A supertanker can hold 3.00 ✕ 105 m3 of liquid (nearly 300,000 tons of crude oil). (a) How long (in s) would it take to fill the tanker if you could divert a small river flowing at 2600 ft3/s into it? s (b) How long (in s) for the same river at a flood stage flow of 100,000 ft3/s? s
Answers
(a)The time required to fill the supertanker when the speed of the river is 2600 \(ft^3/s\). is \(3.62 \times 10^{4}\)seconds to fill the using a small river flowing at
(b) The time required to fill the supertanker when the speed of the river is 100,000 \(ft^3/s\). is \(1.08 \times 10^5\) seconds.
To determine the time it takes to fill the supertanker, we can use the concept of flow rate, which is the volume of liquid passing through a given point per unit of time. The flow rate can be calculated by dividing the volume by the time.
(a) For the small river flowing at 2600 \(ft^3/s\)., we need to convert the volume of the tanker to the same units. 1 \(m^{3}\) is approximately equal to 35.3147 \(ft^3\). Therefore, the volume of the tanker is \(3.00 \times 10^5 \times 35.3147\) = \(1.06 \times 10^7 \ ft^3\). Dividing the volume by the flow rate, we get the time:
Time = Volume / Flow rate = \(\frac{1.06 \times 10^7 }{2600 }\) ≈ \(3.62 \times 10^4\)seconds.
(b) For the flood stage flow of 100,000 \(ft^3/s\), we can use the same approach. The time to fill the supertanker would be:
Time = Volume / Flow rate = \((1.06 \times 10^7) / (100,000 )\) ≈\(1.08 \times 10^5\) seconds.
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Related Questions
A force of 6.1 N acts on a 18 kg body initially at rest. Compute the work done by the force in (a) the first, (b) the second, and (c) the third seconds and (d) the instantaneous power due to the force at the end of the third second.
Answers
The work done by the force in the first second is 1026J. The work done by the force in the second is 5.125 J. The work done by the force in the third second is 3.080 J. The instantaneous power due to the force at the end of the third second is 87 W.
v = at = (6.1 N) / (18 kg) * 1 s = 0.3389 m/s
The kinetic energy of the body after the first second is
K = (1/2) * m * v² = (1/2) * (18 kg) * (0.3389 m/s)² = 1.026 J
The work done by the force in the first second is:
W = K - 0 = 1.026 J
(b) In the second, the velocity of the body is:
v = at = (6.1 N) / (18 kg) * 2 s = 0.6778 m/s
The kinetic energy of the body after the second is:
K = (1/2) * m * v² = (1/2) * (18 kg) * (0.6778 m/s)² = 6.151 J
The work done by the force in the second is:
W = K - 1.026 J = 5.125 J
(c) In the third second, the velocity of the body is:
v = at = (6.1 N) / (18 kg) * 3 s = 1.0167 m/s
The kinetic energy of the body after the third second is:
K = (1/2) * m * v² = (1/2) * (18 kg) * (1.0167 m/s)² = 9.231 J
The work done by the force in the third second is:
W = K - 6.151 J = 3.080 J
(d) v = at = (6.1 N) / (18 kg) * 3.001 s = 1.0198 m/s
The kinetic energy of the body at that instant is:
K = (1/2) * m * v² = (1/2) * (18 kg) * (1.0198 m/s)² = 9.318 J
The work done by the force in a very short interval of time is:
dW = K - 9.231 J = 0.087 J
Therefore, the instantaneous power due to the force at the end of the third second is:
P = dW / dt = 0.087 J / 0.001 s = 87 W
Work is defined as the energy transferred to or from an object by means of a force acting on the object as it moves along a certain distance. Work is expressed as the product of the force and the displacement of the object in the direction of the force. The SI unit of work is the joule.
Work can be done by various forces, including gravitational, electric, and magnetic forces. For example, work is done when an object is lifted against the force of gravity, when an electric current flows through a circuit, or when a magnetic field changes around a conductor.
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if N is 300moles , pressure is 11.7atm, temperature is 100K, what is volume?
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If N is 300 moles, the gas has a volume of roughly 2029.27 liters at 11.7 atm of pressure and 100 K of temperature.
How is the volume determined?The following formula must be used to determine a gas's volume using the Ideal Gas Law equation:
PV = nRT
Where:
Pressure is P. (in atm)
Volume is V. (in liters)
n = the substance's quantity (in moles)
R = 0.0821 L atm/mol K, or the gas constant.
Temperature is T. (in Kelvin)
Using the supplied parameters as a starting point, we obtain: (11.7 atm) × V = (300 moles) × (0.0821 L-atm/mol-K) × (100 K)
After simplifying and finding V, we arrive at the following equation: V = (300 moles) × (0.0821 L atm/mol K) × (100 K) / (11.7 atm)
V = 2029.27 liters.
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Describe the language of the following PDA (z is the stack end symbol) (the figure can be located under a, z/bbz X, z/z b,6/1 ۸, 2/2 90 91 92 a, b/bbb 1 b,b/1
Answers
The language of the given pushdown automaton (PDA) can be described as follows:
The PDA has a stack alphabet consisting of symbols 'a', 'b', 'z', '6', '1', '۸', '2', '9', '0', 'x', 'y'. 'z' represents the stack end symbol.
The transitions of the pushdown automaton (PDA) are as follows:
(a, z, z) -> (X, z): This transition allows the PDA to replace an 'a' at the input with an 'X' on the stack while maintaining the stack end symbol 'z'.(z, b, z) -> (z, z): This transition allows the PDA to pop a 'b' from the input without modifying the stack.(z, z, b) -> (6, 1): This transition allows the PDA to push '6' and '1' onto the stack when encountering a 'b' on the input.(6, 1, b) -> (۸, 2): This transition allows the PDA to replace the '6' and '1' on the top of the stack with '۸' and '2' respectively when another 'b' is read from the input.(x, y, b) -> (b, b, b): This transition allows the PDA to replace 'x' and 'y' on the top of the stack with 'b', 'b', and 'b' when a 'b' is read from the input.(b, b, b, b) -> (1, b): This transition allows the PDA to replace the 'b', 'b', and 'b' on the top of the stack with '1' and 'b' when another 'b' is encountered.(1, b, b) -> (1, 1): This transition allows the PDA to replace the '1' and 'b' on the top of the stack with '1' and '1' when another 'b' is read from the input.(1, 1, z) -> (z, z): This transition allows the PDA to pop '1' from the stack without modifying the input.Thus, the language accepted by this PDA is characterized by a sequence of 'a's followed by a sequence of 'b's, where the number of 'b's is three times the number of 'a's, and each 'b' is followed by a corresponding sequence of '90', '91', '92', 'a', 'b', 'b', 'b', '1', 'b', 'b', '1', and ending with '1'.
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The scatterplot shows the energy in watts and the light output in lumens per watt of several lightbulbs
Answers
The statements D best represents the relationship between the energy and light output.
What is scatter plot ?A scatter plot is a form of plot displays values for two variables for a collection of data using coordinates system.
The scatter plot depicts the energy input and light output in lumens per lightbulb. The statements D best represents the relationship between the energy and light output.
There is no apparent association between the energy and light output, shows the relationship between the energy and light output.
Hence,statement D is correct.
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Consider the autonomous equation y
′
=sin(y/2). (a) Find all the critical points. (b) Draw the phase line with range (−3π,3π) and determine the stability of each eritical point involved.
Answers
(a) The critical points are all values of y that can be written as 2nπ, where n is an integer.
(b) The solution is decreasing for y < -3π and y > 3π, and increasing for -3π < y < 3π.
The critical points of an autonomous equation are the points where the derivative is equal to zero. In this case, we have the autonomous equation y' = sin(y/2).
(a) To find the critical points, we set the derivative equal to zero and solve for y.
sin(y/2) = 0
Since sin(θ) = 0 when θ is a multiple of π, we have:
y/2 = nπ
where n is an integer. Solving for y, we get:
y = 2nπ
Therefore, the critical points are all values of y that can be written as 2nπ, where n is an integer.
(b) To draw the phase line with the given range (-3π, 3π) and determine the stability of each critical point, we can consider the sign of the derivative at different points.
For y < -3π, sin(y/2) is negative, so the derivative y' is negative. This means that the solution is decreasing in this interval.
For -3π < y < 3π, sin(y/2) is positive, so the derivative y' is positive. This means that the solution is increasing in this interval.
For y > 3π, sin(y/2) is negative, so the derivative y' is negative. This means that the solution is decreasing in this interval.
Based on this information, we can conclude that the critical points at y = 2nπ are all stable, as the solution approaches these points from both directions.
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The resistance of a filament that carries 2 A when a 10-V potential difference across it is. A) 2 ohms. B) 5 ohms. C) 10 ohms. D) 20 ohms.
Answers
The resistance of the filament which carries the current of 2 A and has a potential difference of 10 V is b) 5 ohms
The current passing through the filament = 2A
The potential difference in the filament = 10 V
The resistance of the filament can be found using the formula,
R = V/I
where R is the resistance of the filament
V is the potential difference between the filament
I is the current passing through the filament
Let us substitute the known values in the above equation, we get
R = 10 / 2
= 5 ohms
Therefore, the resistance of the filament is 5 ohms
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The quantity y appearing in Bernoulli's equation MUST be measured:
A)upward from the center of the Earth
B)upward from the surface of the Earth
C)upward from the lowest point in the flow
D)downward from the highest point in the flow
E)upward from any convenient level
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The quantity y appearing in Bernoulli's equation can be measured upward from any convenient level. There is flexibility in choosing a reference point as long as the measurements are consistent.
Bernoulli's equation relates the pressure, velocity, and elevation of a fluid flowing in a streamline. The equation is given as:
P + (1/2)ρv² + ρgh = constant
where P is the pressure, ρ is the density of the fluid, v is the velocity of the fluid, g is the acceleration due to gravity, and h is the elevation or height above a reference point.
The quantity y in Bernoulli's equation represents the elevation or height above the chosen reference point. It is the vertical distance from the reference point to the location in the fluid flow where the measurements are being taken.
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Moe is playing with a yo-yo. He throws the yo-yo down and then pulls it back up. The motion of the yo-yo is represented by the equation y=2x^2−4.8x
, where x represents the number of seconds since the yo-yo left Moe's hand, and y represents the vertical height in inches of the yo-yo with respect to Moe's hand. Note that when the yo-yo is in Moe's hand, y = 0, and when the yo-yo is below his hand, y is negative. a. How long is Moe's yo-yo in the air before it comes back to Moe's hand? Write and solve a quadratic equation to find the times that the yo-yo is in Moe's hand. b. How long does it take for the yo-yo to turn around, that is, to start its return to his hand? Use what you know about parabolas to help you. c. How long is the yo-yo's string? That is, what is y when the yo-yo changes direction? d. Draw a sketch of the graph representing the motion of Moe's yo-yo. On the sketch, label the important points: when the yo-yo is in Moe's hand and when it changes direction.
Answers
Answer:
30is the answer but it wa seasy
What force causes you to move upward when you jump into the air?
a. You push down on Earth with a force greater than your weight, so Earth pushes back on you with a force that is equal in magnitude but opposite in direction
b. You push down on Earth, and because Earth is so large, it pushes back on you with a force that is greater than the downward force you exerted on it
c.You push down on Earth, but because Earth is stationary, Earth cannot exert an upward force on you so internal forces from your muscles act on your body itself.
d.You push down on Earth, which causes it to move slightly down under your feet and give you the sensation that you have risen.
Answers
For the following six questions, match the descriptions to the below people (A-J)
A) Eratosthenes B) Aristarchus C) Isaac Newton D) Aristotle E) Ptolemy F) Galileo G) Hipparchus H) Kepler I) Nicolaus Copernicus J) Tycho Brahe
23. Discovered the phases of Venus using a telescope.
24. First to consider ellipses as orbits.
25. Foremost ancient Greek philosopher.
26. Ancient Greek who believed in a sun-centered universe.
27. First to measure the size of the Earth to good accuracy.
28. Developed the first predictive model of the solar system.
Answers
The correct match of the descriptions to the below people are 23 - F, 24 - H, 25 - D, 26 - I, 27 - A, 28 - B.
23 - F Galileo: Galileo Galilei is credited with discovering the phases of Venus using a telescope. Through his observations, he observed that Venus went through a series of phases similar to those of the Moon, which supported the heliocentric model of the solar system.
24 - H Kepler: Johannes Kepler was the first to consider ellipses as orbits. He formulated the laws of planetary motion, known as Kepler's laws, which stated that planets move in elliptical paths with the Sun at one of the foci. Kepler's work revolutionized our understanding of celestial mechanics.
25 - D Aristotle: Aristotle, the ancient Greek philosopher, is considered one of the foremost thinkers in history. While his contributions span various fields, including philosophy and natural sciences, his views on astronomy were geocentric. He believed that the Earth was the center of the universe and that celestial bodies moved in perfect circles around it.
26 - I Nicolaus Copernicus: Nicolaus Copernicus was an astronomer who proposed the heliocentric model of the solar system, in which the Sun, rather than the Earth, was at the center. Copernicus's revolutionary idea challenged the prevailing geocentric view and laid the foundation for modern astronomy.
27 - A Eratosthenes: Eratosthenes was an ancient Greek mathematician and astronomer who made significant contributions to geography and astronomy. He is known for his accurate measurement of the Earth's circumference. By measuring the angle of the Sun's rays at two different locations, he estimated the Earth's circumference with remarkable accuracy.
28 - B Aristarchus: Aristarchus of Samos is credited with developing the first predictive model of the solar system. He proposed a heliocentric model centuries before Copernicus, suggesting that the Sun was at the center of the universe, with the Earth and other planets orbiting it. Aristarchus's model was a significant departure from the prevalent geocentric view of the time.
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A flywheel with a radius of 0.600 m starts from rest and accelerates with a constant angular acceleration of 0.200 rad/s2 . Part A: Compute the magnitude of the tangential acceleration of a point on its rim at the start.; Part B: Compute the magnitude of the radial acceleration of a point on its rim at the start.; Part C: Compute the magnitude of the resultant acceleration of a point on its rim at the start.; Part D: Compute the magnitude of the tangential acceleration of a point on its rim after it has turned through 60.0 ∘; Part E: Compute the magnitude of the radial acceleration of a point on its rim after it has turned through 60.0 ∘.; Part F: Compute the magnitude of the resultant acceleration of a point on its rim after it has turned through 60.0 ∘.; Part G: Compute the magnitude of the tangential acceleration of a point on its rim after it has turned through 120.0 ∘.; Part H: Compute the magnitude of the radial acceleration of a point on its rim after it has turned through 120.0 ∘.; Part I: Compute the magnitude of the resultant acceleration of a point on its rim after it has turned through 120.0 ∘.
Answers
Part A:
Radius of flywheel = 0.600 m
Angular acceleration of flywheel (αₐ) = 0.200 rad/s²
The tangential acceleration of a point at the start (αₓ) =
= αₓ = αₐ × r
= αₓ = 0.200 × 0.600
= αₓ = 0.12 m/s²
Part B:
Radius of flywheel = 0.600 m
Angular acceleration of flywheel (αₐ) = 0.200 rad/s²
Angular speed = ω = 0 m/s²
Magnitude of radial acceleration of a point on rim at the start (αₙ)=
= (angular speed)² × r
= 0 × 0.600
= 0 m/s²
Part C:
Radius of flywheel = 0.600 m
Angular acceleration of flywheel (αₐ) = 0.200 rad/s²
Resultant acceleration of a point on the rim at the start =
= α =√(αₙ² + αₓ²)
= α = √ (0² + 0.12²)
= α = 0.12 m/s²
Part D:
Radius of flywheel = 0.600 m
Angular acceleration of flywheel (αₐ) = 0.200 rad/s²
Angular speed = ω = 0 m/s²
The tangential acceleration of a point after 60° turn (αₓ₁) = The tangential acceleration of a point at the start (αₓ)
= αₓ₁ = αₐ × r
= αₓ₁ = 0.200 × 0.600
= αₓ₁ = 0.12 m/s²
Part E:
Radius of flywheel = 0.600 m
Angular acceleration of flywheel (αₐ) = 0.200 rad/s²
Angular speed = ω = 0 m/s²
Angular speed after 60° turn = ω₁ = √(ω² + (2×α×θ))
To find θ,
= θ = 60Π / 180
= θ = Π/30
= θ = 1.04 rad
Thus, ω₁ = √(0 + 2 × 1.04 × 0.2)
= ω₁ = 0.644 rad/s
The radial acceleration of a point after 60° turn (αₓ₂) =
= αₓ₂ = r × ω₁²
= αₓ₂ = 0.600 × 0.644²
= αₓ₂ = 0.248 m/s²
Part F:
Radius of flywheel = 0.600 m
The tangential acceleration of a point after 60° turn (αₓ₁) = 0.12 m/s²
The radial acceleration of a point after 60° turn (αₓ₂) = 0.248 m/s²
The magnitude of resultant acceleration of a point on the rim after 60° turn (α₃) =
= α₃ = √ (αₓ₂² + αₓ₁²)
= α₃ = √ (0.12² + 0.248²)
= α₃ = 0.275 m/s²
Part G:
Radius of flywheel = 0.600 m
Angular acceleration of flywheel (αₐ) = 0.200 rad/s²
Angular speed = ω = 0 m/s²
The tangential acceleration of a point after 120° turn (αₓ₁) = The tangential acceleration of a point at the start (αₓ)
= αₓ₁ = αₐ × r
= αₓ₁ = 0.200 × 0.600
= αₓ₁ = 0.12 m/s²
Part H:
Radius of flywheel = 0.600 m
Angular acceleration of flywheel (αₐ) = 0.200 rad/s²
Angular speed = ω = 0 m/s²
Angular speed after 120° turn = ω₁ = √(ω² + (2×α×θ))
To find θ,
= θ = 120Π / 180
= θ = 2Π/3
= θ = 2.09 rad
Thus, ω₁ = √(0 + 2 × 2.09 × 0.2)
= ω₁ = 0.836 rad/s
The radial acceleration of a point after 120° turn (αₓ₂) =
= αₓ₂ = r × ω₁²
= αₓ₂ = 0.600 × 0.836²
= αₓ₂ = 0.502 m/s²
Part I:
Radius of flywheel = 0.600 m
The tangential acceleration of a point after 120° turn (αₓ₁) = 0.12 m/s²
The radial acceleration of a point after 120° turn (αₓ₂) = 0.502 m/s²
The magnitude of resultant acceleration of a point on the rim after 120° turn (α₃) =
= α₃ = √ (αₓ₂² + αₓ₁²)
= α₃ = √ (0.12² + 0.502²)
= α₃ = 0.515 m/s²
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A cup of coffee has a hydroxide ion concentration of 1. 0 × 10−10 m. What is the ph of this coffee?.
Answers
A cup of coffee has a hydroxide ion concentration of 1. 0 × 10−10 m. The pH of this coffee will be 10
pH = - log [H+]
= - log [1. 0 × \(10^{-10}\)]
= - log 1 - ( -10 log 10 )
= 0 + 10 = 10
The pH of this coffee will be 10
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A boy rode a bike and covered 90 km at an average speed equals 36 km/h but he covered the first thirty km in two hours. What is the average speed at which the remaining distance was covered ?
Answers
Answer:
40
Explanation:
because his increasing speed
a chair is several feet from a fireplace. The side facing the fireplace is warmer due to:
a. Conduction b. Convection c. radiation
Answers
c.
Explanation:!(*´∇`)ﾉ
What is high frequency blue light?
Answers
A range of biological effects, including those on the eye, are caused by high-energy visible light (HEV light), which is high-frequency, high-energy light in the violet/blue region of the visible spectrum between 400 and 450 nm.
Age-related macular degeneration may be brought on by HEV light.
When it comes to the visible light spectrum, blue light is categorized as having wavelengths between 400 and 525 nm. This covers the spectrum's violet to cyan wavelengths. Having a wavelength between 400 and 450 nm, narrow-spectrum blue light, also known as blue LED light or short-wavelength LED light, is a form of high-energy visible light. This light is typical in LEDs as a holdover from computer-screen technology (even when employed in lighting goods).
White light must have blue light as a component. Either narrow-spectrum or broad-spectrum blue can be used to create white. For instance, while other technologies tend to use more cyan and red, LED technology typically combines narrow-spectrum blue and yellow. Violet and cyan spikes, as well as red spikes, are produced by fluorescent coatings.
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A 13000 kg truck initially moving at a constant speed of 127/km/h suddenly applies brakes and stops in 13m. How much work has been done on the truck over this distance?
Answers
1,647,530 J work has been done on the truck over this distance.
The work done on the truck over this distance can be calculated using the equation W = Fs, where W is the work, F is the force, and s is the distance.
W = (13000 kg) x (9.81 m/s2) x (13 m)
W = 1,647,530 J
What is work done?
Work done is a term used to describe the amount of energy transferred or converted from one form to another. It is typically measured in joules (J) and is the product of a force applied to an object over a certain distance. Work done can be used to calculate the amount of energy expended by a system, such as a machine or engine.
Therefore, 1,647,530 J work has been done on the truck over this distance.
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From Earth stars appear very dim, but in space they are quite bright. From Earth we are seeing the stars
A) absolute magnitude
B) apparent magnitude
C) available magnitude
D) absent magnitude
Answers
Answer:
B
Explanation:
A student makes the following claim,
"Scalar and vector quantities are the same thing, so they can be used interchangeably."
Explain the error in the student's claim. Provide an example of each quantity to support your answer.
Please answer quickly!
Answers
Answer:
Explanation:
A scalar quantity is a physical quantity that has only one characteristic - a numerical value.
The scalar value can be positive or negative.
Examples of scalar quantities: temperature, mass, volume, time, density.
A vector quantity is a physical quantity that has two characteristics:
1) a numerical value that is always positive (vector modulus);
2) direction.
Examples of vector physical quantities: speed, acceleration, force.
compared to spiral galaxies, elliptical galaxies are:
Answers
flat disk like shape
________________
o0o0o0o0o0o0o0o0
certain white dwarf star once was like our sun, now it has the size of the moon with the mass of the sun. how much would a 65 kg person weigh on this star?
Answers
A white dwarf is an extremely dense object, with the mass of the sun compressed into the size of the moon. This means that the gravitational force on the surface of the white dwarf is much greater than the gravitational force on the surface of the sun or the moon.
The weight of an object is equal to its mass multiplied by the gravitational force at the surface of the celestial body. The formula for weight is W = mg, where W is weight, m is mass, and g is the gravitational force at the surface of the celestial body.
To find the weight of a 65 kg person on the white dwarf, we need to know the gravitational force at the surface of the white dwarf. The gravitational force at the surface of a celestial body is given by the formula g = GM/R^2, where G is the gravitational constant, M is the mass of the celestial body, and R is the radius of the celestial body.
For a white dwarf with the mass of the sun (M = 1.989 x 10^30 kg) and the size of the moon (R = 1.737 x 10^6 m), the gravitational force at the surface is:
g = (6.674 x 10^-11 N m^2/kg^2)(1.989 x 10^30 kg)/(1.737 x 10^6 m)^2 = 1.67 x 10^12 N/kg
The weight of a 65 kg person on the white dwarf is:
W = (65 kg)(1.67 x 10^12 N/kg) = 1.09 x 10^14 N
So, a 65 kg person would weigh 1.09 x 10^14 N on the white dwarf. This is equivalent to about 1.09 x 10^11 kg or 109 billion kg. This is an extremely large weight, and it would be impossible for a human to survive on the surface of a white dwarf.
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A battery is connected to a light bulb, lighting the bulb. Where in the circuit is the current the greatest?.
Answers
Answer:
No where
Explanation:
Current is the same everywhere in the circuit
Simplicity of conducting the study is to ________ as ability to test large numbers of participants is to ________.
Answers
Simplicity of conducting the study is to archival research as ability to test large numbers of participants is to surveys.
What is Archival research?Archival research is a type of research which involves seeking out and extracting the evidence from archival records. These archival records may be held together either in the collection of institutions, such as libraries and museums, or in the custody of the organization that originally generated or accumulated them, or in that of a successor object.
Archival research can be contrasted with the secondary research, which involves identification and consulting the secondary sources related to the topic of enquiry; and with the other types of primary research and empirical investigation such as fieldwork and experiment.
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We have a 10kg ball and a 15kg ball. We drop them from the top of the science building while objects are at rest and there is no wind resistance. Which one hits the ground first?
Answers
What type of energy is due to motion potential energy or kinetic energy
Answers
Answer:
Kinetic Energy
Explanation:
Potential Energy is where something is building up energy to move.
Answer:
Kinetic Energy
Explanation:
Write the formula for a partially inelastic collision and explain what all 6 variables are and the unit each is measured in. State Newton’s 3 rd Law of Motion. State the Law of Conservation of Momentum and explain it in your own way.
Answers
Answer:
i don`t get it
Explanation:
what is the degeneracy of the n = 2 shell of atomic hydrogen considering (n, l, ml) and no magnetic field?
Answers
In the absence of a magnetic field, the degeneracy of the n = 2 shell of atomic hydrogen is 4. The degeneracy of an energy level refers to the number of distinct quantum states that have the same energy.
The n = 2 shell of atomic hydrogen has four possible values for the quantum numbers (n, l, ml), which correspond to the four orbitals present in this shell. The possible values of l for the n = 2 shell are 0 and 1, meaning that the possible values of ml are 0, +1, 0, and -1, respectively.
The degeneracy of an energy level refers to the number of distinct quantum states that have the same energy. In the case of the n = 2 shell of atomic hydrogen, the degeneracy is 4, since there are four distinct orbitals with the same energy.
It is important to note that this calculation does not take into account the effects of a magnetic field, which can split the energy levels and change the degeneracy. However, in the absence of a magnetic field, the degeneracy of the n = 2 shell of atomic hydrogen is 4.
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In Ancient Greece, athletes competing in the long jump used handheld weights called halteres to lengthen their jumps. You are a 78 kg history major on the track and field team and decide to try this strategy. You jump at 10.3 m/s, 22.8 degrees above the horizontal. At the peak of your jump you throw two 5.5 kg masses horizontally behind you such that their velocity is zero in the ground's reference frame. What distance do the halteres add to the jump in m
Answers
The halter add the distance to the jump in meters is 0.55 m.
What is projectile?When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion is called the projectile.
The magnitude of velocity u =10.3 m/s, angle of jumping θ = 22.8 degrees.
Components of velocity in x and y direction are
Vx = 10.3 cos 22.8 = 9.5 m/s
Vy = 10.3 sin 22.8 = 4 m/s
Maximum Range of athlete achieved using halter is given by
R = u²sin2θ /g
where, u = initial velocity, θ is the angle of projection and g is the gravitational acceleration.
Substituting the values, we get
R = (10.3)² sin(2 x 22.8 °) / 2 x 9.81
R = 7.75m
At the peak of jump you throw two 5.5 kg masses horizontally behind you such that their velocity is zero in the ground's reference frame.
The momentum is conserved in this situation,
(M+2m)Vxo =MVx'
Vx' = (M+2m)/M x Vxo'
Change in x component of velocity ΔVx = Vx' -Vxo
Vxo = 2m/M x Vx
Vxo = 2 x 5.5 /78 x 9.5
Vxo = 1.34 s
Maximum height gained when final velocity is zero
Vy = 0 = Vyo -gt
time t = Vyo/g = 4/9.8 = 0.41s'
Increase in range by using of halters is
ΔR = ΔVx' x t
ΔR = 1.34 x 0.41
ΔR =0.55m
Thus, the halter add the distance to the jump in meters is 0.55 m.
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For a parallel-plate capacitor the energy density is uniform in the region between the plates, except near the edges of the plates. Is this also true for a spherical capacitor?.
Answers
No. this is not true for a spherical capacitor.
A spherical capacitor can be thought of as what is shown in the attached figure.That phenomenon for the parallel plate capacitor occurs because of the sudden change in the electric field intensity. Between the plates, the electric field intensity is constant and uniform and that uniformity changes at the edges.Such a uniform field intensity is not found in a spherical capacitor.The field at a given distance from the sphere is a function of that distance so it changes from place to place.So the energy density is not uniform in a region between 2 spherical surfaces.Therefore, what is true for spherical capacitors is that the energy density is like that at the edges of a parallel plate capacitor.#SPJ4
A rescue pilot drops a survival kit while her plane is flying at an altitude of 2000.0 m with a forward velocity of 100.0 m/s.
If air friction is disregarded, how far in advance of the starving explorer's drop zone should she release the package? (round
to the nearest 10s place)
meters
Answers
2. What is the kinetic energy of a 25 kg object traveling at 15 m/s?
Answers
Answer:
The kinetic energy is
= 1250 j
Explanation:
The kinetic energy is
KE = 1/2mv²
The mass is m = 25kg
The velocity is v = 10ms^1
So, KE = 12.25.10² = 1250 j