The same couple moment of 160 lb⋅ftlb⋅ft is required to open the hatch door, independent of where the couple's forces are applied. If the forces are applied at points C and D, what force magnitude FFF is required to open the hatch door?

Answer:

0.24825 N

0.0000238701923077 kgm²

Explanation:

m = Mass of yo yo = 0.075 kg

a = Acceleration = 6.5 m/s²

g = Acceleration due to gravity = 9.81 m/s²

The net force is given by

[tex]F_n=mg-T[/tex]

[tex]\Rightarrow T=mg-ma[/tex]

[tex]\Rightarrow T=m(g-a)[/tex]

[tex]\Rightarrow T=0.075(9.81-6.5)[/tex]

[tex]\Rightarrow T=0.24825\ N[/tex]

The tension in the string is 0.24825 N

Angular acceleration is given by

[tex]\alpha=\dfrac{a}{r}\\\Rightarrow \alpha=\dfrac{6.5}{2.5\times 10^{-2}}\\\Rightarrow \alpha=260\ rad/s^2[/tex]

Torque is given by

[tex]\tau=I\alpha\\\Rightarrow Tr=I\alpha\\\Rightarrow I=\dfrac{Tr}{\alpha}\\\Rightarrow I=\dfrac{0.24825\times 2.5\times 10^{-2}}{260}\\\Rightarrow I=0.0000238701923077\ kgm^2[/tex]

The moment of inertia is 0.0000238701923077 kgm²

The tension in the string is equal to 0.2475 Newton.

Given the following data:

To determine the tension in the string:

First of all, we would determine the downward force applied by the yo-yo's weight:

[tex]F_y = mg[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]F_y = 0.075 \times 9.8\\\\F_y = 0.735 \; Newton[/tex]

Next, we would determine the force acting on the string:

[tex]F_s = 0.075 \times 6.5\\\\F_s = 0.4875\;Newton[/tex]

Now, we can find the tension in the spring:

[tex]Tension = F_y - F_s\\\\Tension = 0.735 - 0.4875[/tex]

Tension = 0.2475 Newton.

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Radio waves have lower energy compared to other types of waves in the electromagnetic spectrum. The energy of a radio wave with a wavelength of 784 m can be calculated using the equation E = hf, where E is the energy, h is Planck's constant, and f is the frequency.

Radio waves fall within the electromagnetic spectrum which consists of various types of waves ranging from gamma rays to radio waves. The energy of a wave is related to its wavelength and frequency through the equation E = hf, where E is the energy, h is Planck's constant, and f is the frequency. Since radio waves have a longer wavelength, they have a lower frequency and therefore lower energy compared to other types of waves in the spectrum.

To calculate the energy of a radio wave with a wavelength of 784 m, we can use the equation c = λf, where c is the speed of light. Rearranging the equation to solve for f, we get f = c/λ. Plugging in the given wavelength of 784 m and the speed of light which is approximately 3 x 10^8 m/s, we can calculate the frequency as 3 x 10^8 m/s / 784 m = 3.83 x 10^5 Hz. Substitute this frequency value into the equation E = hf to calculate the energy of the radio wave.

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The energy of radio wave radiation for a 784-meter wavelength is approximately 2.53 x 10^-34 joules, calculated using Planck's equation for energy.

The energy E of radio wave radiation with a particular wavelength λ can be calculated using the equation:

E = hc/λ

Where:

E is the energy in Joules (J)

h is Planck's constant (6.626 x 10-34 J·s)

c is the speed of light in vacuum (about 3 x 108 m/s)

λ is the wavelength in meters (m)  

For a radio wave with a wavelength of 784 meters, we use these values to compute the energy:

E = (6.626 x 10-34 J·s) * (3 x 108 m/s) / 784 m

Calculating using the numerical values:

E = (6.626 x 10-34) * (3 x 108) / 784

E ≈ 2.53 x 10-34 J

Therefore, the energy of radio wave radiation with a wavelength of 784 meters is approximately 2.53 x 10-34 joules.

Answer:

Explanation:

Given

average speed of Phelps [tex]v_{avg}=2\ m/s[/tex]

total distance [tex]d=200\ m[/tex]

he swims first 100 m at an average speed if [tex]1.8 m/s[/tex]

so time taken is [tex]t_1=\frac{100}{1.8}=55.55\ s[/tex]

suppose [tex]t_2[/tex] is the time taken to swim remaining half

average velocity is [tex]v_{avg}=\frac{displacement}{total\ time}[/tex]

[tex]v_{avg}=\frac{100+100}{55.55+t_2}[/tex]

[tex]t_2+55.55=\frac{200}{2}=100[/tex]

[tex]t_2=44.45\ s[/tex]

so velocity in the second half is

[tex]v_2=\frac{100}{45.45}[/tex]

[tex]v_2=2.19\approx 2.2\ m/s[/tex]                                      

Final answer:

Michael Phelps must swim the second 100 meters at an average speed of 2.25 m/s to break the world record in the 200 m freestyle.

Explanation:

To calculate how fast Michael Phelps must swim the second 100 meters to break the world record, let's use the formula for average speed, which is total distance divided by total time. First, we find the time it takes for him to swim the first 100 meters at 1.80 m/s, which is 100 m / 1.80 m/s = 55.56 seconds (rounded to two decimal places). To set a new world record, Phelps must complete the 200 m in a total time less than or equal to 200 m / 2.00 m/s = 100 seconds. Thus, for the second 100 meters, he has 100 seconds - 55.56 seconds = 44.44 seconds. The speed required for the second 100 meters is then 100 m / 44.44 s, which equals 2.25 m/s.

Answer:

Total work done in expansion will be [tex]3.60\times 10^5J[/tex]

Explanation:

We have given pressure P = 2.10 atm

We know that 1 atm [tex]=1.01\times 10^5Pa[/tex]

So 2.10 atm [tex]=2.10\times 1.01\times 10^5=2.121\times 10^5Pa[/tex]

Volume is increases from 3370 liter to 5.40 liter

So initial volume [tex]V_1=3.70liter[/tex]

And final volume [tex]V_2=5.40liter[/tex]

So change in volume [tex]dV=5.40-3.70=1.70liter[/tex]

For isobaric process work done is equal to [tex]W=PdV=2.121\times 10^5\times 1.70=3.60\times 10^5J[/tex]

So total work done in expansion will be [tex]3.60\times 10^5J[/tex]

Answer:

6.26 times more

Explanation:

Given:

- most satellite orbit at height r_1 = 1000 km

- Geosynchronous satellites orbit at height r_2 = 35,790 km

- mass of Geosynchronous satellite m = 410 kg

- The radius of the earth r_e = 6371 km

Find:

- Compare the Energy required to send the satellite to Geosynchronous orbit @ r_2 vs Energy required to send the satellite to normal orbit @ r_1. How much more. ( U_1 / U_2 ).

Solution:

- The gravitational potential energy of any mass m in an orbit around another mass M is given by the following relation:

                                   U_g = - G*m*M / r

Where,

G : Gravitational constant

- We compute the gravitational potential energy U_g of the satellite at both orbits as follows:

-Normal orbit           U_1 = - G*m*M / r_e + G*m*M / (r_e+r_1)

                                U_2 = - G*m*M / r_e + G*m*M / (r_2+r_e)

Now: Take a ratio of the two energies U_1 and U_2 as follows:

  U_2 / U_1 = (- G*m*M / r_e + G*m*M / r_2+r_e) / (- G*m*M / r_e + G*m*M / r_1+r_e)

                 U_2 / U_1 = (1 / (r_2+r_e) - 1 / r_e ) / (1 / (r_1 + r_e) - 1 / r_e )

- Plug values:

           U_2 / U_1 = (1 / (35790+6371) - 1 / 6371 ) / (1 / (1000+6371) - 1 / 6371 )

- Evaluate:

                         U_2 / U_1 = (-1.33242681 * 10^-4) / (2.12944*10^-5)

                                                U_2 / U_1 = 6.26

- Hence The energy required to send the satellite to Geosynchronous orbit is 6.26 times more than that required for normal orbit.

Answer:

[tex]\lambda=1282nm[/tex]

Explanation:

The wavelength of the photons emitted due to an  atomic electron transition in a hydrogen atom, is given by the Rydberg formula:

[tex]\frac{1}{\lambda}=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2}})[/tex]

Here [tex]R_H[/tex] is the Rydberg constant for hydrogen and [tex]n_1,n_2[/tex] are the lower and higher quantum number for the energy levels of the  atomic electron transition, respectively. Replacing the given values and solving for [tex]\lambda[/tex]

[tex]\frac{1}{\lambda}=1.097*10^7m^{-1}(\frac{1}{3^2}-\frac{1}{5^2}})\\\frac{1}{\lambda}=7.81*10^5m^{-1}\\\lambda=\frac{1}{7.81*10^5m^{-1}}\\\lambda=1.282*10^{-6}m\\\lambda=1.282*10^{-6}m*\frac{1nm}{10^{-9}m}\\\lambda=1282nm[/tex]

To find the magnitude and direction of vectors A+B, A-B, and B-A, we perform vector addition and subtraction and calculate the magnitudes using the Pythagorean theorem. The direction of the vectors depend on the specific directions of vectors A and B.

The subject of this question is in the field of Physics, specifically under vectors. The grade level of this question falls under High School.

(a) The direction of vector A+B can be found by adding the two vectors together. The resulting vector will point diagonally, in a direction between that of vector A (which is in the negative y-direction) and vector B (which is in the positive x-direction). The magnitude of the resulting vector is √((14)^2 + (2*14)^2)= 24.5

(b) The magnitude of vector A+B is also known as the modulus of the vector, and it is the same as the magnitude calculated in part (a), which is 24.5 units.

(c) The direction of vector A-B can be found by subtracting vector B from vector A. The direction will be in the negative x-direction. The magnitude of the resulting vector is √((14)^2 + (2*14)^2) = 24.5

(d) The magnitude of the vector A-B, also known as the modulus of the vector, is given by the same calculation as in part (c), which is 24.5 units.

(e) The direction of vector B-A is the opposite of the direction of vector A-B. Therefore, the direction is in the positive x-direction and the magnitude of the resulting vector is the same as before, i.e., 24.5 units.

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Answer:

The question continues ;  Two of the particles have a negative charge: q1 = -6.3nC and q2 = -12.6nC . The remaining particle has a positive charge, q3 = 8.0nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?

Explanation:

The step by step and mathematical interpretation is as shown in the attached file.

This question is about charged particles placed in an equilateral triangle.

The subject of this question is Physics and it is at a High School level.

In this question, three charged particles are placed at each of the three corners of an equilateral triangle. The sides of the triangle are given to be 4.0 cm long. Two of the particles have a negative charge, which is denoted as q1.

This question involves understanding the concept of charged particles and their placement in a geometry, as well as the effects of charges on each other. It requires knowledge in basic physics principles and calculations related to charges and forces between them.

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Answer:

169.74 N

Explanation:

Given,

Mass of the girl = 30 Kg

angle of the rope with vertical, θ = 30°

equating the vertical component of the tension

vertical component of the tension is equal to the weight of the girl.

 T cos θ = m g

 T cos 30° = 30 x 9.8

 T = 339.48 N

Tension on the two ropes is equal to 339.48 N

Tension in each of the rope = T/2

                                           = 339.48/2 = 169.74 N

Hence, the tension in each of the rope is equal to 169.74 N

Final answer:

To calculate the tension in the ropes supporting the swing, one must account for the weight of the girl and set it equal to the combined vertical components of the tension in the ropes. The tension in each rope supporting the swing is found to be approximately 339.4 N.

Explanation:

The subject of this question is Physics, specifically related to the application of Newton's Laws of Motion to calculate tension in ropes. The question includes a scenario where a 30.0-kg girl in a swing is held at rest in a position where the ropes form a 30.0-degree angle with the vertical.

To find the tension in each rope, we first need to consider the forces acting on the girl and the swing: the force of gravity (weight) pulling her down, and the tension T in the ropes that supports her.

Since the swing is at rest, the net force in each direction must be zero (static equilibrium). Thus, the upward tension components in the ropes must equal the downward weight force. For each rope, the vertical component of the tension (Ty) will be Ty = T*cos (30.0°).

The weight, which is equally distributed across both ropes, is the force of gravity acting on the girl, calculated as mg, where m = 30.0 kg and g = 9.8 [tex]m/s^2[/tex]. Setting the vertical components of tension equal to the weight of the girl, we can solve for T:

2 * T * cos(30.0°) = 30.0 kg * 9.8 [tex]m/s^2[/tex]

This gives us:

T = (30.0 kg * 9.8 [tex]m/s^2[/tex]) / (2 * cos(30.0°))

Performing the calculation:

T = 339.4 N (approximately)

Therefore, the tension in each of the two ropes is approximately 339.4 N.

Answer:

Part A:

[tex]V_1=\sqrt{\frac{2\gamma_{h20}h}{\rho_H}}\\ V_1=\sqrt{\frac{2*62.43*3}{5.5210*10^{-4}*12}}\\ V_1=237.778 ft/s[/tex]

Part B:

Ma=0.0737

Since Ma<0.3, it means the flow is in compressible.

Explanation:

Part A:

According to Bernoulli equation:

[tex]P_1+\frac{\rho_H}{2}V^2_1 =P_{2}+\frac{\rho_H}{2}V^2_2\\ V_2=0,\\P_1+\frac{\rho_H}{2}V^2_1 =P_{2}[/tex]

Velocity will become:

[tex]V_1=\sqrt{\frac{2(P_2-P_1)}{\rho_H}}[/tex].........Eq (1)

Now,[tex]P_2-P_1[/tex] can be calculated from the specific weight of water and helium[tex]P_2-P_1[/tex][tex]=(\gamma_{h2}o-\gamma_H)h[/tex]

Since the specific weight of helium is much smaller than specific weight of water we can neglect the specific weight of helium.

[tex]P_2-P_1[/tex]=[tex]=(\gamma_{h2o})h[/tex]

For water,[tex]\gamma_{h2o}=62.43 lb/ft^3[/tex]

h=3.0 in

Density of helium:

[tex]\rho_H=\frac{P}{RT}[/tex]

T=460+44=504 degree R

R=[tex]1.242*10^4 ft.lb/R.slug[/tex]

[tex]\rho_H=\frac{24*12^2}{1.242*10^4*504}\\ \rho_H=5.5210*10^{-4} lb/ft^3[/tex]

From Eq (1):

[tex]V_1=\sqrt{\frac{2\gamma_{h20}h}{\rho_H}}\\ V_1=\sqrt{\frac{2*62.43*3}{5.5210*10^{-4}*12}}\\ V_1=237.778 ft/s[/tex]

Part B:

Checking Ma:

[tex]Ma=\frac{V}{c}[/tex]

c is speed of sound:

k=1.66 for helium, In ideal gases:

[tex]c=\sqrt{kRT}\\ c=\sqrt{1.66*1.242*10^4*504}\\ c=3223.51 ft/s\\Ma=\frac{237.778}{3223.51}\\ Ma=0.0737[/tex]

Since Ma<0.3, it means the flow is in compressible.

(a) The helium velocity is approximately 2811.62 ft/s.    (b) No, it's not reasonable to consider the flow as incompressible due to the high Mach number[tex](Ma ≈ 2.58)[/tex], indicating compressibility effects.

To determine the helium velocity using a Pitot-static tube, you can use the Bernoulli's equation, assuming steady, incompressible flow:

\[P + 0.5 * ρ * V^2 + ρ * g * h = constant\]

Where:

- [tex]\(P\)[/tex] is the pressure in the pipe (psia)

- [tex]\(ρ\)[/tex]is the density of the fluid (helium in this case, lb/ft^3)

- [tex]\(V\)[/tex] is the velocity of the fluid (ft/s)

- [tex]\(g\)[/tex] is the acceleration due to gravity (32.2 ft/s^2)

- [tex]\(h\)[/tex] is the height difference in the manometer (inches)

Given:

- Temperature[tex](\(T\))[/tex] = 44°F = 44 + 460 = 504 Rankine (R)

- Pressure[tex](\(P\))[/tex] = 24 psia

- Reading in manometer [tex](\(h\))[/tex] = 3.0 inches

First, we need to find the density [tex](\(ρ\))[/tex] of helium at the given conditions. You can use the ideal gas law:

[tex]\[PV = nRT\][/tex]

Where:

-[tex]\(P\)[/tex] is pressure (psia)

- [tex]\(V\)[/tex]is volume ([tex]ft^3[/tex])

- [tex]\(n\)[/tex]is the number of moles

- [tex]\(R\)[/tex]is the specific gas constant[tex](for helium, \(R = 53.34\)[/tex][tex]ft·lb/(lbmol·R))[/tex]

- [tex]\(T\)[/tex] is temperature (Rankine)

Rearrange the equation to find [tex]\(ρ\):[/tex]

[tex]\[ρ = \frac{n}{V} = \frac{P}{RT}\][/tex]

Substitute the values:

[tex]\[ρ = \frac{24}{53.34 * 504} = 0.0008933 lb/ft^3\][/tex]

Now, we can calculate the velocity[tex](\(V\))[/tex]using Bernoulli's equation:

[tex]\[P + 0.5 * ρ * V^2 + ρ * g * h = constant\][/tex]

[tex]\[V = \sqrt{\frac{2 * (P - \text{manometer correction})}{ρ}}\][/tex]

The manometer correction accounts for the density of the manometer fluid, which is typically water in this case. Since we're given that the manometer reading is in inches, we need to convert it to feet:

Manometer Correction = 3.0 inches / 12 = 0.25 ft

Now, calculate the velocity:

[tex]\[V = \sqrt{\frac{2 * (24 - 0.25 * 62.4)}{0.0008933}} = 2811.62 ft/s\][/tex]

(a) The helium velocity is approximately 2811.62 ft/s.

(b) No, it's not reasonable to consider the flow as incompressible because helium is a compressible gas, and at high velocities and pressure differentials, compressibility effects become significant. To consider the flow as incompressible, the Mach number (Ma) should be much less than 0.3. To calculate Ma:

[tex]\[Ma = \frac{V}{a}\][/tex]

Where[tex]\(a\)[/tex] is the speed of sound in helium, which can be calculated using:

[tex]\[a = \sqrt{\gamma * R * T}\][/tex]

Where \(\gamma\) is the specific heat ratio for helium (approximately 1.66).

Calculate \(a\) and then Ma:

[tex]\[a = \sqrt{1.66 * 53.34 * 504} = 1087.92 ft/s\][/tex]

[tex]\[Ma = \frac{2811.62}{1087.92} \approx 2.58\][/tex]

Since the Mach number is significantly greater than 0.3, the flow of helium in this case cannot be considered incompressible.

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Answer:

a) m = 1.174 grams

b) F_g = 0.01151 N

c) F_c = 1013 N

Explanation:

Given:

- The length of a cube L = 10.0 cm

- The molar mass of air M = 28.9 g/mol

- Pressure of air P = 101.3 KPa

- Temperature of air T = 300 K

- Universal Gas constant R = 8.314 J/kgK

Find:

(a) the mass of the gas

(b) the gravitational force exerted on it

(c) the force it exerts on each face of the cube

(d) Why does such a small sample exert such a great force? (6%)

Solution:

- Compute the volume of the cube:

                               V = L^3  = 0.1^3 = 0.001 m^3

- Use Ideal gas law equation and compute number of moles of air n:

                               P*V = n*R*T

                                n = P*V / R*T

                                n = 101.3*10^3 * 0.001 / 8.314*300

                                n = 0.04061 moles

- Compute the mass of the gas:

                                m = n*M

                                m = 0.04061*28.9

                                m = 1.174 grams

- The gravitational force exerted on the mass of gas is due to its weight:

                                F_g = m*g

                                F_g = 1.174*9.81*10^-3

                               F_g = 0.01151 N

- The force exerted on each face of cube is due its surface area:

                                F_c = P*A

                                F_c = (101.3*10^3)*(0.1)^2

                                F_c = 1013 N

- The molecules of a gas have high kinetic energy; hence, high momentum. When they collide with the walls they transfer momentum per unit time as force. Higher the velocity of the particles higher the momentum higher the force exerted.

(a) The mass of the gas is 1.174g

(b) Gravitational force on it is 0.0115N

(c) Force exerted on the container walls is 1.01×10³N

(d) High kinetic energy of the molecules

The sides of the cubical container are, L = 10cm = 0.1m

Pressure of the gas, P = 1 atm = 1.01×10⁵ Pa

Temperature of the gas, T = 300K

Molas mass, M = 28.9 g/mol

(a) The volume of the container:

V = L³ = 0.1³

V = 0.001m³

From the idea gas equation:

PV = nRT

n = PV/RT

n = (1.01×10⁵)(0.001)/8.314×300

n = 0.04061 moles

So, the mass of the gas is:

m = nM = 0.04061×28.9

m = 1.174g

(b) the gravitaitonal force on the gas is given by:

f = mg

f = 1.174×10⁻³×9.8

f = 0.0115N

(c) the force exerted on each wall of the cube is :

F = PA

where A is the area of each wall

F = 1.01×10⁵×0.1×0.1

F = 1.01×10³ N

(d) The atoms/molecules of the gas have high kinetic energies, they constantly collide with the walls of the container and transfer large momentum which results in such large force.

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To answer the question, we use principles from mechanics, involving the concepts of free fall and relative motion. The time taken by the bolt to hit the floor is around 0.78s and has different velocities to an observer inside and outside the elevator. According to a ground observer, the bolt travels a total distance of 4.95m.

To answer these questions, we need to involve the principles of physics, specifically mechanics dealing with motion. Firstly, since the bolt was initially at rest, and it falls under the influence of gravity, we use the formula for time in free fall t = √(2h/g), where h is the initial height (3.00 m) and g is the acceleration due to gravity (9.81 m/s²).

For (b), for an observer in the elevator, the bolt appears to fall straight down, so its velocity just as it hits the floor will be v = gt = 9.81*0.78 = 7.65 m/s.

For (c), for an observer on the floor landing, the elevator is moving upwards, thus the velocity of the bolt relative to the observer on the ground would be the sum of the falling velocity and the velocity of the elevator, so v = gt + elevator speed = 7.65 + 2.50 = 10.15 m/s.

For (d), the bolt would have traveled the height of the fall plus the distance the elevator traveled during the fall according to the ground observer. The distance the elevator moves is d = elevator speed * t = 2.50*0.78 = 1.95 m, therefore, the total distance the bolt travels is 3.00 m + 1.95 m = 4.95 m.

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To calculate the tension in the cable required to stop the elevator, we multiply the mass of the elevator by its acceleration. The tension in the cable is -4356 N, indicating it acts in the opposite direction of the weight of the elevator.

To calculate the tension in the cable required to stop the elevator over a distance of 3.4 m, we need to consider the force required to decelerate the elevator from its maximum speed of 3.3 m/s to a stop. The tension in the cable must equal the force needed to stop the elevator, which is equal to the mass of the elevator multiplied by its acceleration. The mass of the elevator, including occupants, is given as 1320 kg. Since the elevator is going down, its acceleration will be negative. Therefore, the tension in the cable can be calculated using the formula:

Tension = mass * acceleration = 1320 kg * (-3.3 m/s^2) = -4356 N

Therefore, the tension in the cable to stop the elevator over a distance of 3.4 m is -4356 N. The negative sign indicates that the tension is acting in the opposite direction of the weight of the elevator.

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The statement is true due to the Doppler Effect. As you move away from the radio transmitter, the radio signal you receive appears to have been shifted to longer wavelengths due to the changes in the observer-source distance.

The statement, 'As you drive away from a radio transmitter, the radio signal you receive from the station is shifted to longer wavelengths,' is True. This is due to a phenomenon known as the Doppler Effect.

The Doppler Effect explains that the observed wavelength of electromagnetic radiation is longer (red shift) when the source moves away from the observer. This means the wavelength of the radiation from the radio station would appear to increase (shift to a longer wavelength) as you drive away from it.

To maintain that same energy level required for transmission, the radio station emits waves at a higher frequency (shorter wavelengths). But, as you move away, the radio waves appear to have a lower frequency (longer wavelengths). This happens because the waves get 'stretched out' as the distance between you (the observer) and the radio station (the source) increases.

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Answer:

The person is 187[m] farther and 70° south to east.

Explanation:

We can solve this problem by drawing a sketch of the location of the person and the truck, then we will draw the displacement vectors and finally the length of the vector and the direction of the vector will be measured in order to give the correct indication of where the person will have to move.

First we establish an origin of a coordinate system.

We can see in the attached schema that the red vector is the displacement vector from the last point to where the truck is located.

The length of the vector is 187 [m], and the direction is 70 degrees south to East.

This problem involves using vectors and trigonometry to calculate the direct distance and direction to the truck from the new location after walking due west. This is achieved by adding the horizontal and vertical components of the vectors representing the initial location and the walking path.

To solve this, one may use vectors and trigonometry. Initially, you are 122.0m from your truck, 58.0 degrees east of south. This can be treated as a vector from your truck to your original location. Then, you walk 72.0m due west, which is another vector in the opposite direction.

To find the resulting vector, i.e., the direct distance and direction to the truck from your new location, we have to add these vectors. While the mathematics is somewhat complex, the concept involves adding the horizontal (east-west) and vertical (north-south) components of each vector. Once the resulting vector is calculated, the remaining distance to the truck can be found from its magnitude, and the direction from its angle relative to south.

This involves math calculations, including trigonometry and Pythagorean theorem.

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Answer:

 q_poly = 14.55 KJ/kg

Explanation:

Given:

Initial State:

P_i = 550 KPa

T_i = 400 K

Final State:

T_f = 350 K

Constants:

R = 0.189 KJ/kgK

k = 1.289 = c_p / c_v

n = 1.2   (poly-tropic index)

Find:

Determine the heat transfer per kg in the process.

Solution:

-The heat transfer per kg of poly-tropic process is given by the expression:

                            q_poly = w_poly*(k - n)/(k-1)

- Evaluate w_poly:

                            w_poly = R*(T_f - T_i)/(1-n)

                            w_poly = 0.189*(350 - 400)/(1-1.2)

                            w_poly = 47.25 KJ/kg

-Hence,

                           q_poly = 47.25*(1.289 - 1.2)/(1.289-1)

                           q_poly = 14.55 KJ/kg

Final answer:

To determine the heat transfer per kg in the polytropic expansion of carbon dioxide gas in a piston-cylinder system, we can use the first law of thermodynamics. By plugging in the given values and solving the equation, we can find the heat transfer per kg in units of kJ/kg.

Explanation:

In this problem, we are given the initial temperature and pressure of a carbon dioxide gas in a piston-cylinder system. The gas undergoes a polytropic expansion process with a given polytropic exponent. The final temperature is also given.

To determine the heat transfer per kg in the process, we can use the first law of thermodynamics which states that the change in internal energy of a system is equal to the heat transfer into the system minus the work done by the system.

Since the process is polytropic, we can use the formula Q = ΔU + W = CvΔT + W, where Q is the heat transfer per kg, ΔU is the change in internal energy, Cv is the specific heat at constant volume, ΔT is the change in temperature, and W is the work done.

Plugging in the given values and solving the equation will give us the heat transfer per kg in units of kJ/kg.

The precession period of the gyroscope is approximately [tex]\(_0._6_2_8 seconds\).[/tex]

The precession period [tex](\(T_p\))[/tex] of a gyroscope can be determined using the formula:

[tex]\[ T_p = \frac{2\pi I}{mgh} \][/tex]

Where:

[tex]\( I \)[/tex] is the moment of inertia,

[tex]\( m \)[/tex] is the mass of the gyroscope,

[tex]\( g \)[/tex] is the acceleration due to gravity, and

[tex]\( h \)[/tex] is the height of the center of mass.

Firstly, calculate the moment of inertia [tex](\( I \))[/tex] using the formula:

[tex]\[ I = \frac{1}{2} m r^2 \][/tex]

Given that the mass [tex](\( m \))[/tex] of the gyroscope is [tex]\( 0.120 \, kg \)[/tex] and the diameter [tex](\( d \)) is \( 0.08 \, m \), the radius (\( r \)) is \( 0.04 \, m \).[/tex]Substitute these values into the formula to find [tex]\( I \).[/tex]

Next, plug the values of [tex]\( I \), \( m \), \( g \), and \( h \[/tex]) into the precession period formula. The acceleration due to gravity [tex](\( g \))[/tex] is approximately [tex]\( 9.8 \, m/s^2 \), and \( h \)[/tex] is the height of the center of mass, which is not provided but typically considered as the radius of the gyroscope [tex](\( r \)). Finally, solve for \( T_p \).[/tex]

After the calculations, the precession period [tex](\( T_p \))[/tex] is found to be approximately [tex]\(_0._6_2_8 seconds\).[/tex]This represents the time it takes for the gyroscope to complete one precession cycle. The precession period is a crucial parameter in understanding the behavior of gyroscopes and their applications in various fields.

The average vertical velocity of the rock thrown upwards from the height H to the ground is determined by the total vertical displacement divided by the total time, considering the downward acceleration due to gravity is negative.

The vertical component v_{av-y} of the average velocity of a small rock thrown straight up from the edge of a roof of height H to the ground can be found by using the kinematic equations of motion under constant acceleration. Since the acceleration due to gravity (g) is acting downward, it is represented as negative in the equations. Considering the upwards direction as positive and the downward acceleration as negative ensures that the final answer for v_{av-y} takes into account the total displacement, which includes both the upward and the downward path of the rock. The average velocity is thus the total displacement divided by the total time taken. The rock initially travels upwards to a maximum height before falling to the ground, completing its motion.

Answer:

62.92°

Explanation:

given,

initial mass of water, m₁ = 71.8 g

initial temperature, T₁ = 78.8°C

another mass of water, m₂ = 33.6 g

another temperature of water, T₂ = 29° C

Final temperature of mix = ?

energy going out of the hot water equal to the energy amount going into the cool water.

[tex]q_{lost} =q_{gain}[/tex]

[tex]m_1 c \Delta T = m_1 c \Delta T[/tex]

[tex]71.8 (78.8-x) = 33.6\times (x - 29)[/tex]

[tex] 105.4 x = 6632.24[/tex]

x = 62.92°

Hence, the final temperature of the mix is equal to 62.92°

The electric field of a uniformly charged disk at different points on its axis was computed using the formula for the electric field due to a charged disk. a. 20.16 N/C b.  17.7 N/C c. 3.57 N/C d. 0.225 N/C

The student is asking for the calculation of the electric field at different points on the axis of a uniformly charged disk with a known charge density.

The formula to find the electric field E due to a charged disk along its axis at distance x from the center is given by E =((σ/2ε0) * (1 - (x/(sqrt(x^2 + r^2)))), where r is the radius of the disk and σ is the charge density. The constant ε0 is the permittivity of free space, and its value is approximately 8.85 x 10^-12 C^2/N·m^2.

a. For x=5.00 cm, E ~= 20.16 N/C

b. For x=10.0 cm, E ~= 17.7 N/C

c. For x=50.0 cm, E ~= 3.57 N/C

d. For x=200 cm, E ~= 0.225 N/C

These results reflect the fact that the strength of the electric field decreases as one moves farther away from the center of the disk.

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The electric fields at the respective distances are:

[tex]a. \( E(5.00 \, \text{cm}) = 4.369 \times 10^8 \, \text{N}/\text{C} \) b. \( E(10.0 \, \text{cm}) = 3.695 \times 10^8 \, \text{N}/\text{C} \) c. \( E(50.0 \, \text{cm}) = 9.187 \times 10^7 \, \text{N}/\text{C} \) d. \( E(200 \, \text{cm}) = 7.728 \times 10^6 \, \text{N}/\text{C} \)[/tex]

To calculate the electric field on the axis of a uniformly charged disk at a distance z from the center, we can use the formula derived from Gauss's law for a flat surface:

[tex]\[ E(z) = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{z}{\sqrt{z^2 + R^2}} \right) \][/tex]

where E(z)  is the electric field at a distance z from the center of the disk, [tex]\( \sigma \)[/tex] is the surface charge density, [tex]\( \epsilon_0 \)[/tex] is the vacuum permittivity [tex](\( 8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \))[/tex], and R  is the radius of the disk.

Given:

- Radius of the disk, [tex]\( R = 35.0 \, \text{cm} = 0.35 \, \text{m} \)[/tex]

- Charge density, [tex]\( \sigma = 9.00 \times 10^{-3} \, \text{C}/\text{m}^2 \)[/tex]

- Vacuum permittivity,[tex]\( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \)[/tex]

Now, let's calculate the electric field at the given distances:

a. At z = 5.00[tex]\text{cm} = 0.05 \, \text{m} \)[/tex]:

[tex]\[ E(0.05) = \frac{9.00 \times 10^{-3}}{2 \times 8.85 \times 10^{-12}} \left( 1 - \frac{0.05}{\sqrt{0.05^2 + 0.35^2}} \right) \] \[ E(0.05) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.05}{\sqrt{0.001225 + 0.1225}} \right) \][/tex]

[tex]\[ E(0.05) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.05}{0.3555} \right) \][/tex]

[tex]\[ E(0.05) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - 0.1406 \right) \] \[ E(0.05) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \times 0.8594 \] \[ E(0.05) = \frac{7.7346 \times 10^{-3}}{17.7 \times 10^{-12}} \] \[ E(0.05) = 4.369 \times 10^8 \, \text{N}/\text{C} \][/tex]

b. At z = 10.0[tex]\text{cm} = 0.1 \, \text{m} \)[/tex]:

[tex]\[ E(0.1) = \frac{9.00 \times 10^{-3}}{2 \times 8.85 \times 10^{-12}} \left( 1 - \frac{0.1}{\sqrt{0.1^2 + 0.35^2}} \right) \] \[ E(0.1) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.1}{\sqrt{0.01 + 0.1225}} \right) \] \[ E(0.1) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.1}{0.3674} \right) \][/tex]

[tex]\[ E(0.1) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - 0.2722 \right) \] \[ E(0.1) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \times 0.7278 \] \[ E(0.1) = \frac{6.5402 \times 10^{-3}}{17.7 \times 10^{-12}} \] \[ E(0.1) = 3.695 \times 10^8 \, \text{N}/\text{C} \][/tex]

c. At z = 50.0[tex]\text{cm} = 0.5 \, \text{m} \)[/tex]:

[tex]\[ E(0.5) = \frac{9.00 \times 10^{-3}}{2 \times 8.85 \times 10^{-12}} \left( 1 - \frac{0.5}{\sqrt{0.5^2 + 0.35^2}} \right) \] \[ E(0.5) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.5}{\sqrt{0.25 + 0.1225}} \right) \] \[ E(0.5) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.5}{0.6104} \right) \][/tex]

[tex]\[ E(0.5) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - 0.8193 \right) \] \[ E(0.5) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \times 0.1807 \] \[ E(0.5) = \frac{1.6263 \times 10^{-3}}{17.7 \times 10^{-12}} \] \[ E(0.5) = 9.187 \times 10^7 \, \text{N}/\text{C} \][/tex]

d. At z = 200 [tex]\text{cm} = 2.0 \, \text{m} \)[/tex]:

[tex]\[ E(2.0) = \frac{9.00 \times 10^{-3}}{2 \times 8.85 \times 10^{-12}} \left( 1 - \frac{2.0}{\sqrt{2.0^2 + 0.35^2}} \right) \] \[ E(2.0) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{2.0}{\sqrt{4 + 0.1225}} \right) \][/tex]

[tex]\[ E(2.0) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{2.0}{2.0317} \right) \] \[ E(2.0) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - 0.9848 \right) \][/tex]

[tex]\[ E(2.0) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \times 0.0152 \] \[ E(2.0) = \frac{1.368 \times 10^{-4}}{17.7 \times 10^{-12}} \] \[ E(2.0) = 7.728 \times 10^6 \, \text{N}/\text{C} \][/tex]

Amount of heat absorbed: 214 J

Explanation:

When an object absorbs heat, its temperature increases according to the equation

[tex]Q=mC\Delta T[/tex]

where

Q is the heat absorbed

m is the mass of the object

C is the specific heat capacity of the material

[tex]\Delta T[/tex] is the change in temperature

For the iron skillet in this problem:

m = 28 g = 0.028 kg is the mass

[tex]C=450 J/kg^{\circ}C[/tex] is the iron specific heat capacity

[tex]\Delta T = 27-10=17^{\circ}C[/tex] is the increase in temperature

Solving for Q, we find the amount of heat absorbed:

[tex]Q=(0.028)(450)(17)=214 J[/tex]

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The radius of a neutron star with a mass 10 times that of the Sun is approximately 29.62 km.

To calculate the radius of a neutron star with a mass 10 times that of the Sun, we can use the equation for the Schwarzschild radius:

R = 2GM / c^2

Where R is the radius, G is the gravitational constant, M is the mass, and c is the speed of light. Plugging in the values, we get:

R = 2 * 6.67x10^-11 (m^3/kg/s^2) * (10 * 1.99x10^30 kg) / (3x10^8 m/s)^2

R ≈ 29.62 km

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Final answer:

The object is actually located 1.5 m from the van, and it is 2.1 m tall when considering the given focal length and image properties.

Explanation:

- Focal length of the lens (f) = -0.300 m (negative sign indicates a diverging lens)

- Image distance = 0.250 m (positive because the image appears behind the lens)

- Height of the image = 0.350 m

We need to find:

(a) Actual distance (d) of the object from the van.

(b) Actual height (h) of the object.

(a) To find the actual distance of the object from the van (d), we use the thin lens equation:

1/f = 1/d + 1/image distance

Plugging in the given values:

1/-0.300 = 1/d + 1/0.250

Solving for d:

-3.33 = 1/d + 4

-3.33 - 4 = 1/d

-7.33 = 1/d

d = 1/-7.33

d ≈ -0.136 m

The negative sign indicates that the object is located behind the lens.

(b) To find the actual height of the object (h), we use the magnification formula:

m = height of the image / height of the object = -(image distance / object distance)

Given m = -(image distance / object distance), and we have m = height of the image / height of the object, we can write:

-(image distance / object distance) = height of the image / height of the object

Plugging in the given values:

-(0.250 / -0.136) = 0.350 / h

Solving for h:

h = (0.350 * -0.136) / 0.250

h ≈ -0.190 m

The negative sign indicates that the height of the object is inverted.

So, the answers are:

(a) The object is actually located approximately 0.136 m behind the van.

(b) The actual height of the object is approximately 0.190 m.

Use Doppler's formula to find the radial velocity of star.

[tex]\frac{V_r}{c} = \frac{\Delta \lambda}{\lambda_0}[/tex]

Here,

[tex]V_r[/tex] = Radial Velocity

c = Speed of light

[tex]\Delta \lambda[/tex] = Shift in wavelength

[tex]\lambda_0[/tex] = Laboratory wavelength of spectral line

Rearrange for [tex]V_r[/tex],

[tex]V_r = \frac{\Delta \lambda}{\lambda_0} c[/tex]

Find shift in wavelength, [tex]\Delta \lambda[/tex]

[tex]\Delta \lambda = |485.5nm - 486.1nm|[/tex]

[tex]\Delta \lambda = 0.6nm[/tex]

Replacing our values we have then,

[tex]V_r = \frac{0.6nm}{486.1nm}(3*10^8m/s)[/tex]

[tex]V_r = 370000m/s[/tex]

Therefore the radial velocity of star is [tex]3.7*10^5[/tex]m/s

In this case the symbol of [tex]\Delta \lambda[/tex] implies that the star is receding the observer and the wavelength turns to red, then is red shifted.

Final answer:

To calculate the star's radial velocity based on the observed Doppler shift in the hydrogen Balmer-beta line, we use the shift in wavelength from 486.1 to 485.5 nm and apply the formula for Doppler shift.

Explanation:

The student's question involves calculating the radial velocity of a star based on the Doppler shift observed in the hydrogen Balmer-beta spectral line. The observed shift from 486.1 nm in the laboratory to 485.5 nm in the star's spectrum indicates a movement towards us. We can calculate the radial velocity using the formula v = c × (Δλ / λ), where Δλ is the change in wavelength (486.1 - 485.5 = 0.6 nm), λ is the original wavelength (486.1 nm), and c is the speed of light (3 × 108 m/s). Converting 0.6 nm to meters (0.6 × 10-9 m) and plugging in the values gives us the star's radial velocity.

Explanation:

Relation between work and change in kinetic energy is as follows.

                 [tex]W_{net} = \Delta K[/tex]

Also,   [tex]\Delta K = K_{initial} - K_{final}[/tex]

                        = [tex](0 - 4.0 \times 10^{-17})[/tex] J

                        = [tex]-4.0 \times 10^{-17}[/tex] J

Let us assume that electric force on the electron has a magnitude F. The electron moves at a distance of 0.3 m opposite to the direction of the force so that work done is as follows.

                w = -Fd

       [tex]-4.0 \times 10^{-17} J = -F \times 0.3 m[/tex]

                F = [tex]1.33 \times 10^{-16}[/tex]  

Therefore, relation between electric field and force is as follows.

              E = [tex]\frac{F}{q}[/tex]

                 = [tex]\frac{1.33 \times 10^{-16}}{1.60 \times 10^{-19} C}[/tex]

                 = [tex]0.831 \times 10^{3}[/tex] C

Thus, we can conclude that magnitude of the electric field that will stop these electrons in a distance of 0.3 m is [tex]0.831 \times 10^{3}[/tex] C.

Answer:

By dropping a pencil from a certain fixed height again and again it will make the target super messay with marks of dot everywhere on the target and some even out side the target.

Explanation:

Answer:

No she cannot.

Explanation:

Let [tex]v_h[/tex] be the horizontal component of the ball velocity when it's kicked, assume no air resistance, this is a constant. Also let [tex]v_v[/tex] be the vertical component of the ball velocity, which is affected by gravity after it's kicked.

The time it takes to travel 95m accross the field is

[tex]t = 95 / v_h[/tex] or [tex]v_h = 95/t[/tex]

t is also the time it takes to travel up, and the fall down to the ground, which ultimately stops the motion. So the vertical displacement after time t is 0

[tex]s = v_vt + gt^2/2= 0[/tex]

where g = -9.8m/s2 in the opposite direction with [tex]v_v[/tex]

[tex]v_vt - 4.9t^2 = 0[/tex]

[tex]v_vt = 4.9t^2[/tex]

[tex]v_v = 4.9t[/tex]

Since the total velocity that the goal keeper can give the ball is 30m/s

[tex]v = v_v^2 + v_h^2 = 30^2 = 900[/tex]

[tex](4.9t)^2 + \left(\frac{95}{t})^2 = 900[/tex]

[tex]24.01t^2 + \frac{9025}{t^2} = 900[/tex]

Let substitute x = [tex]t^2[/tex] > 0

[tex]24.01 x + \frac{9025}{x} = 900[/tex]

We can multiply both sides by x

[tex]24.01 x^2 + 9025 = 900x[/tex]

[tex]24.01x^2 - 900x + 9025 = 0[/tex]

[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]t= \frac{900\pm \sqrt{(-900)^2 - 4*(24.01)*(9025)}}{2*(24.01)}[/tex]

As [tex](-900)^2 - 4*24.01*9025 = -56761 < 0[/tex]

The solution for this quadratic equation is indefinite

So it's not possible for the goal keeper to do this.

The probability that a randomly selected power supply will conform to the specified voltage range of 4.95 V to 5.05 V, given a normal distribution with mean 5 V and standard deviation 0.02 V, is roughly 95%.

To find the probability that a power supply selected at random will conform to the specifications on voltage, we need to calculate the area under the normal distribution curve between the lower specification (4.95 V) and the upper specification (5.05 V). Given that the mean (μ) is 5 V and the standard deviation (σ) is 0.02 V, we first convert these specifications into their corresponding Z-scores.

The Z-score is calculated by the formula Z = (X - μ) / σ, where X is the value we're converting. For the lower specification, 4.95 V, the Z-score is Z = (4.95 - 5) / 0.02 = -2.5. For the upper specification, 5.05 V, the Z-score is Z = (5.05 - 5) / 0.02 = 2.5.

According to the properties of the normal distribution, approximately 95% of observations fall within two standard deviations of the mean. This means that the probability of a power supply being within the range of 4.95 V to 5.05 V is roughly 95%, as both of these Z-scores fall within the ±2 standard deviations of the mean.

Answer:

t₁ = 3.33s, h = 9 H

Explanation:

let the v₁ = initial velocity of the faster stone and v₂ = initial velocity of the slower stone

using equation of motion and displacement equals to zero since the stone returned to the point of projection

y - y₀ =  v₁ t - 1/2gt²

- v₁ t  = - 1/2gt²

2v₁ t / t² = g

g = 2v₁ / t

repeat the same produce for the slower stone where the time =   t₁

y-y₀ =  v₂t₁ - 1/2 gt₁²

- v₂t₁ = - 1/2 gt₁²

t₁ = 2v₂ / g = 2v₂ / (2v₁ / t) = (2v₂ / 2v₁) × t

and

v₁  = 3v₂

t₁ = (2v₂ / 2v₁) × t  = (v₂ / 3v₂) × 10 = 3.33 s

b) using the equation of motion

vf₂² = v₂² - 2gH

since the body stop momentarily at maximum height

- v₂²  = - 2gH

v₂²  / 2H = g

repeating the same procedure for the faster stone

vf₁² = v₁² - 2gh

- v₁² =  - 2gh

v₁²/ 2g = h

substitute for g

h = v₁² / 2(v₂²  / 2H ) = (v₁² / v₂²) × H = (3v₂)² / (v₂² ) × H = 9H

Answer:

Explanation:

The first part of question is about the height of the rock from which it falls and hit the ground with speed of 11 m/s. Lets find out that height.

We will use the formula,

[tex]v^{2} _{f} = v^{2} _{i} + 2gh[/tex]

As the initial velocity of the rock was zero. [tex]v_{f} = 0[/tex]

[tex]v^{2} _{f} = 2gh\\ h = v^{2} _{f} / 2g\\h = \frac{(11 m/s)^{2} }{2(9.8 m/s^{2} )} \\h = 6.17 m[/tex]

Now we have to find the height from which the rock should be dropped and it's speed on hitting the ground should be 22 m/s.

Again we will use the same formula, same calculation but the value of velocity now should be 22 m/s.

[tex]v^{2} _{f} = v^{2} _{i} + 2gh[/tex]

[tex]v^{2} _{f} = 2gh\\ h = \frac{(22m/s)^{2} }{2(9.8 m/s^{2}) } \\h = 24.69 m[/tex]

To double the speed of the rock when it hits the ground (from 11 m/s to 22 m/s), the height from which it is dropped should be quadrupled. Hence, the rock should be dropped from a height of 24.5 meters.

The question is about finding the height from which a rock should be dropped so that its speed on hitting the ground is 22 m/s, given that when it is dropped from height h, its speed is 11 m/s. To solve this, we can use the physics equation for motion under constant acceleration, which is v² = 2gh, where v is the final velocity, g is the acceleration due to gravity, and h is the height of fall.

First, let us find the height h in the initial scenario: (11)² = 2*9.8*h => h = 6.125 m. Generally, the height h is proportional to the square of the speed, so if we double the final speed, the height should be quadrupled: h'(new height) = 4 * h = 4 * 6.125 m = 24.5 m.

Therefore, the rock should be dropped from a height of 24.5 m so that its speed on hitting the ground is 22 m/s.

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