Stretched 1 cm beyond its natural length, a rubber band exerts a restoring force of magnitude 2 newtons. Assuming that Hooke's Law applies, answer the following questions:(a) How far (in units of meters) will a force of 3 newtons stretch the rubber band?

To solve this problem we will apply the concept related to kinetic energy based on the ideal gas constant and temperature. From there and with the given values we will find the temperature of the system. As the temperature is the same it will be possible to apply the root mean square speed formula that is dependent on the element's molar mass, the ideal gas constant and the temperature, this would be:

[tex]KE = \frac{3}{2} RT[/tex]

Where,

KE = Average kinetic energy of an ideal gas

[tex]R = 8.314JK^{-1}mol^{-1}[/tex]= Ideal gas constant

T = Temperature

Replacing we have,

[tex]KE = \frac{3}{2} RT[/tex]

[tex]5930J/mol = \frac{3}{2}(8.314JK^{-1}mol^{-1})T[/tex]

[tex]T = 475.503K[/tex]

Therefore the temperature is 475.5K

RMS velocity of [tex]F_2[/tex] gas is

[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]

Where,

M = Molar mass of [tex]F_2[/tex]

[tex]M = 38.00g/mol[/tex]

[tex]M = 38.00*10^{-3} kg/mol[/tex]

[tex]T = 475.5K[/tex]

[tex]R = 8.314JK^{-1}mol^{-1}[/tex]

Replacing we have,

[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]

[tex]v_{rms} = \sqrt{\frac{3(8.314JK^{-1}mol^{-1})(475.5K )}{38.00*10^{-3} kg/mol}}[/tex]

[tex]v_{rms} = 558.662m/s[/tex]

Therefore, the RMS velocity of [tex]F_2[/tex] gas is 558.6m/s

To find the net electric force on an electron due to two point charges at the origin and at x=0.8m, one has to first calculate the electric field due to each charge at the location of the electron. After finding the total electric field it's multiplied with the charge of the electron to obtain the force.

The subject of this question is Physics, specifically the concept of Electric Force. In this problem, we need to find the net force on an electron located on the x-axis at two different points because of two point charges at the origin and at x=0.8m.

Step 1: Calculate the electric field at the given point due to each point charge by using the formula E= KQ/r² where k = 9 x 10⁹ N m²/C²(Q is the charge and r is the distance from the charge to the point in question).

Step 2: Once the electric field due to each charge is found, sum these together to get the total electric field at the point.

Step 3: Electric force that an electron experiences in the field can be found using the electric field (E) and the charge of an electron (e) by the formula F= eE where e is 1.602 x 10⁻¹⁹ Coulomb.

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The net electric force on an electron at x = 0.200 m is [tex]1.66 * 10^{-17}[/tex] N towards the origin, due to both charges. At x = 1.20 m, the net force is [tex]6.95 * 10^{-18}[/tex] N towards q1, away from the origin.

A -4.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.

1. Find the net electric force that the two charges would exert on an electron placed at a point on the x-axis at x = 0.200 m.

To find the net electric force on an electron at x = 0.200 m, we will use Coulomb's Law:

[tex]F = (k * |q_1 * q_2| )/ r^2[/tex]

Here, [tex]q_1[/tex] and [tex]q_2[/tex] are the charges, r is the distance between them, and k is Coulomb's constant,[tex]8.99 * 10^9 N.m^2/C^2.[/tex]

The electron at x = 0.200 m is 0.200 m from q1 and 0.600 m from [tex]q_2[/tex]. Both forces are attractive since all charges are negative.

Force due to [tex]q_1[/tex]:

[tex]F_1 = k * |e * q_1| / (0.200)^2 = 8.99 * 10^9 N.m^2/C^2 * |-1.6 x 10^-{19} C * -4.00 * 10^{-9} C| / (0.200 m)^2 = 1.44 * 10^{-17} N[/tex]

Force due to [tex]q_2[/tex]:

[tex]F_2 = k * |e * q_2| / (0.600)^2 = 8.99 * 10^9 N.m^2/C^2 * |-1.6 x 10^{-19} C * -5.50 x 10^{-9} C| / (0.600 m)^2 = 2.20 * 10^{-18} N[/tex]

The net force will be the sum of these forces, taking directions into account.

Net force = F1 (towards -x) + F2 (towards -x) [tex]= 1.44 * 10^{-17} N + 2.20 * 10^{-18} N = 1.66 * 10^{-17} N[/tex]

This force is towards the origin.

2. Find the net electric force that the two charges would exert on an electron placed at a point on the x-axis at x = 1.20 m.

We repeat the procedure for the electron at x = 1.20 m:

The electron at x = 1.20 m is 1.20 m from q1 and 0.400 m from q2.

Force due to q1:

[tex]F1 = k * |e * q1| / (1.20)^2 = 8.99 * 10^9 N.m^2/C^2 * |-1.6 x 10^{-19} C * -4.00 * 10^-9 C| / (1.20 m)^2 = 2.00 * 10^{-18} N[/tex]

Force due to q2:

[tex]F2 = k * |e * q2| / (0.400)^2 = 8.99 * 10^9 N.m^2/C^2 * |-1.6 x 10^{-19} C * -5.50 * 10^{-9} C| / (0.400 m)^2 = 4.95 * 10^{-18} N[/tex]

The net force is:

Net force = F1 (towards +x) - F2 (towards -x) [tex]= 2.00 * 10^{-18} N + 4.95 * 10^{-18} N = 6.95 * 10^{-18}[/tex] N towards q1, away from the origin

Answer:

9.25 x 10^-4 Nm

Explanation:

number of turns, N = 8

major axis = 40 cm

semi major axis, a = 20 cm = 0.2 m

minor axis = 30 cm

semi minor axis, b = 15 cm = 0.15 m

current, i = 6.2 A

Magnetic field, B = 1.98 x 10^-4 T

Angle between the normal and the magnetic field is 90°.

Torque is given by

τ = N i A B SinФ

Where, A be the area of the coil.

Area of ellipse, A = π ab = 3.14 x 0.20 x 0.15 = 0.0942 m²

τ = 8 x 6.20 x 0.0942 x 1.98 x 10^-4 x Sin 90°

τ = 9.25 x 10^-4 Nm

thus, the torque is 9.25 x 10^-4 Nm.

The magnitude of the torque on the coil is 9.25 x 10⁻⁴ Nm

According to the question we have the following data:

Number of turns of the coil N = 8

Semi-major axis of the ellipse a = 40/2 cm = 0.2 m

Semi-minor axis, b = 30/2 cm = 0.15 m

Current in the coil, i = 6.2 A

Magnetic field, B = 1.98 x 10⁻⁴ T

The angle between the normal and the magnetic field is 90°.

So the torque on the coil is given by:

τ = NiABsinθ

Now, the area of ellipse:

A = πab

A = 3.14 x 0.20 x 0.15 = 0.0942 m²

Thus,

τ = 8 x 6.20 x 0.0942 x 1.98 x 10⁻⁴ x sin 90°

τ = 9.25 x 10⁻⁴ Nm

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Answer:

Air, water, rock and soil

Explanation:

The environment where the general properties of wave motion can be observed is the wave medium which is any substance or particle that carries the wave, or through which the wave travels.

Explanation:

Using Newtons second law on each block

F = m*a

Block 1

[tex]T_{1} - u*g*M_{1} = M_{1} *a \\\\T_{1} = M_{1}*(a + u*g) ... Eq1[/tex]

Block 2

[tex]T_{2} - u*g*M_{2} = M_{2} *a \\\\T_{2} = M_{2}*(a + u*g) ... Eq2[/tex]

Block 3

[tex]- (T_{1} + T_{2} ) + g*M_{3} = M_{3} *a \\\\T_{1} + T_{2} = M_{3}*( -a + g) ... Eq3[/tex]

Solving Eq1,2,3 simultaneously

Divide 1 and 2

[tex]\frac{T_{1} }{T_{2}} = \frac{M_{1}*(a+u*g)}{M_{2}*(a+u*g)} \\\\\frac{T_{1} }{T_{2}} = \frac{M_{1} }{M_{2} }\\\\ T_{1} = \frac{M_{1} *T_{2} }{M_{2} } .... Eq4[/tex]

Put Eq 4 into Eq3

[tex]T_{2} = \frac{M_{3}*(g-a) }{1+\frac{M_{1} }{M_{2} } } ...Eq5[/tex]

Put Eq 5 into Eq2 and solve for a

[tex]a = \frac{M_{3}*g -u*g*(M_{1} + M_{2}) }{M_{1} + M_{2} + M_{3} } .... Eq6[/tex]

Substitute back in Eq2 and use Eq4 and solve for T2 & T1

[tex]T_{2} = M_{2}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\T_{1} = M_{1}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\[/tex]

Answer:

d =  -26 mm

Explanation:

given,

initial position of paramecium, x = 65 mm

final position of paramecium, y = 39 mm

displacement of the paramecium's = ?

displacement  = final position - initial position

 d =  y - x                      

d = 39 - 65                

d =  -26 mm              

The paramecium's displacement comes out to be -26 mm

Answer:

The final speed of the car in the first 10 seconds will be 20 m/s.

Explanation:

Given that,

Initial speed of the car, u = 0

Acceleration of the car, [tex]a=2\ m/s^2[/tex]

Time, t = 10 s

We need to find the speed of the car in the first 10 seconds. It can be calculated using first equation of motion. It is given by :

[tex]v=u+at[/tex]

v is the final speed of the car

[tex]v=at[/tex]

[tex]v=2\ m/s^2\times 10\ s[/tex]

v = 20 m/s

So, the final speed of the car in the first 10 seconds will be 20 m/s. Hence, this is the required solution.

Final answer:

The horizontal plank on a pivot supported by a spring exhibits simple harmonic motion with angular frequency ω = √(3k/m). When the mass is 5.00 kg and the spring constant is 100 N/m, the frequency is approximately 1.95 Hz.

Explanation:

A horizontal plank of mass m and length L is pivoted at one end and supported by a spring of force constant k at the other. This setup is displaced by a small angle θ from its horizontal equilibrium position and released.

Part A: Derivation of Simple Harmonic Motion

To show the plank moves with simple harmonic motion (SHM) and to find the angular frequency ω, we analyze the forces acting on the plank when displaced. The restoring force F exerted by the spring is F = -kx, where x is the linear displacement of the spring. For small angles, θ, x ≈ Lθ. Thus, F ≈ -kLθ. Applying Newton's second law for rotational motion, τ = Iα, where τ is the torque, I is the moment of inertia of the plank, and α is the angular acceleration. The torque caused by the spring is τ = -kL²θ, and the moment of inertia of the plank about the pivot is I = (1/3)mL². From τ = Iα, we get α = -3k/mθ, indicating SHM with an angular frequency ω = √(3k/m).

Part B: Calculating the Frequency

Given m = 5.00 kg and k = 100 N/m, ω = √(3k/m) = √(3*100/5) = √60. Hence, the frequency f is f = ω/(2π) = √60/(2π) ≈ 1.95 Hz.

To solve this problem we will apply the concept of frequency in a string from the nodes, the tension, the linear density and the length of the string, that is,

[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]

Here

n = Number of node

T = Tension

[tex]\mu[/tex] = Linear density

L = Length

Replacing the values in the frequency and value of n is one for fundamental overtone

[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]

[tex]f = \frac{1}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})[/tex]

[tex]\mathbf{f = 281.2Hz}[/tex]

Similarly plug in 2 for n for first overtone and determine the value of frequency

[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]

[tex]f = \frac{2}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})[/tex]

[tex]\mathbf{f = 562.4Hz}[/tex]

Similarly plug in 3 for n for first overtone and determine the value of frequency

[tex]f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})[/tex]

[tex]f = \frac{3}{2(62*10^{-2})}\bigg (\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}} \bigg)[/tex]

[tex]\mathbf{f= 843.7Hz}[/tex]

The procedures involves using short pulses of high intensity laser beams to repair detached parts of the retina. The energy delivered per pulse, average pressure exerted by laser, new wavelength and frequency of laser light inside the eye, and the maximum electric and magnetic field in the laser beam are calculated. The details of these calculations have been given.

A) The energy delivered to the retina with each pulse can be calculated using the formula Energy = Power x Time. Here, power is 250mW or 0.250 joules/second and time is 1.50 ms or 0.00150 seconds. So, the energy is 0.250 joules/second x 0.00150 seconds = 0.000375 joules or 375 μJ per pulse.

B) The average pressure exerted by the laser beam can be calculated using the formula Pressure = Power/Area where the area of the laser spot is the area of a circle of diameter 510 μm. The laser light is completely absorbed upon hitting a surface, therefore the pressure the beam exerts is 2* Power/Area of beam.

C) Inside the vitreous humor of the eye, the wavelength of the laser light decreases as it is inversely proportional to the refractive index of the medium. So the new wavelength in the vitreous humor is 810 nm/ 1.34 = 604.48 nm.

D) The frequency of light doesn't change when it enters a different medium so it remains the same in the vitreous humor as it was when in air. It can be calculated using the formula Frequency = Speed of Light / Wavelength.

E) The maximum electric field in the laser beam can be computed using the formula for energy density.

F) The maximum value of the magnetic field in the laser beam is proportional to the maximum electric field.

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The energy delivered to the retina per pulse is 0.375 mJ. The pressure exerted is approximately 3.68 Pa, and the wavelength inside the vitreous humor is 604.5 nm. The frequency is 3.70 × 10¹⁴ Hz, with maximum electric and magnetic field values of 1.53 × 10⁵ V/m and 5.10 × 10⁻⁴ T, respectively.

A) Energy Delivered to the Retina Per Pulse

The power of the laser beam is 250 mW, which is 0.250 W. The duration of each pulse is 1.50 ms, which is 1.50 × 10⁻³ s.

Calculate the energy delivered during each pulse using the formula:

Energy (J) = Power (W) × Time (s).

Energy = 0.250 W × 1.50 × 10⁻³ s

= 3.75 × 10⁻⁴ J (or 0.375 mJ).

B) Average Pressure Exerted by the Pulse

If the beam is fully absorbed, the pressure exerted is given by:

Pressure (P) = Power (P) / Area (A) × c,

where c is the speed of light (approximately 3 × 10⁸ m/s).

Calculate the area of the circular spot:

Area (A) = π (d/2)²

where diameter

d = 510 μm = 510 × 10⁻⁶ m.

Area = π (510 × 10⁻⁶ / 2)²

≈ 2.04 × 10⁻⁷ m².

P = 0.250 W / (2.04 × 10⁻⁷ m²) × 3 × 10⁸ m/s

≈ 3.68 N/m² (Pa).

C) Wavelength of Laser Light Inside the Vitreous Humor

The wavelength inside the medium is given by: λ' = λ / n, where λ is the wavelength in air (810 nm) and n is the refractive index (1.34).

λ' = 810 nm / 1.34

≈ 604.5 nm.

D) Frequency of the Laser Light Inside the Vitreous Humor

The frequency of light remains constant through different media and is given by:

f = c / λ,

where c is the speed of light

λ is the wavelength in air.

Frequency f = 3 × 10⁸ m/s / 810 × 10⁻⁹ m

≈ 3.70 × 10¹⁴ Hz.

E) Maximum Value of the Electric Field in the Laser Beam

The maximum electric field E₀ is given by:

E₀ = √(2I/ε₀c),

where I is the intensity,

ε₀ is the permittivity of free space,

c is the speed of light.

Intensity I = P / A

= 0.250 W / 2.04 × 10⁻⁷ m²

≈ 1.23 × 10⁶ W/m².

Using ε₀ = 8.85 × 10⁻¹² F/m: E₀

≈ √(2 × 1.23 × 10⁶ W/m² / 8.85 × 10⁻¹² F/m × 3 × 10⁸ m/s)

≈ 1.53 × 10⁵ V/m.

F) Maximum Value of the Magnetic Field in the Laser Beam

The magnetic field B₀ is related to the electric field E₀ by:

B₀ = E₀ / c.

B₀ = 1.53 × 10⁵ V/m / 3 × 10⁸ m/s

≈ 5.10 × 10⁻⁴ T.

These calculations help us understand how laser pulses interact with biological tissues, aiding in precision medical procedures.

Therefore, the energy delivered to the retina per pulse is 0.375 mJ. The pressure exerted is approximately 3.68 Pa, and the wavelength inside the vitreous humor is 604.5 nm. The frequency is 3.70 × 10¹⁴ Hz, with maximum electric and magnetic field values of 1.53 × 10⁵ V/m and 5.10 × 10⁻⁴ T, respectively.

Answer:

F' = 169.45N

This is a vector addition involving two vectors. In order to do this correctly, we need to resolve each of those forces into their vertical and horizon components and sum them up accordingly (all vertical components summed together and all horizontal components summed together). Then the magnitude of the summation is found by taking the square root of the sun of the squares of the summations along the vertical and the horizontal.

Explanation:

See the attachment below for the full solution to the problem.

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Answer:

ω = 2.55 rad/sec

Explanation:

Assuming no other external forces acting in the horizontal plane, the only force keeping the  rider in a circular path of a radius equal to his distance to the center of rotation, is the centripetal force.

According to Newton's 2nd law, in the horizontal direction, we have:

F = Fc = m*a = m*ω²*r

We know that a = ac = 10*g = 98.0 m/s², and that r = 15.0 m.

Replacing these values in (1), and solving for ω, we get:

ω = √98.0m/s²/ 15.0 m = 2.55 rad/sec

The centrifugal force in a rocket.

The centrifugal force is the force that is related to the outwards or away from the body. The larger force is used to expose the aspiring astronomers to accelerations that are the same as those experienced by the rockets that are launched and the air reentries.

Thus the answer is  ω = 2.55 rad/sec

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Answer:

1.686 m

Explanation:

From coulomb's law,

F = kq1q2/r² ...................................... Equation 1

Where F = electrostatic force  between the two charges, q1 = first charge, q2 = second charge, r = distance between the charges.

making r the subject of the equation,

r = √(kq1q2/F).......................... Equation 2

Given: F = 5.05 N, q1 = 28.0 μC = 28×10⁻⁶ C, q2 = 57.0 μC = 57.0×10⁻⁶ C

Constant: k = 9.0×10⁹ Nm²/C².

Substituting into equation 2

r = √(9.0×10⁹×28×10⁻⁶×57.0×10⁻⁶/5.05)

r = √(14364×10⁻³/5.05)

r = √(14.364/5.05)

r = √2.844

r = 1.686 m

r = 1.686 m.

Thus the distance must be 1.686 m

Answer:

Vector C = -1.56i^ +1.07j^

This question requires that we use the properties of the scalar product of two vectors to find the required x and y component.

The dot product of two perpendicular vectors is equal to and the product of two vectors that are not parallel is equal to a nonzero value.

These are the properties that have been used in solving this problem alongside solving the simultaneous questions generator.

Explanation:

The full solution can be found in the attachment below.

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To solve this problem it is necessary to apply the concepts related to the voltage depending on the electric field and the distance, as well as the load depending on the capacitance and the voltage. For the first part we will use the first mentioned relationship, for the second part, we will not only define the load as the capacitance by the voltage but also place it in terms of the Area, the permittivity in free space, the voltage and the distance.

PART A ) Voltage in function of electric field and distance can be defined as,

[tex]V = Ed[/tex]

Our values are,

[tex]E = 1.2*10^5 V/m[/tex]

[tex]d = 3.0mm = 3*10^{-3}[/tex]

Replacing,

[tex]V = (1.2*10^5)(3*10^{-3})[/tex]

[tex]V = 360v[/tex]

Therefore the potential difference across the capacitor is 360V

PART B) The charge can be defined as,

[tex]Q = CV = \frac{\epsilon AV}{d}[/tex]

Here,

[tex]\epsilon = 8.85*10^{-12} F/m[/tex], Permittivity of free space

[tex]A = s^2[/tex], area of each capacitor plate

s = Length of capacitor plate

Replacing,

[tex]Q = \frac{\epsilon AV}{d}[/tex]

[tex]Q = \frac{(8.85*10^{-12})(0.03)^2(240)}{2.0*10^{-8}m}[/tex]

[tex]Q = 9.558*10^{-10}C[/tex]

Therefore the charge on each plate is [tex]9.558*10^{-10}C[/tex]

The potential difference across the parallel-plate capacitor is 360 V. The charge on each plate of the capacitor is approximately 0.95 x 10^-8 C.

The potential difference across a parallel-plate capacitor is calculated using the formula V = Ed, where E is the electric field strength and d is the distance (or spacing) between the plates. As given, E is 1.2 x 10^5 V/m, and d is 3.0 mm (or 3.0 x 10^-3 m). Therefore, the potential difference V across the plates is given by V = 1.2 x 10^5 V/m * 3.0 x 10^-3 m = 360 V.

The amount of charge Q on each plate of the capacitor can be found using the formula Q = εEA, where ε is the permittivity of free space (ε = 8.85 x 10^-12 F/m), E is the electric field strength, and A is the area of the plate. Substituting the values given, we have A = 3.0 cm * 3.0 cm = 9 cm^2 = 9 x 10^-4 m^2, E = 1.2 x 10^5 V/m, and ε = 8.85 x 10^-12 F/m. Therefore, Q = εEA = 8.85 x 10^-12 F/m * 1.2 x 10^5 V/m * 9 x 10^-4 m^2 ≈ 0.95 x 10^-8 C.

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Answer:

2630250 N/C, horizontally left

0

[tex]1.67\times 10^{-8}\ s[/tex]

1434.825 N/C, horizontally left

Explanation:

m = Mass of particle

u = Initial velocity = [tex]4.2\times 10^6\ m/s[/tex]

v = Final velocity = 0

t = Time taken

s = Displacement = 3.5 cm

q = Charge of particle = [tex]1.6\times 10^{-19}\ C[/tex]

Force is given by

[tex]F=qE[/tex]

Acceleration is given by

[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times E}{1.67\times 10^{-27}}\\\Rightarrow a=95808383.23353E[/tex]

[tex]v^2-u^2=2as\\\Rightarrow v^2-u^2=2\times 95808383.23353Es\\\Rightarrow E=\dfrac{0^2-(4.2\times 10^6)^2}{2\times 95808383.23353\times 0.035}\\\Rightarrow E=-2630250\ N/C[/tex]

Magnitude of electric field is 2630250 N/C

Direction is horizontally to the left

The angle counterclockwise from left is zero.

[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times -2630250}{1.67\times 10^{-27}}\\\Rightarrow a=-2.52\times 10^{14}\ m/s^2[/tex]

[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-4.2\times 10^6}{-2.52\times 10^{14}}\\\Rightarrow t=1.67\times 10^{-8}\ s[/tex]

The time taken is [tex]1.67\times 10^{-8}\ s[/tex]

Acceleration is given by

[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times E}{9.11\times 10^{-31}}\\\Rightarrow a=175631174533.4797E[/tex]

[tex]v^2-u^2=2as\\\Rightarrow v^2-u^2=2\times 175631174533.4797Es\\\Rightarrow E=\dfrac{0^2-(4.2\times 10^6)^2}{2\times 175631174533.4797\times 0.035}\\\Rightarrow E=-1434.825\ N/C[/tex]

Magnitude of electric field is 1434.825 N/C

Direction is horizontally to the left

Answer: Electron affinity of F equals

275.8kJ/mol

Explanation: Electron affinity is the energy change when an atom gains an electron.

Let's first calculate the energy required -E(r)to dissociate KF into ions not neutral atom which is given.

E(r) = {z1*z2*e²}/{4π*permitivity of space*r}

z1 is -1 for flourine

z2 is +1 for potassium

e is magnitude of charge 1.602*EXP{-9}C

r is ionic bond length of KF(is a constant for KF 0.217nm)

permitivity of free space 8.854*EXP{-12}.

Now let's solve

E(r)= {(-1)*(1)*(1.602*EXP{-9})²} /

{4*3.142*8.854*EXP(-12)*0.217*EXP(-9)

E(r) = - 1.063*EXP{-18}J

But the energy is released out that is exothermic so we find - E(r)

Which is +1.603*EXP{-18}J

Let's now convert this into kJ/mol

By multiplying by Avogadro constant 6.022*EXP(23) for the mole and diving by 1000 for the kilo

So we have,

1.603*EXP(-18) *6.022*EXP(23)/1000

-E(r) = 640.2kJ/mol.

Now let's obtain our electron affinity for F

We use this equation

Energy of dissociation (nuetral atom)= electron affinity of F +(-E(r)) + ionization energy of K.

498kJ/mol

=e affinity of F + 640.2kJ/mol

+(-418kJ/mol)

(Notice the negative sign in ionization energy for K. since it ionize by losing an electron)

Making electron affinity of F subject of formula we have

Electron affinity (F)=498+418-640.2

=275.8kJ/mol.

Answer:b

Explanation:

Given

Force of attraction is F when charges are d distance apart.

Electrostatic force is given by

[tex]F=\frac{kq_1q_2}{d^2}---1[/tex]

where k=constant

[tex]q_1[/tex] and [tex]q_2[/tex] are charges

d=distance between them

In order to double the force i.e. 2F

[tex]2F=\frac{kq_1q_2}{d'^2}----2[/tex]

divide 1 and 2 we get

[tex]\frac{F}{2F}=\frac{d'^2}{d^2}[/tex]

[tex]d'=\frac{d}{\sqrt{2}}[/tex]

Answer:

A) The total distance traveled when t = 8.3 s is 234 ft.

B) The average velocity of the particle is 4.1 ft/s.

C) The average speed at t = 8.3 s is 28 ft/s.

D) The instantaneous velocity at t = 8.3 s is 92 ft/s.

E) The acceleration of the particle at t = 8.3 s is 37.8 ft/s²

Explanation:

Hi there!

A) The position of the particle at a time "t", in feet, is given by the function "s":

s(t) = t³ - 6t² - 15t + 7

First, let´s find at which time the particle changes direction. The sign of the instantaneous velocity indicates the direction of the particle. We will consider the right direction as positive. The origin of the frame of reference is located at s = 0 and t = 0 so that the particle at t = 0 is located 7 ft to the right of the origin.

The instaneous velocity (v(t)) of the particle is the first derivative of s(t):

v(t) = ds/dt = 3t² - 12t - 15

The sign of v(t) indicates the direction of the particle. Notice that at t = 0,

v(0) = -15. So, initially, the particle is moving to the left.

So let´s find at which time v(t) is greater than zero:

v(t)>0

3t² - 12t - 15>0

Solving the quadratic equation with the quadratic formula:

For  every t > 5 s, v(t) > 0 (the other solution of the quadratic equation is -1. It is discarded because the time can´t be negative).

Then, the particle moves to the left until t = 5 s and, thereafter, it moves to the right.

To find the traveled distance at t= 8.3 s, we have to find how much distance the particle traveled to the left and how much distance it traveled to the right.

So, let´s find the position of the particle at t = 0, at t = 5 and at t = 8.3 s

s(t) = t³ - 6t² - 15t + 7

s(0) = 7 ft

s(5) = 5³ - 6 · 5² - 15 · 5 + 7 = -93 ft

s(8.3) = 8.3³ - 6 · 8.3² - 15 · 8.3 + 7 = 40.9 ft

So from t = 0 to t = 5, the particle traveled (93 + 7) 100 ft to the left, then from t = 5 to t = 8.3 the particle traveled (93 + 40.9) 134 ft to the right. Then, the total distance traveled when t = 8.3 s is (134 ft + 100 ft) 234 ft.

B) The average velocity (AV) is calculated as the displacement over time:

AV = Δs / Δt

Where:

Δs = displacement (final position - initial position).

Δt = elapsed time.

In this case:

final position = s(8.3) = 40.9 ft

initial position = s(0) = 7 ft

Δt = 8.3 s

So:

AV = (s(8.3) - s(0)) / 8.3 s

AV = (40.9 ft - 7 ft) / 8.3 s

AV = 4.1 ft/s

The average velocity of the particle is 4.1 ft/s (since it is positive, it is directed to the right).

C) The average speed is calculated as the traveled distance over time. The traveled distance at t = 8.3 s was already obtained in part A: 234 ft. Then, the average speed (as) will be:

as = distance / time

as = 234 ft / 8.3 s

as = 28 ft/s

The average speed at t = 8.3 s is 28 ft/s

D) The instantaneous velocity at any time t was obtained in part A:

v(t) = 3t² - 12t - 15

at t = 8.3 s

v(8.3) = 3(8.3)² - 12(8.3) - 15

v(8.3) = 92 ft/s

The instantaneous velocity at t = 8.3 s is 92 ft/s.

E) The particle acceleration at any time t, is obtained by derivating the velocity function:

v(t) = 3t² - 12t - 15

dv/dt = a(t) =  6t - 12

Then at t = 8.3 s

a(8.3) = 6(8.3) - 12

a(8.3) = 37.8 ft/s²

The acceleration of the particle at t = 8.3 s is 37.8 ft/s²

Given the missing information, we can only calculate the particle's instantaneous velocity and acceleration at 8.3 s, which can be obtained by taking derivatives of the position function.

This is a physics problem involving kinematics and calculus. Let's address each part of the question:

A. To find the total distance, we would need to know the initial position of the particle. Without this information, we cannot accurately calculate the total distance.

B. Average velocity is defined as the displacement divided by the time interval, which is irrelevant in this case because we do not know the displacement.

C. Similar to B, without displacement information, we cannot calculate average speed.

D. Instantaneous velocity is given by the first derivative of the position function. By taking the derivative of s, we can get the velocity function: v(t) = 3t^2 - 12t - 15. Plug in t = 8.3 s, we can then get the instantaneous velocity.

E. Acceleration is given by the first derivative of the velocity function or the second derivative of the position function. With our velocity function, v(t), we can take its derivative to find a(t) = 6t - 12. Plugging in t = 8.3 s will give the particle's acceleration at that time.

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To solve this problem we will proceed to find the density from the specific gravity. Later we will find the specific volume as the inverse of the density. Finally with the data obtained we will find the total weight in the bottle.

a) [tex]\rho = \gamma * 1000[/tex]

Here,

[tex]\rho[/tex] = Density

[tex]\gamma[/tex] = Specific gravity

[tex]\rho = 1.1 * 1000[/tex]

[tex]\rho = 1100 kg/m3[/tex]

b)

[tex]\text{Specific volume}= \frac{1}{\rho}[/tex]

[tex]\upsilon = \frac{1}{1100}[/tex]

[tex]\upsilon = 0.00090909 m^3/kg[/tex]

From the equivalences of meters to feet and kilograms to pounds, we have to

[tex]1m = 3.280839895 ft[/tex]

[tex]1 kg = 2.2046 lbm[/tex]

Converting the previous value to British units:

[tex]\upsilon = 0.00090909 m^3/kg (\frac{3.280839895^3 ft^3}{1m^3} )(\frac{1kg}{2.2046 lbm})[/tex]

[tex]\upsilon= 0.0145757 ft^3 / lbm[/tex]

c)

[tex]V = 2*10^{-3} m^3[/tex]

Mass of the liquid in bottle is

[tex]m = V\rho[/tex]

[tex]m= (2*10^{-3} m^3 )(1100kg/m^3)[/tex]

[tex]m = 2.2kg = 2200g[/tex]

Therefore the Total weight

[tex]W= 157 + 2200 = 2357 g[/tex]

There are some placeholders in the expression, but they can be safely assumed

Answer:

(a) [tex]f=1617.9\ Hz[/tex]

(b) [tex]T=0.618\ ms[/tex]

(c) [tex]A=20 \ Volts[/tex]

(d) [tex]\varphi=60^o[/tex]

Explanation:

Sinusoidal Waves

An oscillating wave can be expressed as a sinusoidal function as follows

[tex]V(t)&=A\cdot \sin(2\pi ft+\varphi )[/tex]

Where

[tex]A=Amplitude[/tex]

[tex]f=frequency[/tex]

[tex]\varphi=Phase\ angle[/tex]

The voltage of the question is the sinusoid expression  

[tex]V(t)=20cos(5\pi\times 103t+60^o)[/tex]

(a) By comparing with the general formula we have

[tex]f=5\pi\times 103=1617.9\ Hz[/tex]

[tex]\boxed{f=1617.9\ Hz}[/tex]

(b) The period is the reciprocal of the frequency:

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

[tex]\displaystyle T=\frac{1}{1617.9\ Hz}=0.000618\ sec[/tex]

Converting to milliseconds

[tex]\boxed{T=0.618\ ms}[/tex]

(c) The amplitude is

[tex]\boxed{A=20 \ Volts}[/tex]

(d) Phase angle:

[tex]\boxed{\varphi=60^o}[/tex]

Answer:

d = 0.247 mm

Explanation:

given,

λ = 633 nm

distance from the hole to the screen = L = 4 m

width of the central maximum = 2.5 cm

                                             2 y = 0.025 m

                                               y = 0.0125 m

For circular aperture

  [tex]sin \theta = 1.22\dfrac{\lambda}{d}[/tex]

using small angle approximation

  [tex]\theta = \dfrac{y}{D}[/tex]

now,

   [tex]\dfrac{y}{D} = 1.22\dfrac{\lambda}{d}[/tex]

   [tex]y = 1.22\dfrac{\lambda\ D}{d}[/tex]

   [tex]d = 1.22\dfrac{\lambda\ D}{y}[/tex]

   [tex]d = 1.22\dfrac{633\times 10^{-9}\times 4}{0.0125}[/tex]

         d =0.247 x 10⁻³ m

         d = 0.247 mm

the diameter of the hole is equal to 0.247 mm

The diameter of the hole is approximately 0.0201 mm.

The width of the central maximum in a diffraction pattern can be determined using the formula:

w = (2 * λ * D) / x

Where w is the width of the central maximum, λ is the wavelength of the light, D is the distance between the aperture and the screen, and x is the diameter of the hole. Rearranging the formula, we can solve for x:

x = (2 * λ * D) / w

Plugging in the given values, we get:

x = (2 * 633 * 10^-9 * 4.0) / 0.025

x ≈ 0.0201 mm

To find the volume flow rate in ideal flow between two horizontal tubes, use the equation A1v1 = A2v2, where A1 and A2 are the cross-sectional areas of the tubes and v1 and v2 are the velocities of the liquid. To find the volume flow rate as a function of DP, substitute the values of A1 and v1 into the equation Q = A1v1. For specific values of DP, substitute the values of A1 and v1 into the equation Q = A1v1.

For liquids, the volume flow rate can be determined using the equation A1v1 = A2v2. In this equation, A1 and A2 are the cross-sectional areas of the first and second tubes, and v1 and v2 are the velocities of the liquid flowing through the tubes. Since the first tube is larger in radius, it has a larger cross-sectional area. Therefore, the velocity of the liquid in the first tube is smaller than in the second tube. The volume flow rate can be expressed as:

Q = A1v1

where Q is the volume flow rate and is equal to the product of the cross-sectional area and velocity of the liquid in the first tube.

(a) The volume flow rate as a function of DP can be found by substituting the values of A1 and v1 into the equation Q = A1v1. Since the radii of the tubes are given, the cross-sectional areas can be calculated using the formula A = πr2.

(b) To evaluate the volume flow rate for DP = 6.00 kPa, substitute the values of A1 and v1 into the equation Q = A1v1.

(c) To evaluate the volume flow rate for DP = 12.0 kPa, substitute the values of A1 and v1 into the equation Q = A1v1.

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To solve this problem we will start by calculating the distance traveled while relating the first acceleration in the given time. From that acceleration we will calculate its final speed with which we will calculate the distance traveled in the second segment. With this speed and the acceleration given, we will proceed to calculate the last leg of its route.

Expression for the first distance is

[tex]s_1 = ut +\frac{1}{2} at^2[/tex]

[tex]s_1 = 0+\frac{1}{2} (3.8)(6)^2[/tex]

[tex]s_1 = 68.4m[/tex]

The expression for the final speed is

[tex]v = v_0 +at[/tex]

[tex]v = 0+(3.8)(6)[/tex]

[tex]v = 22.8m/s[/tex]

Then the distance becomes as follows

[tex]s_2 = vt[/tex]

[tex]s_2 = (22.8)(1.6)[/tex]

[tex]s_2 = 36.48m[/tex]

The expression for the distance at last sop is

[tex]v_1^2=v_0^2 +2as_3[/tex]

[tex]22.8^2 = 0+2(3.3)s_3[/tex]

[tex]s_3 =78.7636m[/tex]

Therefore the required distance between the signs is,

[tex]S = s_1+s_2+s_3[/tex]

[tex]S = 68.4+36.48+78.76[/tex]

[tex]S = 183.64m[/tex]

Therefore the total distance between signs is 183.54m

Answer:

m=57.65 kg

Explanation:

Given Data

Ricardo mass m₁=80 kg

Canoe mass m₂=30 kg

Canoe Length L= 3 m

Canoe moves x=40 cm

When Canoe was at rest the net total torque is zero.

Let the center of mass is at x distance from the canoe center and it will be towards the Ricardo cause. So the toque around the center of mass is given as

[tex]m_{1}(L/2-x)=m_{2}x+m_{2}(L/2-x)[/tex]

We have to find m₂.To find the value of m₂ first we need figure out the value of.As they changed their positions the center of mass moved to other side by distance 2x.

so

2x=40

x=40/2

x=20 cm

Substitute in the above equation we get

[tex]m_{x}=\frac{m_{1}(L/2-x)-m_{2}x }{L/2+x}\\m_{x}=\frac{80(\frac{3}{2}-0.2 )-30*0.2}{3/2+0.2}\\m_{x}=57.65 kg[/tex]

The wave speed and wavelength will differ across strings of different thicknesses held at the same tension; the wave frequency remains the same.

When two strings of different thicknesses are made of the same material and held at the same tension, the wave speed and wavelength will differ, whereas the wave frequency will remain the same. This is because the speed of a wave on a string is determined by the tension (T) and the linear mass density (μ), where v = (T/μ)¹/2; as the tension is constant and the material is the same, it is the difference in thickness (hence, different densities) that causes a variance in wave speed. Since wave frequency (f) is related to the speed (v) and wavelength (λ) by the equation v = fλ, and the oscillator creates waves at a fixed frequency, a change in wave speed inherently impacts the wavelength.

The magnitude of the electric force of attraction between the iron nucleus and its innermost electron is approximately 2.645 × 10^-3 Newtons.

To calculate the magnitude of the electric force of attraction between an iron nucleus and its innermost electron, we can use Coulomb's law:

Given:

Charge of the iron nucleus, q1 = +26e

Charge of the electron, q2 = -e (where e is the elementary charge, 1.6 × 10^-19 C)

Distance between them, r = 1.5 × 10^-12 m

Convert the charge of the nucleus from elementary charges to coulombs:

q1 = +26e × 1.6 × 10^-19 C/e = +4.16 × 10^-18 C

Calculate the electric force using Coulomb's law:

F = (8.9875 × 10^9 N m^2/C^2) × |(+4.16 × 10^-18 C) × (-1.6 × 10^-19 C)| / (1.5 × 10^-12 m)^2

F ≈ (8.9875 × 10^9) × | -6.656 × 10^-37 | / (2.25 × 10^-24)

F ≈ (8.9875 × 10^9) × 2.947 × 10^-13

F ≈ 2.645 × 10^-3 N

So, the magnitude of the electric force of attraction between the iron nucleus and its innermost electron is approximately 2.645 × 10^-3 Newtons.

Answer:

d) F

Explanation:

According to columb's law:

"The magnitude of electrostatic force between two charges is directly proportional to the product of magnitude of two charges and inversly proportional to separation between them."

If q₁ and q₂ are magnitude of two charges, d is distance between them and k is dielectric constant, then force F is given by

                                              [tex]F=\frac{kq_{1}q_{2}}{d^2}[/tex]

According to this force exerted on point charge Q is same as that of 3Q, so force point 3Q charge experience is also F

Answer:

      T = 6.0 N

Explanation:

given,

mass of the cord = 0.46 Kg

length of the supports = 7.2 m

time taken to travel = 0.74 s

tension in the chord = ?

using formula for tension calculation

[tex]T = \dfrac{v^2.m}{l}[/tex]

[tex]v = \dfrac{l}{s}[/tex]

[tex]v = \dfrac{7.2}{0.74}[/tex]

v = 9.73 m/s

now, calculation of tension

[tex]T = \dfrac{9.73^2\times 0.46}{7.2}[/tex]

      T = 6.0 N

The tension in the cord is equal to 6.0 N.

Answer:

a. Electric field lines can never cross each other.

d. Wider spacing between electric field lines indicates a lower magnitude of electric field.

Explanation:

a. The electric field lines cannot be crossed, since this would mean that there would be more than one electric field vector for the same point at the place where the crossing occurs.

d. The space between the field lines is inversely proportional to the intensity of the electric field.

Answer: "a" and "d" are correct

Explanation:

(a) Field lines can never cross. Since a field line represents the direction of the field at a given point, if two field lines crossed at some point, that would imply that the electric field was pointing in two different directions at a single point.

(b) Wider spacing between electric field lines indicates a lower magnitude of electric field.