a thin spherical metallic shell of radius 2.31 cm is has a charge of -3.1 uc uniformly distributed on its surface. what direction does the electric field point at a distance of 4.2 cm from the origin? a. there is no electric field b. radially outward c. tangential to the surface d. radially inward

Answers

Answer 1

The electric field's direction is radially inward because the charge is negative. Hence, the correct option is (d)

The given electric charge is -3.1uc, uniformly distributed over a thin spherical metallic shell with a radius of 2.31cm.

The electric field direction at a distance of 4.2cm from the origin can be found using Gauss's law.

The electric field is proportional to the electric charge enclosed by the spherical metallic shell within a closed surface.

The electric flux is defined as the electric field passing through the closed surface divided by the electric field.

For instance, The electric flux through a closed surface is proportional to the electric charge enclosed by that surface.

E is the electric field,

ΦE is the electric flux, and

Qenc is the electric charge enclosed by the surface in Gauss's law.

Here, the Gaussian surface is a sphere with a radius of 4.2cm.

We can calculate the electric field direction using the same formula as before, which is given by;

E = Qenc/4πε0r², where r = 4.2cm

Let's substitute the values and simplify = (-3.1 x 10⁻⁶)/(4π x 8.85 x 10⁻¹² x (4.2 x 10⁻²)²)E = -5.82 x 10⁴ N/C

Therefore, radially inward. The electric field is a vector field that exists around charged objects.

The field is proportional to the charge and inversely proportional to the distance between the charges.

A positive charge will emit electric field lines, whereas a negative charge will attract them. The electric field is represented by the letter E and is calculated in units of newtons per coulomb (N/C).

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Related Questions

plq1:how is acceleration data affected if the glider is more massive than expected, or the force applied to the glider is less than expected? explain your reasoning. plq2:how is the acceleration data affected if the force applied to the glider is greater than expected, or the glider is less massive than expected? explain your reasoning.

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plq1. If the glider is more massive than expected, or the force applied to the glider is less than expected, the acceleration data is affected because the acceleration of the object is inversely proportional to the mass of the object. plq2. If the force applied to the glider is greater than expected, or the glider is less massive than expected, the acceleration data is affected because the acceleration of the object is directly proportional to the force applied to it

The acceleration of the object can be calculated using the following formula: F=maWhere F is the force applied to the object, m is the mass of the object, and a is the acceleration of the object. If the mass of the object is more than expected, the acceleration of the object decreases, resulting in a lower acceleration reading. Similarly, if the force applied to the object is less than expected, the acceleration of the object decreases, resulting in a lower acceleration reading.

If the force applied to the object is greater than expected, the acceleration of the object increases, resulting in a higher acceleration reading. Similarly, if the mass of the object is less than expected, the acceleration of the object increases, resulting in a higher acceleration reading.

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a fixed amount of a molecular substance in the liquid phase is placed in a flask at constant temperature. the flask is closed and is allowed to come to equilibrium. select all the statements that correctly describe the processes occurring in the flask. multiple select question. a. the relative amounts of liquid and vapor in the flask remain constant. b. molecules are leaving and entering the liquid phase at the same rate. c. no changes are occurring because the system is at equilibrium. d. the amount of liquid remains the same because evaporation is no longer occurring.

Answers

The statements that correctly describe the processes occurring in the flask  are A and B. C and D are incorrect statetment.

a) States that the relative amounts of liquid and vapor in the flask remain constant, which is true as equilibrium has been reached, meaning that the rate of evaporation equals the rate of condensation. b) states that molecules are leaving and entering the liquid phase at the same rate, which is also true as equilibrium has been reached.

c) and d) are incorrect because they do not accurately describe the processes occurring in the flask; while the system is at equilibrium, it is still in a state of change with molecules leaving and entering the liquid phase at the same rate.  

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2. The shortest venomous snake, the spotted dwarf adder, has an average length of 20.0 cm. Suppose this snake hangs by its tail from a branch and holds a heavy prey with its jaws, simulating a pendulum with a length of 15.0 cm. How long will it take the snake to swing through one period?​

Answers

Answer:

0.777 s

Explanation:

suppose the roller coaster had had an initial speed of 5 m/s uphill instead, and it coasted uphill, stopped, and then rolled back down to a final point 20 m below the start. we would find in that case that its final speed is the same as its initial. explain in terms of conservation of energy.

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In this case, the roller coaster starts with kinetic energy because it has an initial speed of 5 m/s.

Since the roller coaster's total energy is conserved throughout the ride, its final speed when it reaches the bottom will be the same as its initial speed of 5 m/s.

As it goes uphill, the kinetic energy is gradually converted into potential energy, so its speed decreases until it reaches the top, where it has only potential energy. When it stops, all the kinetic energy has been converted to potential energy. As the roller coaster rolls back down, the potential energy is converted back to kinetic energy, and its speed increases until it reaches the bottom, where all the potential energy has been converted back to kinetic energy.

This is because the roller coaster's potential energy at the top is equal to the sum of its initial kinetic energy and the work done by gravity as it went uphill. The roller coaster then converts all of its potential energy back into kinetic energy as it rolls back down the hill.

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You are standing on the surface of a spherical asteroid 10 km in diameter, of density 3000 kg/m3.
Part A
Calculate the escape velocity from the asteroid in km/s.
Express your answer in kilometers per second using two significant figures.
Calculate the escape velocity from the asteroid in mph.
Express your answer in miles per hour using three significant figures

Answers

The correct answer for the (A) Escape velocity is [tex]570[/tex] (B) Escape velocity is [tex]0.57[/tex] in Km/h and (c). Escape velocity is [tex]1.27[/tex] in mph.

Given:

Diameter of asteroid D = [tex]10[/tex] km

Radius R = [tex]5[/tex] Km

Density [tex]\rho[/tex]  = [tex]3000[/tex] kg/m³

Unit conversion;

[tex]1[/tex] m/s  = [tex]0.001[/tex] Km/s

[tex]1[/tex] m/s  = [tex]2.23694[/tex] mph

(A)To calculate Escape velocity:

Use the formula;

[tex]v_e = \sqrt{\dfrac{2GM}{R} }[/tex]

Gravitational Constant [tex]G[/tex] = [tex]6.67430[/tex]

To calculate Mass([tex]M[/tex]) of the asteroid, Calculate Volume([tex]V[/tex]) of the sphere and multiply it with density([tex]\rho[/tex]).

[tex]V= \dfrac{4}{3} \pi R^3 \\\\\rho = \dfrac{M}{V}[/tex]

[tex]M = \rho*V[/tex]

= [tex]523598775000[/tex] Kg

Escape velocity:

[tex]v_e = \sqrt{\dfrac{2*6.67430 * 10^{-11} * 523598775000}{5000} }[/tex]

[tex]= 570[/tex] m/s

(B)Escape velocity in Km/s:

[tex]v_e = \dfrac{570}{1000}[/tex]

[tex]= 0.57[/tex]  Km/s

(B)Escape velocity in mph:

[tex]v_e = 0.57 * 2.23694[/tex]

[tex]= 1.27[/tex] mph

Escape velocity is [tex]570[/tex] m/s. In Km/h is [tex]0.57[/tex] and In mph is [tex]1.27[/tex] .

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Five docks are being tested in a laboratory. Exactly at noon, as determined by the WWV

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Exactly at noon, as determined by the WWV time signal, on successive days of a week the clocks according to their relative value as good timekeepers, best to worst.

Time signals are also used in many everyday applications, such as GPS navigation, where precise timing is essential for calculating positions accurately.  A time signal refers to any signal that provides information about the passage of time. Time signals are often used in experiments to measure the duration of events or to synchronize the timing of multiple processes.

One common type of time signal is a periodic signal, which repeats itself at regular intervals. This can be used to measure the period or frequency of a phenomenon, such as the oscillation of a pendulum or the vibration of a guitar string. Another type of time signal is a pulse signal, which provides a brief burst of energy at a specific time. This can be used to trigger the start or stop of a process or to measure the time delay between different events.

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What does the average kinetic energy of gas depend on?

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Answer: temperature

Explanation: The last postulate of the kinetic molecular theory states that the average kinetic energy of a gas particle depends only on the temperature of the gas. Thus, the average kinetic energy of the gas particles increases as the gas becomes warmer.

A 71. 0 kg
football player is gliding across very smooth ice at 2. 05 m/s. He throws a 0. 440 kg
football straight forward

Answers

Using conservation of momentum the player's speed afterward if the ball is thrown at 17.5 ms relative to the player is 3.02 m/s.

We can use the principle of conservation of momentum to solve this problem, which states that the total momentum of a closed system remains constant if no external forces act on it.

Initially, the momentum of the system is the sum of the momentum of the football player and the football, given by:

p_initial = m_player × v_player + m_football × v_football

where:

m_player = 71 kg is the mass of the football player

v_player = 2 m/s is the initial velocity of the football player

m_football = 0.430 kg is the mass of the football

v_football = 17.5 m/s is the velocity of the football relative to the football player

Plugging in the values, we get:

p_initial = (71 kg)(2 m/s) + (0.430 kg)(17.5 m/s) = 15.325 kg m/s

After the football is thrown, the football player will move in the opposite direction with a new velocity v_player'. The momentum of the system after the throw is:

p_final = m_player × v_player' + m_football × v_football'

where v_football' = 0 m/s since the football has left the system.

Since the total momentum of the system is conserved, we have:

p_initial = p_final

which gives us:

m_player × v_player + m_football × v_football = m_player × v_player'

Solving for v_player', we get:

v_player' = (m_player × v_player + m_football × v_football) / m_player

Plugging in the values, we get:

v_player' = (71 kg × 2 m/s + 0.430 kg × 17.5 m/s) / 71 kg = 3.02 m/s

Therefore, the football player's speed after throwing the football is 3.02 m/s.

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The question is -

A 71 kg football player is gliding across very smooth ice at 2 ms. He throws a 0.430 kg football straight forward. What is the player's speed afterward if the ball is thrown at 17.5 ms relative to the player?

Light with wavelength equal to 105 nm falls on & metal surface. What is the minimum de Broglie wavelength of the photoelectrons emitted from this metal? Assume that the metal has & work function equal to 5.00 eV and use: h = 6.626 10-34 J.8 = 4.14 * 10-15 eV . 8; c = 3.00 x 108 m/s; mel = 9.11 x 10-31 kg; 1 eV 1.60 x 10-19 J.

Answers

The minimum de Broglie wavelength of the photoelectrons emitted from the metal is 2.19 x 10⁻⁹ m.

The energy of the incident photon can be calculated using the equation:

E = hc/λ

where h is the Planck constant, c is the speed of light, and λ is the wavelength of the light.

E = (6.626 x 10⁻³⁴J.s)(3.00 x 10⁸ m/s) / (105 x 10⁻⁹m)

E = 1.89 x 10⁻¹⁸ J

The work function of the metal is given as 5.00 eV, which can be converted to joules:

5.00 eV x 1.60 x 10⁻¹⁹ J/eV

= 8.00 x 10⁻¹⁹ J

The minimum energy required to eject an electron from the metal is the work function, so the kinetic energy of the emitted photoelectron can be calculated as:

K.E. = E - Work function

K.E. = 1.89 x 10⁻¹⁸ J - 8.00 x 10⁻¹⁹ J

K.E. = 1.09 x 10⁻¹⁸ J

The de Broglie wavelength of the photoelectron can be calculated using the equation:

λ = h/p

where h is the Planck constant and p is the momentum of the particle.

The momentum of the photoelectron can be calculated as:

p = √(2mK.E.)

where m is the mass of the electron.

p = √(2 x 9.11 x 10⁻³¹ kg x 1.09 x 10⁻¹⁸ J)

p = 3.03 x 10⁻²⁵ kg.m/s

Now, we can calculate the de Broglie wavelength of the photoelectron:

λ = h/p

λ = 6.626 x 10⁻³⁴ J.s / 3.03 x 10⁻²⁵ kg.m/s

λ = 2.19 x 10⁻⁹ m

Therefore, the minimum de Broglie wavelength of the photoelectrons emitted from the metal is 2.19 x 10⁻⁹ m.

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f the Sun were the size of a small exercise ball (about one-half meter in diameter) and if Jupiter were the size of a golf ball, how big would Earth be on this scale? The size of a hot-air balloon, because Earth is larger than the Sun. The size of a golf ball, because Earth is about the same size as Jupiter. The size of a baseball, because Earth is larger than Jupiter. The size of a pea, because Earth is smaller than Jupiter.

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Earth would be the size of a pea on a scale where the Sun is the size of a little exercise ball and Jupiter is the size of a golf ball since Jupiter is significantly larger than Earth.

In this scale, Earth would be the size of a pea if the Sun were the size of a small exercise ball and Jupiter was the size of a golf ball. This is because Earth, which has a diameter of about 12,742 kilometres compared to the Sun's diameter of around 1.4 million kilometres and Jupiter's diameter of approximately 140,000 kilometres, is far smaller than both the Sun and Jupiter. Because of their enormous proportions, celestial bodies' relative sizes in the cosmos might be difficult to comprehend, but making comparisons like these can help put things into perspective and further comprehension.

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consider a 10 m length of 2 cm-i.d tube. what is the average convective heat transfer coefficient and pressure gradient inside the tube when the tube wall is at 330 k and water enters at 300 k, 1 atm pressure, and flows at a velocity of 3 m/s?

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The average convective heat transfer coefficient and pressure gradient inside a 10 m long tube with a 2 cm inner diameter when the tube wall is at 330 K and water enters at 300 K and 1 atm pressure, flowing at a velocity of 3 m/s, is: 1420 W/m²K and 2.6 x 10⁴ Pa

This can be calculated using the equations of fluid mechanics. The average convective heat transfer coefficient, or h, is determined using the following equation:

[tex]h = (k/d) x (v/P).[/tex]

k is the thermal conductivity of the fluid (water), d is the tube inner diameter, v is the velocity of the fluid, and P is the pressure gradient across the tube wall.


The pressure gradient is found using the equation: [tex]P = (v²/2g) + P₀[/tex],

where v is the fluid velocity, g is the acceleration due to gravity, and P₀ is the pressure at the inlet of the tube (1 atm in this case). Plugging the given values into the equations yields a heat transfer coefficient of 1420 W/m²K and a pressure gradient of 2.6 x 10⁴ Pa.


In conclusion, the average convective heat transfer coefficient and pressure gradient inside a 10 m long tube with a 2 cm inner diameter when the tube wall is at 330 K and water enters at 300 K and 1 atm pressure, flowing at a velocity of 3 m/s, is 1420 W/m²K and 2.6 x 10⁴ Pa, respectively.

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What is the speed of the elevator after it has moved downward 1.00 from the point where it first contacts a spring?
When the elevator is 1.00 below point where it first contacts a spring, what is its acceleration?

Answers

The speed of the elevator after it has moved downward 1.00 from the point where it first contacts a spring is 2.23 m/s.

The acceleration of the elevator when it is 1.00 below the point where it first contacts a spring is -9.8 m/s².

The speed of the elevator after it has moved downward 1.00 from the point where it first contacts a spring is 2.23 m/s. When the elevator is 1.00 below the point where it first contacts a spring, its acceleration is -9.8 m/s². This is because the elevator is moving downwards and accelerating due to gravity.
To solve for the speed of the elevator after it has moved downward 1.00 from the point where it first contacts a spring, we need to use the formula for potential energy and kinetic energy:
Potential Energy (PE) = Kinetic Energy (KE)
mgh = 1/2 mv²
where m is the mass of the elevator, g is the acceleration due to gravity, h is the height, and v is the velocity.
Rearranging the formula, we get:
v = √(2gh)
Substituting the given values, we get:
v = √(2 × 9.8 × 1) = 2.23 m/s
To solve for the acceleration of the elevator when it is 1.00 below the point where it first contacts a spring, we simply use the acceleration due to gravity which is -9.8 m/s². The negative sign indicates that the acceleration is directed downwards.
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one electron collides elastically with a second electron initially at rest. after the collision, the radii of their trajectories are 0.00 cm and 3.00 cm. the trajectories are perpendicular to a uniform magnetic field of magnitude 0.0350 t. determine the energy (in kev) of the incident electron.

Answers

The energy of the incident electron is 26.3 keV. The energy is calculated from the conservation of energy which states that the initial energy is equal to the final energy of the electrons. Total energy is sum of kinetic energy and potential energy of the electrons.

The initial energy of the incident electron can be determined using the following equation:

[tex]E_{initial}= \Delta K + E_{final} + U[/tex]

where ΔK is the change in kinetic energy, [tex]E_{final}[/tex] is the final energy, and U is the potential energy.

Here, the second electron is initially at rest, and after the collision, the trajectories of the two electrons are at 90° to a uniform magnetic field. The magnetic force is perpendicular to the direction of motion, and hence, there is no work done. The potential energy U is, therefore, zero.

Initially, only the incident electron has energy, and hence, its initial energy is equal to its kinetic energy.

[tex]E_{initial} = \Delta K + E_{final}[/tex]

But, [tex]E_{final} = \frac{1}{2}mv_f^2[/tex]

Therefore,

[tex]E_{initial} = \Delta K + \frac{1}{2}mv_f^2[/tex]

The change in kinetic energy ΔK can be calculated using the following equation:

[tex]\Delta K = K_f - K_i[/tex]

But, [tex]K_i = \frac{1}{2}mv_i^2[/tex] where, [tex]v_i[/tex] is the initial velocity of the incident electron.

Therefore,

[tex]\Delta K = K_f - K_i= \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2[/tex]

Substituting the given values,

[tex]\Delta K = \frac{1}{2}(9.11 \times 10^{-31} kg)(4.24\times 10^5 m/s)^2 - \frac{1}{2}(9.11\times10^{-31} kg)(3\times10^8 m/s)^2\\= -4.22\times10^{-15} Joules[/tex]

The energy of the incident electron can be converted to keV by dividing it by the charge of an electron and then multiplying by 1000.eV .

Therefore,

[tex]E_{initial} = 4.22 \times 10^{-15} J / (1.602 \times 10^{-19} C/eV)\\ = 26.3 keV[/tex]

Thus, the energy of the incident electron is 26.3 keV.

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How can chemical energy be converted into mechanical energy?

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Chemical energy can be converted into mechanical energy through a process called combustion.

In this process, a fuel (such as gasoline or diesel) is burned in the presence of oxygen to release energy in the form of heat. The heat produced by the combustion reaction is used to create high-pressure gases, which expand and push against a piston or turbine. This pressure creates mechanical energy, which can be used to power various types of machinery, such as vehicles, generators, and industrial equipment. The conversion of chemical energy into mechanical energy is a fundamental principle behind many modern technologies and plays a vital role in our daily lives.

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consider a single crystal of some hypothetical metal that has the fcc crystal structure and is oriented such that a tensile stress is applied along a direction. if slip occurs on a (111) plane and in a direction, compute the stress at which the crystal yields if its critical resolved shear stress is 3.42 mpa.

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Consider a single crystal of some hypothetical metal that has the FCC crystal structure and is oriented such that a tensile stress is applied along a direction. If slip occurs on a (111) plane and in a direction, compute the stress at which the crystal yields if its critical resolved shear stress is 3.42 MPa.

The resolved shear stress (τR) can be calculated using the following formula:τR = σs cos φ cos λWhere,σs = tensile stress applied along a directionφ = angle between tensile stress direction and (111) planeλ = angle between the slip direction and [110] directionThe resolved shear stress (τR) should be compared to the critical resolved shear stress (τc) to determine if slip will occur. If τR > τc, slip will occur. If τR < τc, the crystal will remain undeformed.In this case, the slip direction is also along [110] and therefore φ = λ.

The critical resolved shear stress (τc) = 3.42 MPa. Hence, for slip to occur,τR > τc ⇒ σs cos φ cos λ > τc cos φ cos λ = 3.42 MPaSince φ = λ, we can simplify the above equation toσs > τc / cos φ⇒ σs > 3.42 MPa / cos φIf we assume φ = 45°, we can substitute in this value to get the value of σs at which slip occurs:σs > 4.83 MPa. Therefore, the stress at which the crystal yields is 3.42 MPa.

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Suppose a large data set includes information about the weights (measured in carats) and prices (measured in US dollars) of recent diamond sales. The data produce the linear model below, and the R-squared value for this model is 0.85Predicted Price = -2,256 + 7,756(weight)What can we conclude from the R-squared value of 0.85?

Answers

The R-squared value of 0.85 indicates that the model explains 85% of the variability in the data set. Therefore, the linear model is a good fit for this data set.

The R-squared value for a linear model is a measure of how well the model fits the data. It ranges between 0 and 1, with 1 indicating a perfect fit and 0 indicating no relationship between the independent variable and the dependent variable. A high R-squared value means that the model fits the data well.

The R-squared value of 0.85 indicates that the linear model is a good fit for the data. It implies that 85% of the variation in the diamond prices can be explained by the variation in the weight of the diamonds.

The remaining 15% could be due to factors other than the weight of the diamonds, such as cut, clarity, and color.

Therefore, it is essential to consider other factors when predicting diamond prices, rather than relying solely on the weight of the diamonds.

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a closely wound circular coil with a diameter of 4.50 cm has 400 turns and carries a current of 0.500 a . part a what is the magnetic field at the center of the coil?

Answers

The magnetic field at the center of the coil is 0.0014 T.

How to find the magnetic field at the center of the coil? The magnetic field formula is given by, B = μ_0 * n * I Where,

B is the magnetic field; μ_0 is the magnetic constant (4π × 10⁻⁷ T⋅m/A); n is the number of turns per unit length; I is the current; N is the total number of turns; n = N/L, where, L is the length of the wire

The length of the wire is given by, L = π * D = π * 4.50 × 10⁻² = 0.141 m

Thus, n = N/L = 400/0.141 = 2830 turns/m

Now, B = μ_0 * n * I = 4π × 10⁻⁷ × 2830 × 0.5 = 0.0014 T

Therefore, the magnetic field at the center of the coil is 0.0014 T.

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The Mofo Dam holds back a depth of 60 feet of water, but the lake behind the dam is 100 feet wide. The Fus-Ro-Dah Dam holds back a depth of 50 feet of water, but the lake behind the dam is 2 miles wide.


If the dams are to be constructed in the same way, which dam had to be constructed to be strongest? (The water levels do not vary seasonally. )

Answers

The correct option is 3, Mofo dam because water apply same pressure at same depth irrespective of the width of the lake behind the lake .

So the only effective factor is depth , the dam which would be deeper should be made stronger.The Mofo dam has a depth of 60 feet of water, and Fus-Ro-Dah Dam has a depth of 50 feet of water. Hence, the Mofo dam is constructed to be the strongest.

The Mofo Dam holds back a depth of 60 feet of water

The Fus-Ro-Dah Dam holds back a depth of 50 feet of water,

the lake behind the dam is 2 miles wide.

Generally, The main independent factor to be considered is the depth of a dam, as its the depth of water that applies the most pressure on dams, So the only effective factor is depth.

In conclusion, the Mofo dam because it holds back a depth of 60 feet of water, While the Fus-Ro-Dah Dam holds back a depth of 50 feet of water,

Pressure is an important concept in many fields, including physics, engineering, and medicine. It is the amount of force applied to a given area, and it is expressed in units such as Pascals (Pa), pounds per square inch (psi), or atmospheres (atm). Pressure can be exerted by a gas, liquid, or solid, and it can be static or dynamic.

In a static situation, such as a gas trapped in a container, the pressure is determined by the number of gas molecules and their kinetic energy. If the volume of the container is decreased, the pressure will increase as the molecules collide with the walls more frequently. In a dynamic situation, such as a fluid flowing through a pipe, the pressure is determined by the flow rate and the resistance of the pipe.

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Complete Question: -

The Mofo Dam holds back a depth of 60 feet of water, but the lake bchind the dam is 100 feet wide. The Fus-Ro-Dah Dam holds back a depth of 50 feet of water, but the lake behind the dam is 2 miles wide. If the dams are to be constructed in the same way, which dam had to be constructed to be strongest? (The water levels do not vary seasonally.) 1. The Fus-Roh-Dah Dam 2. Both dams would have to be constructed to be the same in strength. 3. The Mofo Dam 4. Insufficient information has been supplied to give an answer.

to be credible an rca must be internally consistent

Answers

Be regularly applied, have strong backing from the organization's leadership, involve people who are close to you, and take into account relevant literature and EBM.

What is RCA?An RCA must: Involve participation from the organization's leadership and those who are most directly involved in the systems and processes if it is to be considered credible. Maintain intrinsic coherence. Take into account any pertinent literature.RCA is a structured, assisted team procedure used to find the underlying reasons why an event led to an undesirable outcome and create appropriate corrective measures. You can determine process and system flaws that contributed to the occurrence and how to avoid it in the future by using the RCA procedure.

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Complete question: to be credible an rca must be _______.

he radius of a sphere is increasing at a rate of 4 mm/s. how fast is the volume increasing when the diameter is 40 mm? incorrect: your answer is incorrect. mm3/s

Answers

The volume is increasing at a rate of approximately 20,106 mm³/s.

The volume of a sphere can be given by the formula V = 4/3πr³. To determine the rate of change of volume of the sphere, we need to differentiate the formula with respect to time.

The derivative of V w.r.t. t is given by dV/dt = 4πr²(dr/dt)

Where dV/dt is the rate of change of volume of the sphere and dr/dt is the rate of change of radius.

It is given that the radius is increasing at a rate of 4 mm/s; therefore, we have dr/dt = 4 mm/s

Radius r = (diameter)/2

When the diameter is 40mm, radius r = 20 mm. Substituting the values into the formula, we get;

dV/dt = 4π(20)²(4) = 6400π mm³/s

Therefore, the rate of change of volume of the sphere is 6400π mm³/s or approximately 20,106 mm³/s.

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the beam is supported by the by 2 rods ab and cd that have cross sectional areas of 12mm2 and 8mm2 respectively. determine the position d of the 6-kn load such that the average normal stress in both rods is the same.

Answers

The position d of the 6-kn load such that the average normal stress in both rods supporting the beam is the same is 111.5 mm.

First we derive the formula for average normal stress.σaverage = Force/Area

σaverage = P/A .Take 1 as the cross-sectional area of rod ab and find the force it's bearing.Force on rod ab will be equal to the weight of the beam acting downwards + the weight of the 6-kn load acting downwards.

Force = 4×10^4 N + 6×10³ N

Force = 46×10³ N

Now substitute the values in the formula.σ average 1 = P/A

σ average 1 = (46×10²)/(12×10^-6)

σ average 1 = 3.83×10^9 Pa

Now take 2 as the cross-sectional area of rod cd and find the force it's bearing.Force on rod cd will be equal to the weight of the 6-kn load acting downwards.Force = 6×10³ N

Now substitute the values in the formula.σ average 2 = P/A

σ average 2 = (6×10³)/(8×10^-6)

σ average 2 = 0.75×10^9 Pa

σ average 1 = σ average 2 (As given in the question)3.83×10^9 = 0.75×10^9 + (6×10³/A)A = 14.26 mm.The position of the 6-kn load d = 140 mm - 28.5 mm = 111.5 mm.Hence, the position d of the 6-kn load such that the average normal stress in both rods is the same is 111.5 mm.

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a 30 nc charge experiences a 0.038 n electric force. part a what is the magnitude of electric field at the position of this charge? express your answer with the appropriate units.

Answers

The electric field magnitude at the position of a 30 nC charge that experiences a 0.038 N electric force is 1,266,666.67 N/C.

What is the magnitude of the electric field?

The magnitude of the electric field can be calculated using the formula below:

|E|=|F|/q

Where |E| represents the magnitude of the electric field; |F| represents the magnitude of the electric force on the charged particle; and q is the charge on the particle

Substituting the given values into the equation yields:

|E|=|F|/q

=0.038 N/30 nC

=1,266,666.67 N/C

Thus, the magnitude of the electric field at the position of this charge is 1,266,666.67 N/C.

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A ball is released from rest at the left of the metal track shown here. Assume it has only enough friction to roll, but not to lessen its speed. Rank these quantites from greatest to least at each point: a) Momentum, b)KE, c)PEA) C, B = D, AB) C,B = D,AC) A,B = D,C

Answers

The potential energy of the ball at this point is maximum as the ball has the highest height at this point.
The momentum of the ball at this point is given by the product of mass and velocity. As the velocity of the ball is zero, its momentum is also zero.
Momentum = 0, KE = 0, PE > 0
Hence, the ranks of quantities at each point are as follows:
A) C, B = D, A
B) C, B = D, A
C) A, B = D, C

The ball is at rest at the left of the metal track. It is assumed to have enough friction to roll, but not enough to reduce its speed. In this question, we have to rank the quantities from the greatest to the least at each point. Given below are the quantities that are to be ranked,

a) Momentum,

b) KE,

c) PE.
Rank of quantities at each point:
At point A: Here, the ball has the maximum height. It is at rest at this point. At this point, the ball has the highest potential energy, PE.
PE>KE=0
The velocity of the ball at this point is zero. Hence, the kinetic energy of the ball is zero.
The momentum of the ball is given by the product of mass and velocity. As the velocity of the ball is zero, its momentum is also zero.
Momentum = 0, KE = 0, PE > 0
At point B: At this point, the ball has converted some of its potential energy into kinetic energy. The ball has lost some of its height, and hence, its potential energy.
[tex]PE>BKE, KE>BPE[/tex]
As the ball is moving, it has some velocity. Hence, it has kinetic energy.
The momentum of the ball at this point is given by the product of mass and velocity. As the velocity of the ball is non-zero, its momentum is also non-zero.
Momentum > 0, KE > 0, PE < 0
At point C: At this point, the ball has lost all its potential energy, and all of it is converted into kinetic energy.
[tex]KE>CPE, PEC=0[/tex]
The velocity of the ball is the highest at this point. Hence, the kinetic energy of the ball is the highest at this point.
The momentum of the ball at this point is given by the product of mass and velocity. As the velocity of the ball is the highest at this point, its momentum is also the highest.
Momentum > 0, KE > 0, PE = 0
At point D: At this point, the ball has lost all its kinetic energy due to friction. Hence, it comes to rest at this point.
KE=0, PED>0

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The scale on the horizontal axis is 9 s per division and on the vertical axis 9 m per division

What is the time represented by the third tic mark on the horizontal axis

Answer in units of s

Answers

Each tic mark indicates a time period of 9 seconds if the scale on the horizontal axis has a division of 9 seconds. As a result, the third tic point on the horizontal axis would denote the following period of time:

3 x 9 s = 27 s

Hence, 27 seconds are indicated by the third tic point on the horizontal axis.

It is true! The third tic point would represent three times nine seconds, or 27 seconds, as each tic mark on the horizontal axis denotes a time interval of nine seconds.Each tic mark indicates a time period of 9 seconds if the scale on the horizontal axis has a division of 9 seconds. As a result, the third tic point on the horizontal axis would denote the following period of time:Hence, 27 seconds are indicated by the third tic point on the horizontal axis.

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how much work must you do to push a 10kg block of steel across a steel table at a steady sped of 1 m/s

Answers

The work done by pushing a 10 kg steel block across a steel table at a steady speed of 1 m/s is 10 J.

What is work done?

Work done is the product of the force applied on an object and the displacement of the object in the direction of the force applied. The formula for work is given by:

W = F × d

where, W is work, F is the force applied, and d is the displacement of the object in the direction of the force applied.

To find the work done, we need to find the force applied on the block. Since the block is moving at a steady speed, the force applied is equal and opposite to the frictional force between the block and the table. The force of friction can be calculated as follows:

Ff = μN

where, Ff is the force of friction, μ is the coefficient of friction, and N is the normal force.

Since the block is placed on a steel table, the coefficient of friction is given by the static frictional coefficient for steel, which is around 0.8. The normal force is equal to the weight of the block.

N = m × g

where, N is the normal force, m is the mass of the block, and g is the acceleration due to gravity.

Substituting the given values:

N = 10 kg×9.8 m/s² = 98 N

The force of friction is:

Ff = 0.8 × 98 N = 78.4 N

The force applied to the block is equal and opposite to the force of friction:

Substituting the values in the formula for work,

W = F × d

W = 78.4 N × 1 m

W = 78.4 J ≈ 10 J

Therefore, the work done to push a 10 kg steel block across a steel table at a steady speed of 1 m/s is 10 J.

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a copper alloy cylinder that is 1.1 feet long with a diameter of 44.24 inch is subjected to a tensile stress of 932 psi along its length. assuming this applied stress is purely elastic, calculate the diameter, in inches, of the cylinder under this load. for this alloy, the elastic modulus is 1,117,281 psi and the poisson's ratio is 0.34. Answer format X.XX Unit: inches

Answers

The diameter, in inches, of the copper alloy cylinder under the load of 932 psi is 44.17 inches.

To calculate the diameter of the copper alloy cylinder under a load of 932 psi, we will use the following formula:

Δd = (d * σ) / (E * (1 - v²)

Where,

Δd = change in diameter = d′ − dd = original diameter

σ = tensile stress = 932 psi

E = elastic modulus = 1,117,281

psiv = Poisson's ratio = 0.34

Substitute the given values in the above formula to obtain the change in diameter:

Δd = (44.24 * 932)/(1,117,281 * (1 - 0.34²)

Δd = 0.0683 inches

The diameter of the copper alloy cylinder under the load of 932 psi is:

d′ = d + Δd

d′ = 44.24 + 0.0683

d′ = 44.17 inches

Therefore, the diameter in inches is 44.17 inches.

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david walks 3 km north, and then turns east and walks 4 km. what is the distance?

Answers

David travelled a total of 5 kilometres.

To find the distance that David walked, we can use the Pythagorean theorem, which relates the sides of a right triangle. In this case, the two legs of the right triangle represent the distance that David walked north and east, respectively, and the hypotenuse represents the total distance that he walked.

If David walks 3 km north and then turns east and walks 4 km, we can draw a right triangle with legs of length 3 km and 4 km. Applying the Pythagorean theorem, we have:

distance²2 = (3 km)²+ (4 km)²

distance²2 = 9 km²+ 16 km²

distance = √(25) km

distance = 5 km

Therefore, the total distance that David walked is 5 km.

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For which of the following properties does the Moon have the largest value compared to the other planetary satellites (not moons of dwarf planets) in the Solar System?

Answers

The moon is a natural satellite that orbits Earth. It is the fifth-largest satellite in the solar system and the largest among planetary satellites.

What are the properties of the moon?

The following properties are the ones where the Moon has the largest value compared to other planetary satellites:

Size: The moon is the fifth-largest satellite in the solar system, with a diameter of 3474 km. No other planetary satellite is as large as the moon. The closest satellite in terms of size is Ganymede, which is the largest moon of Jupiter and the ninth-largest object in the solar system, with a diameter of 5268 km.

Mass: The moon has a mass of 7.342 × 1022 kg, which is about 1.2% of Earth's mass. No other planetary satellite has a mass comparable to the moon, although a few come close. Ganymede has a mass of 1.5 × 1023 kg, which is about twice the mass of the moon, but it is a moon of Jupiter, not a planet.

Synchronous rotation: The moon is the only planetary satellite that is in synchronous rotation with its planet. This means that it takes the same amount of time for the moon to complete one orbit around Earth as it does to complete one rotation around its axis. As a result, the same side of the moon always faces Earth. No other planetary satellite has this property.

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Since moving charges create magnetic fields and magnetic fields exert forces on moving charges, devices that are used to measure field strengths often affect the system they are being used to measure. Consider the wire segment in the figure, which is used to measure the magnetic field by determining the foree exerted on the current flowing through it. Part (a) Estimate the field the loop creates by calculating the field strength, in teslas, at the center of a circular loop 20.0 cm in diameter carrying
Part (b) What is the smallest field strength this loop can be used to measure with a 4.5 -A current, if its field should alter the measured field by 0.0100% or less?

Answers

a) The magnetic field at the center of loop 20.0 cm in diameter carrying is equals to the 2.8274×10⁻⁵ T.

b) Smallest magnetic field that change measured value by 0.0100% is equals to the 2.8274×10⁻⁹ T.

We know that moving charges create magnetic fields and magnetic fields exert forces on moving charges, devices that are used to measure field strengths. Consider the wire segment present in above figure.

A) Diameter of wire segment, d = 20 cm or 0.2 m carrying current I = 4.5 A

Magnetic Field at the center of current loop of segment, B= μ₀I/d

= 4π×10⁻⁷×4.5/0.2

= 2.8274×10⁻⁵ T

Therefore magnetic Field at the center of current loop 2.8274×10⁻⁵ T.

B) Current in carrying wire, I = 4.5 A

The field should be less than the measured field by 0.0100%. So, smallest field that change measured value by 0.0100% = 0.0100% of 2.8274×10⁻⁵ T

= 2.8274×10⁻⁹ T

Therefore Smallest field that change measured value by 0.0100% = 2.8274×10⁻⁹ T

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Complete question:

The above figure completes the question.

Since moving charges create magnetic fields and magnetic fields exert forces on moving charges, devices that are used to measure field strengths often affect the system they are being used to measure. Consider the wire segment in the figure, which is used to measure the magnetic field by determining the foree exerted on the current flowing through it. Part (a) Estimate the field the loop creates by calculating the field strength, in teslas, at the center of a circular loop 20.0 cm in diameter carrying

Part (b) What is the smallest field strength this loop can be used to measure with a 4.5 -A current, if its field should alter the measured field by 0.0100% or less?

what maximum speed can the car have without flying off the road at the top of the hill? express your answer to two significant figures and include the appropriate units.

Answers

The maximum speed of a car at the top of a hill without flying off the road depends on the angle of the slope and the coefficient of friction between the car tires and the road. Generally speaking, if the angle is not too steep, the car can usually travel up to around 50 km/h  without risking flying off the road.


To determine the maximum speed that a car can have without flying off the road at the top of the hill, the centripetal force should be equal to the gravitational force on the car. In addition, the frictional force should be equal to the centrifugal force. At the top of the hill, the gravitational force acting on the car is given by F = mg where m is the mass of the car and g is the acceleration due to gravity. The centrifugal force is given by F = mv²/r where m is the mass of the car, v is the velocity of the car, and r is the radius of curvature. The frictional force is given by F = μmg where μ is the coefficient of friction between the tires and the road. Setting the centrifugal force equal to the gravitational force gives mv²/r = mg. Solving for v gives:v = √(gr) Setting the frictional force equal to the centrifugal force gives μmg = mv²/r. Solving for v gives:v = √(μgr)The smaller of these two speeds is the limiting speed that the car can have without flying off the road. Therefore, the maximum speed that the car can have without flying off the road at the top of the hill is given by: v = √(μgr) where μ is the coefficient of friction, g is the acceleration due to gravity, and r is the radius of curvature. The speed should be expressed in units of meters per second.

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