When an object with an electric charge of −0.080mC is 6.0 m from an object with an electric charge of 0.040mC, the force between them has a strength of 0.7989 N. Calculate the strength of the force between the two objects if they are 30.m apart. Round your answer to 2 significant digits.

Answers

Answer 1

The strength of the force between the two objects when they are 30.0 m apart is approximately 2.877 N (rounded to 2 significant digits).

The strength of the force between two charged objects can be calculated using Coulomb's Law:

F = k * (|q₁| * |q₂|) / r²

where F is the force, k is the electrostatic constant (k = 8.99 x 10^9 N·m²/C²), |q₁| and |q₂| are the magnitudes of the charges, and r is the distance between the charges.

Given:

Charge of object 1, q₁ = -0.080 mC = -0.080 x 10^(-3) C

Charge of object 2, q₂ = 0.040 mC = 0.040 x 10^(-3) C

Distance between the objects, r₁ = 6.0 m

Using the given values, we can calculate the strength of the force at 6.0 m:

F₁ = k * (|q₁| * |q₂|) / r₁²

F₁ = (8.99 x 10^9 N·m²/C²) * (| -0.080 x 10^(-3) C| * |0.040 x 10^(-3) C|) / (6.0 m)²

F₁ = (8.99 x 10^9 N·m²/C²) * (0.080 x 10^(-3) C * 0.040 x 10^(-3) C) / (6.0 m)²

F₁ = (8.99 x 10^9 N·m²/C²) * (0.032 x 10^(-6) C²) / (36.0 m²)

F₁ = (8.99 x 0.032 x 10^3) N

F₁ ≈ 287.68 N

Therefore, the strength of the force between the two objects when they are 6.0 m apart is approximately 287.68 N.

Now, let's calculate the strength of the force when the objects are 30.0 m apart:

Distance between the objects, r₂ = 30.0 m

Using Coulomb's Law, we can calculate the strength of the force at 30.0 m:

F₂ = k * (|q₁| * |q₂|) / r₂²

F₂ = (8.99 x 10^9 N·m²/C²) * (0.080 x 10^(-3) C * 0.040 x 10^(-3) C) / (30.0 m)²

F₂ = (8.99 x 10^9 N·m²/C²) * (0.032 x 10^(-6) C²) / (900.0 m²)

F₂ = (8.99 x 0.032 x 10^3) N

F₂ ≈ 2.877 N

Therefore, the strength of the force between the two objects when they are 30.0 m apart is approximately 2.877 N (rounded to 2 significant digits).

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Related Questions

Calculate the resultant vector C from the following cross product: Č = A x B where X = 3î + 2ỹ – lî and B = -1.5ê + +1.5ź =

Answers

Calculate the resultant vector C from the following cross product: Č = A x B where X = 3î + 2ỹ – lî and B = -1.5ê + +1.5ź

To calculate the resultant vector C from the cross product of A and B, we can use the formula:

C = A x B

Where A and B are given vectors. Now, let's plug in the values:

A = 3î + 2ỹ – lî

B = -1.5ê + 1.5ź

To find the cross product C, we can use the determinant method:

|i j k |

|3 2 -1|

|-1.5 0 1.5|

C = (2 x 1.5)î + (3 x 1.5)ỹ + (4.5 + 1.5)k - (-1.5 - 3)j + (-4.5 + 0)l + (-1.5 x 2)ê

C = 3î + 4.5ỹ + 6k + 4.5j + 4.5l - 3ê

Therefore, the resultant vector C is:

C = 3î + 4.5ỹ + 4.5j + 4.5l - 3ê + 6k

So, the answer is C = 3î + 4.5ỹ + 4.5j + 4.5l - 3ê + 6k.

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113 ft3/min water is to be delivered through a 250 foot long smooth pipe with a pressure drop of 5.2 psi. Determine the required pipe diameter as outlined using the following steps: a) Use 3 inches as your initial guess for the diameter of the pipe and indicate what your next guess would be. b) During design, it is determined that the actual pipeline will include 7 standard elbows and two open globe valves. Show how your calculations for part a) would need to be modified to account for these fittings.

Answers

a) The next guess for the pipe diameter would be Y inches.

b) The modified calculations would include the equivalent lengths of the fittings to determine the required pipe diameter.

To determine the required pipe diameter, we can use the Darcy-Weisbach equation, which relates the pressure drop in a pipe to various parameters including flow rate, pipe length, pipe diameter, and friction factor. We can iteratively solve for the pipe diameter using an initial guess and adjusting it until the calculated pressure drop matches the desired value.

a) Using 3 inches as the initial guess for the pipe diameter, we can calculate the friction factor and the resulting pressure drop. If the calculated pressure drop is greater than the desired value of 5.2 psi, we need to increase the pipe diameter. Conversely, if the calculated pressure drop is lower, we need to decrease the diameter.

b) When accounting for fittings such as elbows and valves, additional pressure losses occur due to flow disruptions. Each fitting has an associated equivalent length, which is a measure of the additional length of straight pipe that would cause an equivalent pressure drop. We need to consider these additional pressure losses in our calculations.

To modify the calculations for part a), we would add the equivalent lengths of the seven standard elbows and two open globe valves to the total length of the pipe. This modified length would be used in the Darcy-Weisbach equation to recalculate the required pipe diameter.

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In a small shire of Birmingham a 0.047 uF capacitor is being held at a potential difference of 32 uV. What is the charge of these Birmingham located plates, in coulombs, on one of the plates?

Answers

In a small shire of Birmingham a 0.047 μF capacitor is being held at a potential difference of 32 μV.  the charge on one of the plates of the capacitor located in Birmingham is approximately 1.504 × 10^-10 coulombs (C).

To find the charge on one of the plates of a capacitor, we can use the formula:

Q = C × V

Where:

Q is the charge on one of the plates,

C is the capacitance of the capacitor,

V is the potential difference across the capacitor.

In this case, the capacitance is given as 0.047 μF (microfarads) and the potential difference is 32 uV (microvolts). However, it is important to note that the unit of voltage used in the SI system is volts (V), not microvolts (uV). Therefore, we need to convert the potential difference to volts before calculating the charge.

1 μV = 1 × 10^-6 V

Therefore, 32 uV = 32 × 10^-6 V = 3.2 × 10^-5 V

Now we can calculate the charge using the formula:

Q = (0.047 μF) × (3.2 × 10^-5 V)

Since the unit of capacitance is microfarads (μF) and the unit of voltage is volts (V), the resulting unit of charge will be microcoulombs (μC).

Q = (0.047 × 10^-6 F) × (3.2 × 10^-5 V)

= 1.504 × 10^-10 C

Therefore, the charge on one of the plates of the capacitor located in Birmingham is approximately 1.504 × 10^-10 coulombs (C).

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A coil is connected in series with a 12.4 kΩ resistor. An ideal 48.8 V battery is applied across the two devices, and the current reaches a value of 2.57 mA after 4.27 ms. (a) Find the inductance of the coil. (b) How much energy is stored in the coil at this same moment?

Answers

(a) The inductance of the coil is approximately 81.33 H.

(b) The energy stored in the coil at this moment is approximately 2.097 × 10^-3 J.

To solve this problem, we can use the formula for the voltage across an inductor in an RL circuit and the formula for the energy stored in an inductor.

(a) The voltage across an inductor in an RL circuit is given by:

V = L * di/dt

where V is the applied voltage, L is the inductance, and di/dt is the rate of change of current with respect to time.

Given:

Applied voltage (V) = 48.8 V

Current (I) = 2.57 mA = 2.57 × 10^-3 A

Time (t) = 4.27 ms = 4.27 × 10^-3 s

Rearranging the formula, we have:

L = V / (di/dt)

Substituting the given values:

[tex]L = 48.8 V / (2.57 × 10^-3 A / 4.27 × 10^-3 s)\\L = 48.8 V / (0.6 A/s)\\L ≈ 81.33 H[/tex]

Therefore, the inductance of the coil is approximately 81.33 H.

(b) The energy stored in an inductor is given by the formula:

E = (1/2) * L * I^2

where E is the energy stored, L is the inductance, and I is the current.

Substituting the given values:

[tex]E = (1/2) * 81.33 H * (2.57 × 10^-3 A)^2\\E = (1/2) * 81.33 H * (6.6049 × 10^-6 A^2)\\E ≈ 2.097 × 10^-3 J[/tex]

Therefore, the energy stored in the coil at this moment is approximately 2.097 × 10^-3 J.

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In a water pistol, a piston drives water through a larger tube of radius 1.10 cm into a smaller tube of radius 1.50 mm as in the figure below. A₂ (i) (a) If the pistol is fired horizontally at a height of 1.40 m, use ballistics to determine the time it takes water to travel from the nozzle to the ground. (Neglect air resistance and assume atmospheric pressure is 1.00 atm. Assume up is the positive y-direction. Indicate the direction with the sign of your answer.) S (b) If the range of the stream is to be 7.70 m, with what speed must the stream leave the nozzle? m/s (c) Given the areas of the nozzle and cylinder, use the equation of continuity to calculate the speed at which the plunger must be moved. m/s (d) What is the pressure at the nozzle? (Give your answer to at least four significant figures.) Pa (e) Use Bernoulli's equation to find the pressure needed in the larger cylinder. Pa Can gravity terms be neglected? O Yes O No (f) Calculate the force that must be exerted on the trigger to achieve the desired range. (The force that ust be exerted is due to pressure over and above atmospheric pressure. Enter magnitude.) N

Answers

Summary:

In order to determine the time it takes for the water to travel from the nozzle to the ground when a water pistol is fired horizontally at a height of 1.40 m, we need to consider ballistics. By neglecting air resistance and assuming atmospheric pressure is 1.00 atm, we can calculate the time using the equations of motion. To achieve a range of 7.70 m, the speed at which the stream must leave the nozzle can be calculated using the range formula. By applying the equation of continuity, we can determine the speed at which the plunger must be moved. The pressure at the nozzle can be calculated using Bernoulli's equation, and the pressure needed in the larger cylinder can be found using the same equation.

Explanation:

(a) To calculate the time it takes for the water to travel from the nozzle to the ground, we can analyze the horizontal motion of the water. Since the water pistol is fired horizontally, the vertical component of the motion can be ignored. The height of the water pistol from the ground is given as 1.40 m. Using the equations of motion, we can determine the time it takes for the water to reach the ground.

(b) To achieve a range of 7.70 m, we can use the range formula for projectile motion. By considering the horizontal motion of the water, neglecting air resistance, and assuming an initial vertical displacement of 1.40 m, we can calculate the initial speed at which the stream must leave the nozzle.

(c) The equation of continuity states that the product of the cross-sectional area and the speed of a fluid is constant along a streamline. By using the areas of the nozzle and the cylinder, we can calculate the speed at which the plunger must be moved in order to maintain continuity.

(d) The pressure at the nozzle can be calculated using Bernoulli's equation, which relates the pressure, velocity, and height of a fluid. By neglecting air resistance and considering the fluid flow, we can determine the pressure at the nozzle.

(e) Bernoulli's equation can also be used to find the pressure needed in the larger cylinder. By considering the change in velocity and height between the nozzle and the larger tube, we can calculate the pressure required.

(f) The force that must be exerted on the trigger to achieve the desired range is due to the pressure difference. By considering the pressure over and above atmospheric pressure, we can calculate the magnitude of the force required.

Gravity terms can generally be neglected in this scenario, as we are primarily concerned with the horizontal and vertical components of motion and the fluid flow within the system.

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A uniform, solid cylinder of radius 7.00 cm and mass 5.00 kg starts from rest at the top of an inclined plane that is 2.00 m long and tilted at an angle of 21.0∘ with the horizontal. The cylinder rolls without slipping down the ramp. What is the cylinder's speed v at the bottom of the ramp? v= m/s

Answers

The speed of the cylinder at the bottom of the ramp can be determined by using the principle of conservation of energy.

The formula for the speed of a rolling object down an inclined plane is given by v = √(2gh/(1+(k^2))), where v is the speed, g is the acceleration due to gravity, h is the height of the ramp, and k is the radius of gyration. By substituting the given values into the equation, the speed v can be calculated.

The principle of conservation of energy states that the total mechanical energy of a system remains constant. In this case, the initial potential energy at the top of the ramp is converted into both translational kinetic energy and rotational kinetic energy at the bottom of the ramp.

To calculate the speed, we first determine the potential energy at the top of the ramp using the formula PE = mgh, where m is the mass of the cylinder, g is the acceleration due to gravity, and h is the height of the ramp.

Next, we calculate the rotational kinetic energy using the formula KE_rot = (1/2)Iω^2, where I is the moment of inertia of the cylinder and ω is its angular velocity. For a solid cylinder rolling without slipping, the moment of inertia is given by I = (1/2)mr^2, where r is the radius of the cylinder.

Using the conservation of energy, we equate the initial potential energy to the sum of translational and rotational kinetic energies:

PE = KE_trans + KE_rot

Simplifying the equation and solving for v, we get:

v = √(2gh/(1+(k^2)))

By substituting the given values of g, h, and k into the equation, we can calculate the speed v of the cylinder at the bottom of the ramp.

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In roughly 30-50 words, including an equation if needed,
explain what a "derivative" is in calculus, and explain what
physical quantity is the derivative of displacement if an object
moves

Answers

In calculus, the derivative represents the instantaneous rate of change. In this case, if an object moves 1449 meters downward in 18 seconds, its velocity is approximately 80.5 meters per second downward.

In calculus, a derivative represents the instantaneous rate of change of a quantity with respect to another. In the context of motion, the derivative of displacement is velocity.

To calculate the velocity, we can use the equation:

velocity (v) = change in displacement (Δx) / change in time (Δt)

Given that the object moves 1449 meters downward in 18 seconds, we can substitute these values into the equation:

v = 1449 meters / 18 seconds

Simplifying the equation, we find that the object has an average velocity of approximately 80.5 meters per second in the downward direction.

The complete question should be:

In roughly 30-50 words, including an equation, if needed, explain what a “derivative” is in calculus, and explain what physical quantity is the derivative of displacement if an object moves 1449 meters downward in 18 seconds.

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A series RLC circuit has resistance R = 65.0 M and inductance L = 0.685 H. The voltage source operates at a frequency of
f = 50.0 Hz and the reactance is Z = R = 65.0 0.
Find the circuit's capacitance C (in F).

Answers

The capacitance C of the series RLC circuit can be determined using the given values of resistance R, inductance L, and reactance Z.

In a series RLC circuit,

the impedance Z is given by Z = √(R^2 + (XL - XC)^2), where XL is the inductive reactance and XC is the capacitive reactance.

Given that Z = R = 65.0 Ω, we can equate the reactances to obtain XL - XC = 0.

Solving for XL and XC individually, we find that XL = XC.

The inductive reactance XL is given by XL = 2πfL, where f is the frequency and L is the inductance.

Plugging in the values, we have XL = 2π(50.0 Hz)(0.685 H).

Since XL = XC, the capacitive reactance XC is also equal to 2πfC, where C is the capacitance.

Equating the two expressions, we can solve for C.

By setting XL equal to XC, we have 2π(50.0 Hz)(0.685 H) = 1/(2πfC). Solving for C, we find that C = 1/(4π^2f^2L).

Substituting the given values, we can calculate the capacitance C in Farads.

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Three children are riding on the edge of a merry-go-round that is 122 kg, has a 1.60 m radius, and is spinning at 19.3 rpm. The children have masses of 22.4, 29.5, and 32.8 kg. If the child who has a mass of 29.5 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm? Approximate the merry-go-round as a solid disk, and each child as a point mass. X Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. rpm 24.91 [2.33/5 Points) DETAILS PREVIOUS ANSWERS MY NOTES C

Answers

Three children are riding on the edge of a merry-go-round that is 122 kg, has a 1.60 m radius, and is spinning at 19.3 rpm.  the new angular velocity in rpm when the child moves to the center of the merry-go-round is 19.3 rpm, which remains unchanged.

To solve this problem, we can apply the principle of conservation of angular momentum. Initially, the total angular momentum of the system is given by:

L_initial = I_initial * ω_initial,

where I_initial is the moment of inertia of the merry-go-round and ω_initial is the initial angular velocity.

When the child with a mass of 29.5 kg moves to the center, the moment of inertia of the system changes, but the total angular momentum remains conserved:

L_initial = L_final.

Let's calculate the initial and final angular velocities using the given information:

Given:

Mass of the merry-go-round (merry) = 122 kg

Radius of the merry-go-round (r) = 1.60 m

Angular velocity of the merry-go-round (ω_initial) = 19.3 rpm

Mass of the child moving to the center (m_child) = 29.5 kg

We'll calculate the initial and final moments of inertia using the formulas:

I_initial = 0.5 * m * r^2,  (for a solid disk)

I_final = I_merry + I_child,

where I_merry is the moment of inertia of the merry-go-round and I_child is the moment of inertia of the child.

Calculating the initial moment of inertia:

I_initial = 0.5 * m_merry * r^2

          = 0.5 * 122 kg * (1.60 m)^2

          = 195.2 kg·m^2.

Calculating the final moment of inertia:

I_final = I_merry + I_child

       = 0.5 * m_merry * r^2 + m_child * 0^2

       = 0.5 * 122 kg * (1.60 m)^2 + 29.5 kg * 0^2

       = 195.2 kg·m^2.

Since the child is at the center, its moment of inertia is zero.

Since the total angular momentum is conserved, we have:

I_initial * ω_initial = I_final * ω_final.

Solving for ω_final:

ω_final = (I_initial * ω_initial) / I_final.

Substituting the values we calculated:

ω_final = (195.2 kg·m^2 * 19.3 rpm) / 195.2 kg·m^2

        = 19.3 rpm.

Therefore, the new angular velocity in rpm when the child moves to the center of the merry-go-round is 19.3 rpm, which remains unchanged.

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a) Two reservoirs are connected to two pipes parallel to each other, as shown below. Pipe 1 has a diameter of 50 mm and length of 100 m, while pipe 2 has a diameter of 100 mm and length of 100 m. Given that the friction factor is 0.015, and minor losses are neglected, prove that discharge is approximately to 0.023. (10 marks)

Answers

The discharge through the parallel pipes can be approximately calculated as 0.023, considering the given parameters and neglecting minor losses.

To calculate the discharge through the parallel pipes, we can use the Darcy-Weisbach equation, which relates the flow rate (Q) to the friction factor (f), pipe diameter (D), pipe length (L), and the pressure drop (ΔP). In this case, we neglect minor losses, so we only consider the frictional losses in the pipes.

Calculate the hydraulic diameter (Dh) for each pipe:

For pipe 1: Dh1 = 4 * (cross-sectional area of pipe 1) / (wetted perimeter of pipe 1)

For pipe 2: Dh2 = 4 * (cross-sectional area of pipe 2) / (wetted perimeter of pipe 2)

Calculate the Reynolds number (Re) for each pipe:

For pipe 1: Re1 = (velocity in pipe 1) * Dh1 / (kinematic viscosity of fluid)

For pipe 2: Re2 = (velocity in pipe 2) * Dh2 / (kinematic viscosity of fluid)

Calculate the friction factor (f) for each pipe:

For pipe 1: f1 = 0.015 (given)

For pipe 2: f2 = 0.015 (given)

Calculate the velocity (v) for each pipe:

For pipe 1: v1 = (discharge in pipe 1) / (cross-sectional area of pipe 1)

For pipe 2: v2 = (discharge in pipe 2) / (cross-sectional area of pipe 2)

Set up the equation for the total discharge (Q) through the parallel pipes:

Q = (discharge in pipe 1) + (discharge in pipe 2)

Use the equation for the Darcy-Weisbach friction factor:

f1 = (2 * g * Dh1 * (discharge in pipe 1)^2) / (π^2 * L * (pipe 1 diameter)^5)

f2 = (2 * g * Dh2 * (discharge in pipe 2)^2) / (π^2 * L * (pipe 2 diameter)^5)

Rearrange the equations to solve for the discharge in each pipe:

(discharge in pipe 1) = √((f1 * π^2 * L * (pipe 1 diameter)^5) / (2 * g * Dh1))

(discharge in pipe 2) = √((f2 * π^2 * L * (pipe 2 diameter)^5) / (2 * g * Dh2))

Substitute the given values and calculate the discharge in each pipe.

Calculate the total discharge by summing the individual discharges from each pipe:

Q = (discharge in pipe 1) + (discharge in pipe 2)

Substitute the given values and calculate the total discharge through the parallel pipes.

By following these steps and considering the given parameters, we can approximate the discharge to be approximately 0.023.

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With help from the preceding rules, verify the answers to the following equations:(4.0 ×10⁸) (9.0 ×10⁹)=3.6 ×10¹⁸

Answers

Comparing the result to the given answer  from the preceding rules, we can see that the given answer is incorrect. The correct answer is 36 × 10¹⁷, not 3.6 × 10¹⁸.

To verify the answer to the equation (4.0 × 10⁸) (9.0 × 10⁹) = 3.6 × 10¹⁸, we can use the rules of multiplication with scientific notation.

Step 1: Multiply the coefficients (the numbers before the powers of 10): 4.0 × 9.0 = 36.

Step 2: Add the exponents of 10: 8 + 9 = 17.

Step 3: Write the product in scientific notation: 36 × 10¹⁷.

Comparing the result to the given answer, we can see that the given answer is incorrect. The correct answer is 36 × 10¹⁷, not 3.6 × 10¹⁸.

In summary, when multiplying numbers in scientific notation, you multiply the coefficients and add the exponents of 10. This helps us express very large or very small numbers in a compact and convenient form.

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A 1.0 kQ resistor is connected to a 1.5 V battery. The current
through the resistor is equal to a.1.5mA
b 1.5KA
d1.5A
c 1.5 μA

Answers

The correct answer is (d) 1.5 A.

The current through a resistor connected to a battery can be calculated using Ohm's Law, which states that the current  (I) flowing through a resistor is equal to the voltage (V) across the resistor divided by its resistance (R). Mathematically, it can be expressed as I = V/R.

In this case, the voltage across the resistor is given as 1.5 V, and the resistance is 1.0 kΩ (which is equivalent to 1000 Ω). Plugging these values into Ohm's Law, we get I = 1.5 V / 1000 Ω = 0.0015 A = 1.5 A.

Therefore, the current through the 1.0 kΩ resistor connected to the 1.5 V battery is 1.5 A.

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A 1.10 kg hollow steel ball is submerged in water. Its weight in water is 8.75 N. Find the volume of the cavity inside the ball is (density of steel is 7.99 g/cc).

Answers

the volume of the cavity inside the ball is 5.3 × 10⁻⁴ m³.

The density of water is 1 g/cc or 1000 kg/m³. The density of steel is 7.99 g/cc or 7990 kg/m³. Therefore, the weight of a 1.10 kg steel ball in water can be expressed as follows;

Weight of steel ball in water = Weight of steel ball - Buoyant force

[tex]W = mg - Fb[/tex]

From the question, weight in water is 8.75 N, and the mass of the steel ball is 1.10 kg. Therefore,  W = 8.75 N and m = 1.10 kg.

Substituting the values in the equation above, we have;

8.75 N = (1.10 kg) (9.8 m/s²) - Fb

Solving for Fb, we have

Fb = 1.10 (9.8) - 8.75

= 0.53 N

The buoyant force is equal to the weight of the water displaced.

Thus, volume = (Buoyant force) / (density of water)

Substituting the values in the equation above, we have;

V = Fb / ρV

= 0.53 N / (1000 kg/m³)

V = 0.00053 m³

= 5.3 × 10⁻⁴ m³

Hence, the volume of the cavity inside the ball is 5.3 × 10⁻⁴ m³.

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Through a resistor connected to two batteries in series of 1.5 V
each, a current of 3 mA passes. How much is the resistance of this
element.
a. 0.5KQ
b. 1.00
c 1.0 MQ
d. 1.0 kQ

Answers

Using Ohm's Law, we find that the resistance of the element is 1.0 kΩ. The correct option is d).

Ohm's Law states that the current passing through a resistor is directly proportional to the voltage across it and inversely proportional to its resistance.

Ohm's Law: V = I * R

Where:

V is the voltage across the resistor (in volts)

I is the current passing through the resistor (in amperes)

R is the resistance of the resistor (in ohms)

In this case, we have two batteries in series, each with a voltage of 1.5V. The total voltage across the resistor is the sum of the voltages of both batteries:

V = 1.5V + 1.5V = 3V

The current passing through the resistor is given as 3 mA, which is equivalent to 0.003 A.

Now, we rearrange Ohm's Law to solve for the resistance:

R = V / I

R = 3V / 0.003A

R = 1000 ohms = 1 kΩ

Therefore, the resistance of the element is 1.0 kΩ. The correct option is d).

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Question 27 1 pts Are cosmic rays a form of light? Yes, they are light waves with higher energy than gamma rays. No, they consist of high-energy subatomic particles, not of electromagnetic waves. Yes, they are the name we give to all rays of light that come from the cosmos. Yes, they are light waves with lower energy than radio waves.

Answers

No, cosmic rays are not a form of light. Cosmic rays consist of high-energy subatomic particles, such as protons, electrons, and atomic nuclei, rather than electromagnetic waves. They are not part of the electromagnetic spectrum like light waves. Cosmic rays originate from various astrophysical sources, such as supernovae, active galactic nuclei, and other high-energy events in the universe. These particles are accelerated to extremely high energies and can travel through space, reaching Earth's atmosphere.

Upon interaction with the atmosphere, they can produce secondary particles, leading to cascades of particles known as air showers. While cosmic rays can have interactions with matter and electromagnetic fields, they are fundamentally distinct from light waves and do not belong to the category of electromagnetic radiation.

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5000 heat calories are added to 7800 g of tungsten at 37.0°C.
What will be the final temperature of the tungsten?

Answers

The final temperature of the tungsten can be determined using the specific heat capacity and the principle of conservation of energy.

To find the final temperature of the tungsten, we need to consider the amount of heat energy added to it and its specific heat capacity. The specific heat capacity of tungsten is 0.032 cal/g°C.

The formula to calculate the heat energy absorbed or released by an object is Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

In this case, the heat energy added is 5000 calories, the mass of the tungsten is 7800 grams, and the initial temperature is 37.0°C. We can rearrange the formula to solve for the change in temperature:

ΔT = Q / (mc)

Substituting the given values, we have:

ΔT = 5000 cal / (7800 g * 0.032 cal/g°C) ≈ 6.41°C

To find the final temperature, we add the change in temperature to the initial temperature:

Final temperature = 37.0°C + 6.41°C ≈ 43.41°C

Therefore, the final temperature of the tungsten will be approximately 43.41°C.

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How much work is needed to bring a + 5.0 µC point charge from infinity to a point 2.0 m away from a + 25 µC charge? (you may assume that it is moved at a constant, controlled velocity so that there is no change in kinetic energy)

Answers

The work required to bring a +5.0 µC point charge from infinity to a point 2.0 m away from a +25 µC charge is 6.38 × 10^-5 joules.

To calculate the work, we can use the equation: Work = q1 * q2 / (4πε₀ * r), where q1 and q2 are the charges, ε₀ is the permittivity of free space, and r is the distance between the charges. Plugging in the given values, we get Work = (5.0 µC * 25 µC) / (4πε₀ * 2.0 m). Evaluating the expression, we find the work to be 6.38 × 10^-5 joules.Therefore, the work required to bring the +5.0 µC point charge from infinity to a point 2.0 m away from the +25 µC charge is 6.38 × 10^-5 joules.

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Aray of light impinges on a mirror as shown in (Eigure 1) = 40" A second mirror is fastened at 90 to the first Part A e Figure s not At what angle above horizontal does the ray emerge after reflecting from both mirrors? Express your answer in degrees. VAX ? 0 - 170 Submit Previous Answers Request Answer

Answers

The angle above horizontal at which the ray emerges after reflecting from both mirrors is 50 degrees.

When a ray of light impinges on the first mirror at an angle of 40 degrees, it reflects at the same angle due to the law of reflection. Now, the second mirror is fastened at a 90-degree angle to the first mirror, which means the ray will reflect vertically upwards.

To find the angle above horizontal at which the ray emerges, we need to consider the angle of incidence on the second mirror. Since the ray is reflected vertically upwards, the angle of incidence on the second mirror is 90 degrees.

Using the principle of alternate angles, we can determine that the angle of reflection on the second mirror is also 90 degrees. Now, the ray is traveling in a vertical direction.

To find the angle above horizontal, we need to measure the angle between the vertical direction and the horizontal direction. Since the vertical direction is perpendicular to the horizontal direction, the angle above horizontal is 90 degrees.

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1. (1 p) A circular loop of 200 turns and 12 cm diameter is designed to rotate 90° in 0.2 sec. Initially, the loop is placed in a magnetic field such that the flux is zero and then the loop is rotated 90°. If the electromotive force induced in the loop is 0.4 mV, what is the magnitude of the magnetic field?

Answers

The magnitude of the magnetic field is determined as 3.64 x 10⁻⁴ T.

What is the magnitude of the magnetic field?

The magnitude of the magnetic field is calculated by applying the following formula as follows;

emf = NdФ/dt

emf = NBA sinθ / t

where;

N is the number of turnsB is the magnetic fieldA is the area of the circular loopθ is orientation anglet is the time

The area of the circular loop is calculated as;

A = πr²

r = 12cm/2 = 6 cm = 0.06 m

A = π x (0.06 m)²

A = 0.011 m²

The magnitude of the magnetic field is calculated as;

emf = NBA sinθ/t

B = (emf x t) / (NA x sinθ)

B = (4 x 10⁻³ V x 0.2 s ) / ( 200 x 0.011 m² x sin (90))

B = 3.64 x 10⁻⁴ T

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D Question 1 20 pts Water is moving at a rate of 4.79 m/s through a pipe with a cross sectional area of 4.00cm2. The water gradually descends 9.56m as the pipe increases in area to 8.50 cm². The pressure at the upper level is 152kPa what is the pressure at the lower level? Give your answer in units of kPa (kilo pascals!)

Answers

Given parameters:Velocity of water, v = 4.79 m/sCross-sectional area of the first pipe, A1 = 4.00 cm²Change in height, h = 9.56 mCross-sectional area of the second pipe, A2 = 8.50 cm²Pressure at the upper level, P1 = 152 kPaTo find: Pressure at the lower level, P2Formula used:Bernoulli's equation states that:P1 + 1/2pv1² + pgh1 = P2 + 1/2pv2² + pgh2Where,p is the density of water;v is the velocity of water;g is the acceleration due to gravity (9.8 m/s²);h is the height difference between the two points.

Substituting the given values:P1 + 1/2ρv₁² + ρgh1 = P2 + 1/2ρv₂² + ρgh2Rearranging the above equation, we get:P2 = P1 + 1/2ρ(v₁² - v₂²) + ρg(h2 - h1)Convert the cross-sectional area of the pipe to m²:1 cm² = 10⁻⁴ m²A1 = 4.00 cm² = 4.00 x 10⁻⁴ m²A2 = 8.50 cm² = 8.50 x 10⁻⁴ m²Convert the pressure to Pa:1 kPa = 1000 PaP1 = 152 kPa = 152 x 1000 PaSubstitute the given values and solve for P2:P2 = 152000 + 1/2 x 1000 x (4.79² - 0) + 1000 x 9.8 x (0 - 9.56)P2 = 152000 + 1/2 x 1000 x 22.9721 + 1000 x 9.8 x (-9.56)P2 = 152000 + 11486.052 - 9380.16P2 = 154105.89 PaTherefore, the pressure at the lower level is 154.106 kPa (rounded to three decimal places).

Explanation:This question is based on Bernoulli's equation, which relates the pressure, velocity, and height of a fluid flowing through a pipe. The Bernoulli's equation states that P1 + 1/2pv1² + pgh1 = P2 + 1/2pv2² + pgh2where P1 and P2 are the pressures at two points in the fluid flow; v1 and v2 are the velocities at these two points; h1 and h2 are the heights of these two points; p is the density of the fluid; and g is the acceleration due to gravity.Using the given parameters, we can substitute the values in the equation and solve for the pressure at the lower level. After substituting the values, we get P2 = 152000 + 1/2 x 1000 x (4.79² - 0) + 1000 x 9.8 x (0 - 9.56). By solving this equation, we get P2 = 154105.89 Pa. Therefore, the pressure at the lower level is 154.106 kPa (rounded to three decimal places).

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A concave lens has a focal length of -f. An object is placed between f and 2f on the axis. The image is formed at
Group of answer choices
A. at 2f.
B. Between f and the lens.
C. at f.
D.at a distance greater than 2f from the lens.

Answers

An object placed between f and 2f on the axis of the concave lens, the image is formed between f and the lens. Thus, the correct answer is Option B.

When an object is placed between the focal point (f) and the centre (2f) of a concave lens, the image formed is virtual, upright, and located on the same side as the object. It will appear larger than the object. This is known as a magnified virtual image.

In this situation, the object is positioned closer to the lens than the focal point. As a result, the rays of light from the object pass through the lens and diverge. These diverging rays can be extended backwards to intersect at a point on the same side as the object. This intersection point is where the virtual image is formed.

Since the virtual image is formed on the same side as the object, between the object and the lens, the correct answer is Option B. Between f and the lens.

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The volume (V) of the cone below is given by: Vrh where: R in the radio and his the beight of the cone What is the absolute error in V? Ah AP P 2AR R SR - - 24 R R Ос AV AR AP - 2AR R + Ah Ов AP

Answers

The volume (V) of the cone below is given by: Vrh where: R in the radio and his the beight of the cone, the absolute error in the volume of the

cone is given by: ΔV = (2/3)πR(|hΔR| + |RΔh|)

To find the absolute error in the volume of the cone, we need to consider the errors in the radius (ΔR) and height (Δh), and then calculate the resulting error in the volume (ΔV).

Given:

Volume of the cone: V = (1/3)πR^2h

Error in the radius: ΔR

Error in the height: Δh

To calculate the absolute error in the volume (ΔV), we can use the formula for error propagation:

ΔV = |(∂V/∂R)ΔR| + |(∂V/∂h)Δh|

First, let's calculate the partial derivatives of V with respect to R and h:

(∂V/∂R) = (2/3)πRh

(∂V/∂h) = (1/3)πR^2

Substituting these values into the formula for the absolute error in V, we have:

ΔV = |(2/3)πRhΔR| + |(1/3)πR^2Δh|

Simplifying further, we can factor out πR from both terms:

ΔV = (2/3)πR(|hΔR| + |RΔh|)

Therefore, the absolute error in the volume of the cone is given by:

ΔV = (2/3)πR(|hΔR| + |RΔh|)

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3. A wheel is spinning at w = 24.62 rad's and is slowing down with an acceleration that has a magnitude of a = 11.24 rad/s². In how many Second does it stop?

Answers

The wheel, initially spinning at a rate of 24.62 rad/s, experiences a deceleration of 11.24 rad/s². We find that the wheel will stop rotating after approximately 2.19 seconds.

The equation of motion for rotational motion is given by:

ω = ω₀ + αt, where ω is the final angular velocity, ω₀ is the initial angular velocity, α is the angular acceleration, and t is the time taken. In this case, the wheel is slowing down, so the final angular velocity ω will be 0.

Plugging in the values, we have:

0 = 24.62 rad/s + (-11.24 rad/s²) * t.

Rearranging the equation, we get:

11.24 rad/s² * t = 24.62 rad/s.

Solving for t, we find:

t = 24.62 rad/s / 11.24 rad/s² ≈ 2.19 s.Therefore, it will take approximately 2.19 seconds for the wheel to stop rotating completely.

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A 2.5-cm-tall object is 13 cm in front of a concave mirror that has a 25 cm focal length.Part A: Calculate the image position.
Express your answer to two significant figures and include the appropriate units.
Part B: Calculate the image height. Type a positive value if the image is upright and a negative value if it is inverted.
Express your answer to two significant figures and include the appropriate units.

Answers

The image height is approximately 5.20 cm, and it is upright. To calculate the image position and height, we can use the mirror equation.

1/f =[tex]1/d_i + 1/d_o[/tex]

where:

f = focal length of the mirror (given as 25 cm)

[tex]d_i[/tex]= image distance

[tex]d_o[/tex] = object distance

[tex]d_o[/tex] = -13 cm (since the object is in front of the mirror)

f = 25 cm

Part A: Calculate the image position.

Substituting the values into the mirror equation:

1/25 = 1/[tex]d_i[/tex] + 1/(-13)

To solve for [tex]d_i[/tex], we can rearrange the equation:

1/[tex]d_i[/tex] = 1/25 - 1/(-13)

1/[tex]d_i[/tex] = (13 - 25)/(25 * (-13))

1/[tex]d_i[/tex] = -12/(-325)

[tex]d_i[/tex] = (-325)/(-12)

[tex]d_i[/tex] ≈ 27.08 cm

Therefore, the image position is approximately 27.08 cm behind the mirror.

Part B: Calculate the image height.

To determine the image height, we can use the magnification formula:

m = -[tex]d_i[/tex]/[tex]d_o[/tex]

where:

m = magnification

[tex]d_i[/tex] = image distance (calculated as 27.08 cm)

[tex]d_o[/tex] = object distance (-13 cm)

Substituting the values:

m = -27.08/(-13)

m ≈ 2.08

The magnification tells us whether the image is upright or inverted. Since the magnification is positive (2.08), the image is upright.

To find the image height, we can multiply the magnification by the object height:

[tex]h_i = m * h_o[/tex]

where:

[tex]h_i[/tex]= image height

[tex]h_o[/tex] = object height

Given:

[tex]h_o[/tex] = 2.5 cm

Substituting the values:

[tex]h_i[/tex] = 2.08 * 2.5

[tex]h_i[/tex] ≈ 5.20 cm

Therefore, the image height is approximately 5.20 cm, and it is upright.

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An advanced lat student is studying the effect of temperature on the resistance of a current carrying wire. She applies a voltage to a tungsten wire at a temperature of 59.0"C and notes that it produces a current of 1.10 A she then applies the same voltage to the same wire at -880°C, what current should she expect in A? The temperature coefficient of resistity for tungsten 450 x 10(°C) (Assume that the reference temperature is 20°C)

Answers

The current that the advanced lat student should expect in A is 9.376 × 10⁻⁷ A.

Given data:

Initial temperature of tungsten wire, t₁ = 59.0°C

Initial current produced, I₁ = 1.10 A

Voltage applied, V = Same

Temperature at which voltage is applied, t₂ = -880°C

Temperature coefficient of resistivity of tungsten, α = 450 × 10⁻⁷/°C

Reference temperature, Tref = 20°C

We can calculate the resistivity of tungsten at 20°C, ρ₂₀, as follows:

ρ₂₀ = ρ₁/(1 + α(t₁ - Tref))

ρ₂₀ = ρ₁/(1 + 450 × 10⁻⁷ × (59.0 - 20))

ρ₂₀ = ρ₁/1.0843925

Now, let's calculate the initial resistance, R₁:

R₁ = V/I₁

Next, we can calculate the final resistance, R₂, of the tungsten wire at -880°C:

R₂ = ρ₁/[1 + α(t₂ - t₁)]

Substituting the values, we get:

R₂ = ρ₂₀ × 1.0843925/[1 + 450 × 10⁻⁷ × (-880 - 59.0)]

R₂ = 1.17336 × 10⁶ ohms (approx.)

Using Ohm's law, we can calculate the current, I₂:

I₂ = V/R₂

I₂ = 1.10/1.17336 × 10⁶

I₂ = 9.376 × 10⁻⁷ A or 0.9376 µA (approx.)

Therefore, the current that the advanced lat student should expect is approximately 9.376 × 10⁻⁷ A or 0.9376 µA.

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1. Two equal-mass hockey pucks undergo a glancing collision. Puck 1 is initially at rest and is struck by puck 2 travelling at a velocity of 13 m/s [E]. After the collision Puck 1 travels at an angle of [E 18° N] with a velocity of 20m/s . what is he velocity and direction of Puck 2 [ 4T] [4A)

Answers

The velocity and direction of Puck 2 after the glancing collision can be determined by solving equations based on conservation of momentum and kinetic energy.

In a glancing collision between two equal-mass hockey pucks, where Puck 1 is initially at rest and is struck by Puck 2 traveling at a velocity of 13 m/s [E], the resulting motion can be determined. After the collision, Puck 1 moves at an angle of [E 18° N] with a velocity of 20 m/s.

To find the velocity and direction of Puck 2 after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.

Since the masses of the pucks are equal, we know that the magnitude of the momentum before and after the collision will be the same.

Let's assume that Puck 2 moves at an angle θ with respect to the east direction. Using vector addition, we can break down the velocity of Puck 2 into its horizontal and vertical components. The horizontal component of Puck 2's velocity will be 13 cos θ, and the vertical component will be 13 sin θ.

After the collision, the horizontal component of Puck 1's velocity will be 20 cos (90° - 18°) = 20 cos 72°, and the vertical component will be 20 sin (90° - 18°) = 20 sin 72°.

To satisfy the conservation of momentum, the horizontal component of Puck 2's velocity must be equal to the horizontal component of Puck 1's velocity, and the vertical components must cancel each other out.

Therefore, we have:

13 cos θ = 20 cos 72° (Equation 1)

13 sin θ - 20 sin 72° = 0 (Equation 2)

Solving these equations simultaneously will give us the value of θ, which represents the direction of Puck 2. By substituting this value back into Equation 1, we can calculate the magnitude of Puck 2's velocity.

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The
weight of an object is 5N. When the object is suspended on a spring
balance and immersed in water, the reading on the balance is 3.5
Find the density of the object.

Answers

The density of the object is  1000 kg/m³ when weight of the object is 5N and  the reading on the balance is 3.5.

Given Weight of the object (W) = 5 N

Reading on the spring balance (S) = 3.5 N

Since the reading on the spring balance is the apparent weight of the object in water, it is equal to the difference between the weight of the object in air and the buoyant force acting on it.

Apparent weight of the object in water (W_apparent) = W - Buoyant force

Buoyant force (B) = Weight of displaced water

To find the density of the object, we need to determine the volume of water displaced by the object.

Since the weight of the object is equal to the weight of the displaced water, we can equate the weights:

W = Weight of displaced water

5 N = Weight of displaced water

The volume of water displaced by the object is equal to the volume of the object.

Now, let's calculate the density of the object:

Density (ρ) = Mass (m) / Volume (V)

Since the weight (W) is equal to the product of mass (m) and acceleration due to gravity (g), we have:

W = mg

Rearranging the formula, we can find the mass:

m = W / g

Given that g is approximately 9.8 m/s², substituting the values:

m = 5 N / 9.8 m/s²

= 0.51 kg

Since the volume of water displaced by the object is equal to its volume, we can calculate the volume using the formula:

Volume (V) = Mass (m) / Density (ρ)

Substituting the known values:

Volume (V) = 0.51 kg / ρ

Since the weight of water displaced is equal to the weight of the object:

Weight of displaced water = 5 N

Using the formula for the weight of water:

Weight of displaced water = ρ_water × V × g

Where ρ_water is the density of water and g is the acceleration due to gravity.

Substituting the known values:

5 N = (1000 kg/m³) × V × 9.8 m/s²

Simplifying the equation:

V = 5 N / ((1000 kg/m³) × 9.8 m/s²)

= 0.00051 m³

Now, we can calculate the density of the object:

Density (ρ) = Mass (m) / Volume (V)

ρ = 0.51 kg / 0.00051 m³

= 1000 kg/m³

Therefore, the density of the object is approximately 1000 kg/m³.

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Problem 2 (30 points) A microscopic spring-mass system has a mass m=7 x 10-26 kg and the energy gap between the 2nd and 3rd excited states is 1 eV. a) (2 points) Calculate in joules, the energy gap between the lst and 2nd excited states: E= J b) (2 points) What is the energy gap between the 4th and 7th excited states: E= ev c) (1 point) To find the energy of the ground state, which equation can be used ? (check the formula_sheet and select the number of the equation) d) (1 point) Which of the following substitutions can be used to calculate the energy of the ground state? 0 (6.582 x 10-16) (1) (6.582 x 10-16) (1) (6.582x10-16) 01 O2 X 1 e) (3 points) The energy of the ground state is: E= eV f) (1 point) To find the stiffness of the spring, which equation can be used ? (check the formula_sheet and select the number of the equation)

Answers

a) The energy gap between the first and second excited states is 9 eV, which is equal to 1.442 × 10^-18 J.

b) The energy gap between the fourth and seventh excited states is 27 eV.

c) The equation used to find the energy of the ground state is E = (n + 1/2) × h × f.

d) The correct substitution to calculate the energy of the ground state is (1/2) × (6.582 × 10^-16 J·s) × 9.

e) The energy of the ground state is E = (1/2) × (6.582 × 10^-16 J·s) × 9 eV.

f) The stiffness of the spring can be found using the equation k = mω^2.

a) To calculate the energy gap between the first and second excited states, we can assume that the energy levels are equally spaced. Given that the energy gap between the second and third excited states is 9 eV, we can conclude that the energy gap between the first and second excited states is also 9 eV. Converting this to joules, we use the conversion factor 1 eV = 1.602 × 10^−19 J. Therefore, the energy gap between the first and second excited states is E = 9 × 1.602 × 10^−19 J.

b) Since we are assuming equally spaced energy levels, the energy gap between any two excited states can be calculated by multiplying the energy gap between adjacent levels by the number of levels between them. In this case, the energy gap between the fourth and seventh excited states is 3 times the energy gap between the second and third excited states. Therefore, the energy gap between the fourth and seventh excited states is 3 × 9 eV = 27 eV.

c) The energy of the ground state can be calculated using the equation E = (n + 1/2) × h × f, where E is the energy, n is the quantum number (0 for the ground state), h is the Planck's constant (6.626 × 10^−34 J·s), and f is the frequency.

d) The correct substitution to calculate the energy of the ground state is (1/2) × (6.582 × 10^−16 J·s) × 9.

e) Substituting the values, the energy of the ground state is E = (1/2) × (6.582 × 10^−16 J·s) × 9 eV.

f) To find the stiffness of the spring, we can use Hooke's law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position. The equation for the stiffness of the spring is given by k = mω^2, where k is the stiffness, m is the mass, and ω is the angular frequency.

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A microscopic spring-mass system has a mass m=7 x 10⁻²⁶ kg and the energy gap between the 2nd and 3rd excited states is 9 eV.

a) Calculate in joules, the energy gap between the lst and 2nd excited states: E=____ J

b) What is the energy gap between the 4th and 7th excited states: E= ____ ev

c) To find the energy of the ground state, which equation can be used ? (check the formula_sheet and select the number of the equation)

d) Which of the following substitutions can be used to calculate the energy of the ground state?

2 x 9

(6.582 × 10⁻¹⁶) (9)

(6.582x10⁻¹⁶)²/2

1/2(6.582 x 10⁻¹⁶) (9)

(1/2)9

e) (The energy of the ground state is: E= ____ eV

f) (1 point) To find the stiffness of the spring, which equation can be used ? (check the formula_sheet and select the number of the equation)

Consider a cube whose volume is 125 cm3. Inside there are two point charges q1 = -24 picoC and q2 = 9 picoC. The flux of the electric field across the surface of the cube is:
Select one:
a. 2.71 N/A
b. -1.69 N/A
c. -5.5N/A
d. 1.02 N/A

Answers

Consider a cube whose volume is 125 cm3. Inside there are two point charges q1 = -24 picoC and q2 = 9 picoC. The electric field's flux across the cube's surface is -1.69 N/A.

An electric field is a vector field produced by electric charges that affect other electrically charged objects in the field. Flux of Electric Field: A measure of the flow of an electric field through a particular surface is referred to as electric flux.

The formula for calculating the electric flux through a surface area S with an electric field E that makes an angle θ to the surface normal is given by; Φ = ES cos θ Where E is the electric field and S is the surface area. If q is the total charge enclosed by a surface S, the electric flux through the surface is given by; Φ = q/ε₀ Where q is the total charge enclosed by the surface, and ε₀ is the permittivity of free space.

Consider a cube whose volume is 125 cm³. Inside there are two point charges q1 = -24 picoC and q2 = 9 picoC.The total charge enclosed by the cube is given by;q = q1 + q2= -24 + 9 = -15 pico C The electric flux through the cube is proportional to the total charge enclosed inside the surface. Hence the electric flux through the cube is given byΦ = q/ε₀ = -15 × 10^-12 / 8.85 × 10^-12= -1.69 N/A Therefore, the correct option is b. -1.69 N/A.

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A string is fixed at both ends. The mass of the string is 0.0010 kg and the length is 3.35 m. The string is under a tension of 195 N. The string is driven by a variable frequency source to produce standing waves on the string. Find the wavelengths and frequencies of the first four modes of standing waves.

Answers

The wavelengths and frequencies of the first four modes of standing waves on the string are approximately: Mode 1 - λ = 6.70 m, f = 120.6 Hz; Mode 2 - λ = 3.35 m, f = 241.2 Hz; Mode 3 - λ ≈ 2.23 m, f ≈ 362.2 Hz; Mode 4 - λ = 3.35 m, f = 241.2 Hz.

To find the wavelengths and frequencies of the first four modes of standing waves on the string, we can use the formula:

λ = 2L/n

Where:

λ is the wavelength,

L is the length of the string, and

n is the mode number.

The frequencies can be calculated using the formula:

f = v/λ

Where:

f is the frequency,

v is the wave speed (determined by the tension and mass per unit length of the string), and

λ is the wavelength.

Given:

Mass of the string (m) = 0.0010 kg

Length of the string (L) = 3.35 m

Tension (T) = 195 N

First, we need to calculate the wave speed (v) using the formula:

v = √(T/μ)

Where:

μ is the linear mass density of the string, given by μ = m/L.

μ = m/L = 0.0010 kg / 3.35 m = 0.0002985 kg/m

v = √(195 N / 0.0002985 kg/m) = √(652508.361 N/m^2) ≈ 808.03 m/s

Now, we can calculate the wavelengths (λ) and frequencies (f) for the first four modes (n = 1, 2, 3, 4):

For n = 1:

λ₁ = 2L/1 = 2 * 3.35 m = 6.70 m

f₁ = v/λ₁ = 808.03 m/s / 6.70 m ≈ 120.6 Hz

For n = 2:

λ₂ = 2L/2 = 3.35 m

f₂ = v/λ₂ = 808.03 m/s / 3.35 m ≈ 241.2 Hz

For n = 3:

λ₃ = 2L/3 ≈ 2.23 m

f₃ = v/λ₃ = 808.03 m/s / 2.23 m ≈ 362.2 Hz

For n = 4:

λ₄ = 2L/4 = 3.35 m

f₄ = v/λ₄ = 808.03 m/s / 3.35 m ≈ 241.2 Hz

Therefore, the wavelengths and frequencies of the first four modes of standing waves on the string are approximately:

Mode 1: Wavelength (λ) = 6.70 m, Frequency (f) = 120.6 Hz

Mode 2: Wavelength (λ) = 3.35 m, Frequency (f) = 241.2 Hz

Mode 3: Wavelength (λ) ≈ 2.23 m, Frequency (f) ≈ 362.2 Hz

Mode 4: Wavelength (λ) = 3.35 m, Frequency (f) = 241.2 Hz

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