tex]\cos[/tex](40°)

= -19.425m

x-component displacement is ([tex]R_{y}[/tex]) = [tex]A_{y}+ B_{y}[/tex]

=  A [tex]\cos[/tex](20°) - B [tex]\sin[/tex](40°)

= 12.0[tex]\cos[/tex](20°) - 20.0[tex]\sin[/tex](40°)

= -1.579

resultant displacement

∴

R = [tex]\sqrt{R_{x}^{2} +R_{y}^{2} } }[/tex]

=[tex]\sqrt{(-19.425)^{2}+(-1.579)^{2} }[/tex]

=19.5m

[tex]\theta[/tex] = [tex]\tan^{-1}\left | \frac{R_{x}}{R_{y}} \right |[/tex]

[tex]\theta[/tex] = [tex]\tan^{-1}\left | \frac{1.579}{19.425} \right |[/tex]

[tex]\theta[/tex] = 4.65° S of W

This problem involves finding the result of two vector displacements using vector addition. Using the Pythagorean theorem, the magnitude of the overall displacement can be determined. The direction of the displacement can be calculated using the tangent formula.

This is a vector addition problem in physics. We can solve it using Pythagorean theorem for finding the magnitude and tangent formula for finding the direction.

For the first leg of the journey(A), you are going 20° W of N, or 70° clockwise from the x-axis. For the magnitude of this vector: Ax = 12.0m cos(70°) and Ay = 12.0m sin(70°).

For the second leg of the journey(B), you are going 40° S of W, or 130° clockwise from the x-axis. For the magnitude of this vector: Bx = 20.0m cos(130°) and By = 20.0m sin(130°).

Summing the x and y components gives: Rx = Ax + Bx and Ry = Ay + By. The magnitude R can then be found with Pythagorean theorem formula: R = sqrt(Rx² + Ry²).

The direction can be found using tangent formula: θ = atan(Ry / Rx). As Rx is negative and Ry is positive, θ will fall in the second quadrant.

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Answer:

32.67255 W

Explanation:

F = Force  = 22 N

r = Radius of crank = 0.26 m

s = Displacement = [tex]2\pi r[/tex]

[tex]\theta[/tex] = Angle = 0

t = Time taken = 1.1 s

Work done is given by

[tex]W=Fscos\theta\\\Rightarrow W=22\times 2\pi\times 0.26\times cos0\\\Rightarrow W=35.93981[/tex]

Power is given by

[tex]P=\dfrac{W}{t}\\\Rightarrow P=\dfrac{35.93981}{1.1}\\\Rightarrow P=32.67255\ W[/tex]

The average power being expended is 32.67255 W

Answer:

Yes, it is reckless. This is because it is the responsibility of the pilot to make sure that the direction of the propeller blast is away from people or other aircraft and in a safe direction.

Explanation:

Yes, it is reckless to let the propeller blast face people and other aircraft. This is because it is the responsibility of the pilot to make sure that the direction of the propeller blast is away from people or other aircraft and in a safe direction. People and other aircraft can be injured by the debris and the rocks that are scattered by the engine of the aircraft.

Final answer:

Causing debris and propeller blast to endanger others can be deemed reckless operation of an aircraft, akin to dangerous driving, and can be punishable by aviation law.

Explanation:

The operation of an aircraft in a manner that causes debris, rocks, and propeller blast to be directed towards people or other aircraft can indeed be considered reckless operation. For perspective, consider that landing an aircraft is described as an ultimate challenge and must ensure safety as if it were a casual drive to a golf course. Pilots are responsible for maintaining control of their aircraft and ensuring the safety of both passengers and bystandiles at all times, much like drivers on the road.

Reckless operation of an aircraft can endanger lives and property, and is typically prohibited by aviation law and regulations. If a pilot operates an aircraft without regard to the potential harm it could cause others, this could be construed as reckless. The key is the intent and awareness of the pilot; if they are knowingly causing potential harm, it is indeed reckless.

Answer:

C. is responsible for sliding friction and contact forces

Answer:

Option D.

Explanation:

The correct answer is Option D.

The shrinkage of the bridge material is because of thermal contraction.

Thermal contraction and thermal expansion are the phenomena of the bridge material which takes place due to the change in temperature of the atmosphere.

When the temperature of the surrounding increases expansion of the bridge material takes place and when temperature decreases the contraction of the material takes place.

This phenomenon sometimes damages the structure because due to continuous expansion and contraction of materials strength of the bridge decreases.

Thermal expansion joints in bridges prevent material breakage due to temperature changes.

Thermal expansion joints are used in bridges to allow for the changing length of the bridge due to temperature fluctuations. Without these joints, the bridge material would break due to thermal stress caused by expansion and contraction.

Answer: it is very difficult to remove the inner electrons from an atom because of the strong electrostatic force of attraction by the nucleus. In contrast, it is very easy to remove the outer electrons from an atom because the electrostatic force of attraction from the nucleus is not strong enough to hold the outer electrons hence it is removed.

Answer:

1. What we calculate when we calculate expected frequency is the predicted frequency that can be obtained from an experiment whose outcome is expected to be true.

2. The reason for using expected frequency is because it is a tool used for doing complex probability calculations and predictions. Also, it serves as a working tool that can be used in place of the observed frequency,when the observed frequency is not available.

3. In probability, expected frequency tell us the possible outcomes of an event.

Say, for example rolling of a fair die.

The probability of getting 1,2....,6 as the outcome is 1/6 but that doesn't mean that when we roll a fair die 6 times, we'll get outcomes of 1-6.

So, the essence of expected frequency is to tell us what to expect in an event (this may or may not turn out to be true).

Final answer:

Expected frequencies estimate how often an event should occur based on a probability distribution, providing a basis for comparing actual outcomes in a data set to determine consistency with expected probabilities.

Explanation:

When we calculate expected frequencies, we are determining how often we anticipate an event will occur based on a probability distribution. To find the expected frequency, one typically multiplies the expected percentage by the total number of observations or the sample size. For instance, if an event has a 10% chance of occurring out of 600 trials, the expected frequency would be 0.10 * 600 = 60.

The reason for calculating expected frequencies in this manner is to provide a basis for comparison with the actual frequencies that occur in a data set. This comparison allows us to determine if the outcomes observed are consistent with the expected probabilities, which can be particularly useful in statistical hypothesis testing, such as the chi-square test.

Expected frequencies give us insight into what we might predict to occur over a long period of trials or observations. They are theoretical estimates that can be appraised against actual outcomes to understand the probability distribution of a given scenario.

Answer:

Flight path angle= 15.12°, maximum range= 5.29× 10*6 km

Explanation:

u= 7200m/s, H= 180km= 180000m

Recall that

Maximum height, H= (u*2sin*2∆)/2g

180000= (7200×7200sin*2∆)/2×9.8

(18000×2×98)/7200×7200= sin*2∆

Sin∆= 0.2609

∆= 15.12°

Maximum range, R= u*2/g

(7200×7200)/9.8

= 5289795.92km

= 5.29× 10*6 km

Answer:

The answer to your question is Mass = 41230.7 g  or 41.23 kg    

Explanation:

Data

density = 0.737 g/ml

mass = ?

volume = 14.9 gal

1 gal = 3.78 l

Process

1.- Convert gallons to liters

                                   1 gal ---------------- 3.78 l

                                  14.8 gal -------------  x

                                    x = 55.94 l

2.- Convert liters to milliliters

                                  1 l -------------------  1000 ml

                              55.94 l ---------------   x

                                   x = (55.94 x 1000) / 1

                                   x = 55944 ml

3.- Calculate the mass

Formula

density = [tex]\frac{mass}{volume}[/tex]

Solve for mass

Mass = density x volume

Substitution

Mass = 0.737 x 55944

Simplification and result

Mass = 41230.7 g  or  41.23 kg    

Answer:

1328.7032 kg

Explanation:

P = Pressure = 112 kPa

T = Temperature = 285 K

V = Volume = 7023 m³

R = Gas constant = 8.314 J/mol K

From the ideal gas law we have

[tex]PV=nRT\\\Rightarrow n=\dfrac{PV}{RT}\\\Rightarrow n=\dfrac{112000\times 7023}{8.314\times 285}\\\Rightarrow n=331960.04203\ moles[/tex]

The mass of gas is given by

[tex]m=n\times MW_{He}\\\Rightarrow m=331960.04203\times 4.0026\times 10^{-3}\\\Rightarrow m=1328.70326\ kg[/tex]

The mass of helium in the blimp is 1328.7032 kg

By applying the ideal gas law, we determined the number of moles of helium in the blimp. By using the molecular weight of helium, we calculated that the mass of helium in the blimp is 12000 kg.

We can use the ideal gas law to solve this problem. The ideal gas law is represented by the formula PV = nRT, where P represents the absolute pressure, V is the volume, n is the number of moles, R is the ideal gas constant and T is the temperature.

Based on the given values, we have P = 112 kPa, V = 7023 m^3, R = 8.31 (J/mol.K), and T = 285 K. We need to calculate n (number of moles) first.

Transforming the formula, we get n = PV/RT. Substituting the given values, n = (112,000 Pa * 7023 m^3) / (8.31 J / (mol. K) * 285 K) which gives us n = 3.0 * 10^6 moles.

Then, knowing that the molecular weight of helium is approximately 4 g/mol, we can multiply the number of moles by the molecular weight to find the mass. So, the mass is m = n * Molecular weight = 3.0 * 10^6 moles * 4 g/mol = 12 * 10^6 g or 12000 kg.

Therefore, the mass of the helium in the blimp is 12000 kg.

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Final answer:

The function for the power output of a solar panel in Ann Arbor on a summer day is f(t) = 10sin(πt/14), and on a winter day, the function is g(t) = 10sin(π(t+2)/9). The power output is greatest at 1 pm during summer with 10 watts. Graphing reveals the variation of power throughout the day.

Explanation:

Writing a Function for Solar Panel Power Output:

To answer part (a), we start by incorporating θ = πt/14 into the original power output equation P = 10sinθ, leading to a new function f(t) = 10sin(πt/14).

For part (b), graphing f(t) between 0 ≤ t ≤ 14 will show a sinusoidal curve that represents the power output throughout the day.

For part (c), the power output is greatest when θ = π/2, which occurs when t = 7 (1 pm), and the power output at this time is P = 10 watts.

Regarding part (d), for a typical winter day with sunlight from 8 am to 5 pm, the angle for t hours after 8 am needs to be adjusted. We can denote it as θ1 = π(t+2)/9. Thus, a new function g(t) = 10sin(π(t+2)/9) represents the power output on a typical winter day.

Explanation:

The Vehicle Assembly Building at the Kennedy Space Center in Florida has a volume of 3,666,500 m³.

Volume = 3,666,500 m³

1 m³ = 1000 L

So volume = 3,666,500 x 1000 = 3666500000 L

Volume of vehicle Assembly Building at the Kennedy Space Center in Florida = 3666500000 L

Volume of vehicle Assembly Building at the Kennedy Space Center in Florida = 3.67 x 10⁹ L

The volume in liters and in exponential notation is [tex]3.67\times10^{23}[/tex].

To convert the given volume to liters it is necessary to use the following relation of values:

                                                    [tex]dm^{3} = L\\ m ^{3} = 1000L[/tex]

Therefore, the following calculation must be performed:

                                   [tex]3,666,500m^{3} \times 1000= xL[/tex]

                                        [tex]x = 3,666,500,000 L[/tex]

Now, to convert to scientific notation, leave the number different from the power of 10 between 1 and 10, so that:

                                  [tex]3,666,500,000 = 3.67 \times 10^{9} L[/tex]

So, the volume in liters and in exponential notation is [tex]3.67\times10^{23}[/tex]

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4.472 tonnes of coal burned

Explanation:

The electricity consumption for a typical house in the United States for a year is approximated as 11,000 kWh.

Given that a typical coal-fired power plant produces 2,460 kWh of electricity  per ton of coal burned then;

2460 kWh = 1 ton of coal burned

11,000 kWh =?

cross-product

(11,000*1)÷ 2460 = 4.472 tonnes of coal burned

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The amount of coal required to be  burned in a typical coal-fired power plant to provide the electricity for a typical house in the United States for a year will be 4.472 tonnes.

Given data:

The electricity produced per ton of coal burn is, E = 2,460 kWh.

The given problem is based on the Energy consumption. The energy consumption is the energy utilized to perform various actions such as manufacturing, welding, inhabiting, and many more.

The electricity consumption for a typical house in the United States for a year is approximated as 11,000 kWh.

And as per the given question,

2460 kWh = 1 ton of coal burned

So for 11,000 kWh, the amount of coal need to be burned is,

= 11000/2460

= 4.472 tonnes of coal  

Thus, we can conclude that amount of coal required to be  burned in a typical coal-fired power plant to provide the electricity for a typical house in the United States for a year will be 4.472 tonnes.

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Answer:

(a) [tex]\omega=2.09*10^{5}\frac{rad}{s}[/tex]

(b) [tex]A_{max}=0.033m[/tex]

Explanation:

(a) The angular frequency is defined as:

[tex]\omega=2\pi f[/tex]

Here f is the frequency of the particle, which is inversely proportional to its period:

[tex]f=\frac{1}{T}[/tex]

Replacing, we have:

[tex]\omega=\frac{2\pi}{T}\\\omega=\frac{2\pi}{3*10^{-5}s}\\\omega=2.09*10^{5}\frac{rad}{s}[/tex]

(b) The maximum displacement is given by:

[tex]A_{max}=\frac{v_{max}}{\omega}\\A_{max}=\frac{7*10^3\frac{m}{s}}{2.09*10^5\frac{rad}{s}}\\A_{max}=0.033m[/tex]

Answer:

The maximum capacity of such cup is

2 × 3.142R^3/9×squareroot3

Explanation:

From the diagram in the attachment,

h= height of the cone, r= radius of the cone,R= radius of the original circular piece of paper

Using pythagorean theorem, we get

r^2+h^2=R^2 => r^2=R^2-h^2

Formular for the volume of a cone is given as V=1/3×(3.142)^2h

Substituting for r^2 when R is a constant

V=1/3(3.142)(R^2-h^2)h

V=1/3(3.142)(R^2-h^3)

V'=0=1/3(3.142)R^2-3h^2)

0=R^2-3h^2

h^2=R^2/3 => h=R/squareroot of 3

V"=1/3(3.142)(0-6)=-2×3.142×h

Negative shows that v is concave

V(R/squareroot of 3)=1/3(3.142)[R^2×R/squareroot 0f 3-(R/squareroot of 3)^3

V=2×3.142×R^3/9× squareroot of 3

To find the maximum capacity of a cone-shaped drinking cup made from a circular piece of paper of radius R, you need to use differential calculus to maximize the volume of the cone, derived from the formula V = (1/3)π*r²h, where r = R*(1- θ/2π) is the radius of the cone and h = sqrt[R² - r²] is its height.

The subject of this question is finding the maximum capacity of a cone-shaped drinking cup made from a circular piece of paper of radius R. To find the maximum capacity or volume of such a cup, we need to formulate the problem in terms of mathematical geometry. The volume of a cone, which in this case represents our drinking cup, is given by the formula V = (1/3)π*r²h, where r is the radius of the base circle (which will also be our R), and h is the height of the cone.

The cone is created by cutting out a sector and joining edges CA and CB on the leftover part of the circular paper. Let θ denote the angle of the sector that is cut out (in radians). Since we have subtracted this sector, the circumference of the top circle of the cone is now 2πR*(1-(θ/2π)) = 2πR*(1 - θ/2π) = 2πR - R*θ.

Setting this equal to the circumference of the top of the cone (2πR - R*θ = 2πr, where r is now the radius of the cone), we can solve for r in terms of R and θ: r = R*(1- θ/2π).

The slant height of the cone will be equal to the radius of the initial circular piece of paper, R. Using Pythagorean theorem, the height h of the cone can be expressed as h = sqrt[(R² - r²)] = sqrt[R²(1 - (1- θ/2π)²)].

Substituting the expressions for r and h into the cone volume formula, we get V = (1/3)π[R²(1- θ/2π)]²*sqrt[R²(1 - (1- θ/2π)²)]

This is an expression for the volume V of the cone-shaped drinking cup in terms of R and θ. The task now is to utilize differential calculus to find the maximum of this function V(θ), i.e., we need to find the value of θ (between 0 and 2π) that maximizes the volume V.

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Answer:

the Architect should use {!$FieldType.lead.accessible} expression within the Visualforce page.

Explanation:

Visualforce is a framework that allows developers to build complex,  user friendly interfaces that can be hosted primarily on the Lightning Platform

Controllers provide access to the data that should be displayed in a page, and can modify component behavior. a number of standard controllers are provided by The Lightning platform that contain  functionality and logic that which are used for standard Salesforce pages

The Architect should Use the expression {!$FieldType.lead.accessible}  within the Visualforce page.

To enforce Field-Level Security on a custom Visualforce page using the Standard Lead Controller, the Architect should define the page with the 'standardController' attribute set to 'Lead' and the 'extensions' attribute set to the appropriate Lead Controller extension. The 'with sharing' keyword should be used in the extension to enforce field-level security.

To ensure Field-Level Security is enforced on a custom Visualforce page using the Standard Lead Controller, the Architect should define the Visualforce page with the 'standardController' attribute set to 'Lead' and the 'extensions' attribute set to the appropriate Lead Controller extension. Within the extension, the Architect can use the 'with sharing' keyword to enforce field-level security.

For example:

The 'with sharing' keyword ensures that the Visualforce page adheres to the Org-wide default settings and user's record-level access. This means that any field-level security restrictions defined for the Lead object will be enforced on the page.

Final answer:

The net charge within the cylinder, given the electric flux of -50 Nm²/C² and the permittivity of free space, is -44 picocoulombs (pC).

Explanation:

The question asks to determine the net charge within a cylindrical surface given the total electric flux and the electrical permittivity of free space. According to Gauss's law, the electric flux (φ) through a closed surface is equal to the net charge (Q) enclosed by the surface divided by the permittivity of free space (ε0). The formula is φ = Q / ε0. Given the electric flux is -50 Nm2/C2 and ε0 = 8.85 x 10-12 C2/Nm2, we can solve for Q:

Q = φ x ε0

Q = (-50 Nm2/C2)(8.85 x 10-12 C2/Nm2)

Q = -4.425 x 10-10 C

To convert this to picocoulombs (pC), we multiply by 1012 pC/C:

Q = -4.425 x 10-10 C x 1012 pC/C = -442.5 pC

The closest answer to -442.5 pC is -44 pC, so the correct answer is (d) -44.

Answer:

c. 3

Explanation:

Given:

Since AMA i.e. actual mechanical advantage is defined as the ratio of the output force to the input force on a simple machine. This takes into account the losses occurred.

Now, mathematically:

[tex]AMA=\frac{F_o}{F_i}[/tex]

[tex]AMA=\frac{30}{10}[/tex]

[tex]AMA=3[/tex]

Answer:

kuh

Explanation:

Answer:

The uncertainty in the location that must be tolerated is [tex]1.163 * 10^{-5} m[/tex]

Explanation:

From the uncertainty Principle,

Δ[tex]_{y}[/tex] Δ[tex]_{p}[/tex] [tex]= \frac{h}{2\pi }[/tex]

The momentum P[tex]_{y}[/tex] = (mass of electron)(speed of electron)

                                = [tex](9.109 * 10^{-31}kg)(995 * 10^{3} m/s)[/tex]

                                = [tex]9.0638 * 10^{-25}kgm/s[/tex]

If the uncertainty is reduced to a 0.0010%, then momentum

                              = [tex]9.068 * 10^{-30}kgm/s[/tex]

Thus the uncertainty in the position would be:

                              Δ[tex]_{y} = \frac{h}{2\pi } * \frac{1}{9.068 * 10^{-30} }[/tex]

                              Δ[tex]_{y} \geq 1.163 * 10^{-5}m[/tex]

The uncertainty in location of the electron can be calculated using Heisenberg's Uncertainty Principle, which stipulates a product of the uncertainties in position and momentum to be at least half of reduced Planck's constant. Uncertainty in momentum is given as 0.0010% of the electron's momentum. Careful consideration of quantities and units is essential.

This question revolves around Heisenberg's Uncertainty Principle. According to this principle, the product of the uncertainty in the position, denoted by Δx, and the uncertainty in momentum, denoted by Δp, has a minimum value dictated by Planck's constant, h. Mathematically, it is represented as Δx * Δp ≥ ħ/2, where ħ is the reduced Planck's constant, ħ=h/(2π). The speed of the electron, v, is related to its momentum, p, through mass, m, as p=mv. Therefore, the uncertainty in momentum, Δp, can be calculated as 0.0010% of p. Once we find Δp, using the Uncertainty Principle, we can find the minimum required uncertainty in the electron's location, Δx. In such problems, considering mass and speed of electron and appropriate units for constants are crucial for

correct computation

. Make sure to consider them appropriately.

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Answer:

2.03873 s

70.38735 m

4.07747 seconds

5.82688 seconds

27.37482 m from the ground

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-20}{-9.81}\\\Rightarrow t=2.03873\ s[/tex]

time needed for the ball to reach its maximum height is 2.03873 s

The time taken to go up and the time taken to reach the point from where it was thrown is the same.

So, time needed for the ball to return to the height from which it was thrown is 2.03873+2.03873 = 4.07747 seconds

[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-20^2}{2\times -9.81}\\\Rightarrow s=20.38735\ m[/tex]

The maximum height the ball will reach is 50+20.38735 = 70.38735 m

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 70.38735=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{70.38735\times 2}{9.81}}\\\Rightarrow t=3.78815\ s[/tex]

Time needed to reach the ground is 2.03873+3.78815 = 5.82688 seconds

The time from the maximum height that is required is 5-2.03873 = 2.96127 seconds

[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 2.96127^2\\\Rightarrow s=43.01253\ m[/tex]

The ball will be 70.38735-43.01253 = 27.37482 m from the ground

The problem is solved using kinematic equations. The ball reaches its maximum height in approximately 2.04 s, with the maximum height reaching about 70.4 m. It returns to the original height in around 4.08 s with a velocity at -39.79 m/s, reaches the ground in about 5.82 s, and at t = 5 s, its position is approximately 49.4 m with velocity of around -29 m/s.

The calculations are based on the physics of kinematics, specifically related to the motion under constant acceleration which in this context is gravity (designated 'g'), typically -9.8 m/s².

(a) You can use the equation v = u + gt to solve for time (t). Since the velocity (v) at the maximum height is 0, initial velocity (u) is 20 m/s and g is -9.8m/s², the equation becomes 0 = 20 - 9.8t. Solving for t, we get t = 20/9.8 ≈ 2.04 s.  

(b) The maximum height can be found using the equation s = ut + 0.5gt². Substituting the known values we get the maximum height to be about 70.4m

(c) The time taken for the ball to return to its original height will be double the time to reach maximum height, so it's 2*2.04s = 4.08s. The velocity is simply gt, yields -39.79m/s (negative because it's moving downwards).

(d)We can utilize the equation s = ut + 0.5gt² to find out the time to hit the ground. The total distance covered is 50m + 20.4m = 70.4m. Solving the equation we get t = 3.78s (above the throwing point) and to find the total time on air, add t = 2.04s and value calculated you get t = 5.82 s.  

(e)Using equations of motion, we can find out that at t=5s, the position of the ball is 49.4 m above the ground and its velocity is -29m/s.

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Final answer:

To find their rates of travel, represent Eric's rate as E and Stacy's as 5E. Since together they cover 336 miles in 4 hours, the equation 4E + 20E = 336 leads to Eric's rate of 14 mph and Stacy's rate of 70 mph.

Explanation:

Stacy and Eric travel in opposite directions and collectively cover 336 miles in 4 hours. Since Stacy travels 5 times as fast as Eric, we can set up equations to represent their speeds and use these to find the rates of travel for both.

Let E represent Eric's rate of travel. Then 5E represents Stacy's rate of travel. Since distance equals rate times time, we can set up the equation E x 4 hours + 5E x 4 hours = 336 miles.

Combining terms, we get 4E + 20E = 336, which simplifies to 24E = 336. Dividing both sides by 24 gives us E = 14. So, Eric travels at 14 miles per hour, and Stacy travels at 5 times that rate, which is 70 miles per hour.