Final answer:
To calculate the work done to compress the gas by a factor of 11.0, we can use the relationship between work and volume changes. The work required to compress the gas by a factor of 11.0 is -11,000 J.
Explanation:
To calculate the work done to compress the gas by a factor of 11.0, we can use the relationship between work and volume changes. The work done on a gas during compression is given by the equation:
Work = Pressure ×Change in Volume
In this case, we know that the work required to compress the gas to half its initial volume is 550 J. Let's assume the initial volume of the gas is V. So the work done to compress the gas by a factor of 11.0 can be calculated as:
Work = Pressure × Change in Volume
550 J = Pressure ×(V/2 - V)
550 J = Pressure ×(-V/2)
Solving for Pressure, we get:
Pressure = -1100 J/V
Therefore, the work required to compress the gas by a factor of 11.0 is -1100 J/V× (11V - V) = -11,000 J (negative sign indicates work is done on the gas).
How many meters will sound travel in 60 seconds if the temperature is 20℃ ? How far in an hour ? How far in 0.002 seconds ?
Answer:
a. 20652 m
b. 1239120 m
c. 0.688 m
Explanation:
First we have to find the speed of sound in air of temperature 20°C.
The speed of sound in a medium is dependent on the temperature of that medium. At any given temperature, the speed of sound is given as:
[tex]v(T) = v(T_{0}) + 0.61T[/tex]
where
v(T) = Speed of sound at any temperature T;
[tex]v(T_{0})[/tex] = Speed of sound at 0°C;
T = temperature of the medium.
The speed of sound in air of 0°C is 332 m/s, hence, in air of temperature 20°C, speed will be:
[tex]v(20) = 332 + (0.61 * 20)\\\\v(20) = 332 + 12.2\\\\v(20) = 344.2 m/s[/tex]
a. Distance traveled in 60 seconds:
distance = speed * time
distance = 344.2 * 60
distance = 20652 m
b. Distance traveled in 1 hour:
1 hour in seconds = 1 * 60 * 60 = 3600 seconds
distance = 344.2 * 3600
distance = 1239120 m
c. Distance traveled in 0.002 second:
distance = 344.2 * 0.002
distance = 0.688 m
A 58.0-kg man is riding an escalator in a shopping mall. The escalator moves the man at a constant velocity from ground level to the floor above, a vertical height of 5.00 m. What is the work done on the man by (a) the gravitational force and (b) the escalator
Answer
a) -2842 J
b) 2842 J
Explanation:
The escalator is moving the man of mass 58 kg till the height of 5 meters
So here;
m=58 kg
distance = height = h= 5 m
gravitational acceleration = g= 9.8 m/sec2
a) We know work is scalar product of force and displacement
i-e Work = F.S= FS cosθ, where θ is the angle between force vector and displacement vector
When escalator is moving up - the gravitational force is acting downward i- opposite to displcacement i-e angle between force and displacement is 180 degrees.
we have data
m= 58 kg
F=mg = 58×9.8=568.4 N
S=5 m
Work = F.S=FS cos 180°=568.4×5×(-1)= -2842 J
So the work done by gravitational force is -2842 J and -ve sign here indicates that distance is traveled in opposite direction of force.
b) When escalator is moving up the, force is exerted by the escalator equal to gravitational force and the displacement is in the same direction so the angle between force and displacement is 0 degrees
Work = F.S=FS cos (0)=568.4×5×1=2842 J
So work done by the gravitational force then will be 2842 J
So in both cases work is same but in opposite directions.
f the earth's average surface temperature were to increase, the amount of radiation emitted from the earth's surface would __________ and the wavelength of peak emission would shift toward __________ wavelengths.
Answer
IF the earth's average surface temperature were to increase, the amount of radiation emitted from the earth's surface would _increase_ and the wavelength of peak emission would shift toward _shorter_ wavelengths.
Explanation:
The Energy of radiation emitted by the earth varies directly with the average surface temperature of the earth and inversely with the wavelength of emissions.
E = hv/λ
That is, E = k/λ
Therefore the most peak emissions (highest energies) would have shorter wavelengths.
Complete question:
if the earth's average surface temperature were to increase, the amount of radiation emitted from the earth's surface would __________ and the wavelength of peak emission would shift toward __________ wavelengths.
Answer:
if the earth's average surface temperature were to increase, the amount of radiation emitted from the earth's surface would Increase and the wavelength of peak emission would shift towards Shorter wavelengths.
Explanation:
Stefan-Boltzmann law, a fundamental law of physics, explains the relationship between an object's temperature and the amount of radiation that it emits. This law states that all objects with temperatures above absolute zero (0K) emit radiation at a rate proportional to the fourth power of their absolute temperature.
Expressed mathematically as; E = σT⁴
From this formula above, temperature is directly proportional to amount of radiation emitted.
Thus, if the earth's average surface temperature were to increase, the amount of radiation emitted from the earth's surface would Increase.Also, Energy of emitted radiation can be related to wavelength in the expression below
E =hc/λ
Where;
E is the energy of the emitted radiation
h is Planck's constant
c is the speed of light
λ is the wavelength of the emitted radiation
From the formula above, Energy of the emitted radiation is inversely proportional to the wavelength of the emitted rays.
Thus, there would be a shift towards shorter wavelengths.The motor winds in the cable with a constant acceleration, such that the 20-kg crate moves a distance s = 6 m in 3 s, starting from rest. Determine the tension developed in the cable. The coefficient of kinetic friction between the crate and the plane is mk = 0.3.
Answer:
85 N
Explanation:
Given that crate mass = 20kg
Distance = 6m
Time = 3 seconds
Coefficient of kinetic friction = 0.3
We begin by calculating for acceleration
Which was gotten as 1.33 m/s sq
SEE THE ATTACHEMENT FOR DETAILS
To determine the tension developed in the cable, we consider forces acting on the crate, including the tension in the cable and the force of kinetic friction. We use Newton's second law of motion and equations for tension and force of kinetic friction to calculate the tension. By substituting values into the equations, we can solve for the tension.
Explanation:To determine the tension developed in the cable, we need to consider the forces acting on the crate. Since the crate is moving with constant acceleration, we can use Newton's second law of motion, which states that the sum of the forces acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the forces acting on the crate are the tension in the cable and the force of kinetic friction. The tension in the cable can be calculated using the equation:
Tension = mass of the crate * acceleration + force of kinetic friction
Given that the mass of the crate is 20 kg and the acceleration is the change in velocity divided by the time taken (6 m / 3 s = 2 m/s), we can calculate the tension using the equation:
Tension = 20 kg * 2 m/s^2 + force of kinetic friction
We also need to determine the force of kinetic friction. The force of kinetic friction can be calculated using the equation:
Force of kinetic friction = coefficient of kinetic friction * normal force
The normal force is equal to the weight of the crate, which can be calculated using the equation:
Normal force = mass of the crate * gravitational acceleration
Given that the coefficient of kinetic friction is 0.3 and the gravitational acceleration is 9.8 m/s^2, we can calculate the force of kinetic friction using the equation:
Force of kinetic friction = 0.3 * (20 kg * 9.8 m/s^2)
Now we can substitute this value back into the equation for tension:
Tension = 20 kg * 2 m/s^2 + (0.3 * (20 kg * 9.8 m/s^2))
Solving this equation will give us the tension developed in the cable.
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A particular type of resistor has a tolerance of 3%. Technician A says this indicates that the resistor’s current value can be 3% above or below its stated specification. Technician B says this indicates the resistor’s resistance value can be 3% above or below its stated specification. Who is right?
Answer:
Technician B is correct.
Explanation:
The given value of resistor having 3% tolerance means that the given quantity of the physical parameter can vary by 3% of what is specified. This variation can be either more or less than the quantified value.
When the variation can occur on both sides of the stated value then it is called bilateral tolerance, usually represented as, [tex]\pm3\%[/tex].When the variation is permissible only in one direction then it is called unilateral tolerance, represented by the sign + for the impressibility on the higher side and [tex]-[/tex] sign for the impressibility on the lower side.For Other Than One And Two Family Dwellings, When Building A New Electrical Service At Least One 125 Volt, Single Phase, 15 Or 20 Ampere Rated Receptacle Outlet Shall Be Located Within At LEAST ________ Feet Of The Electrical Service Equipment.
Answer and Explanation
It was initially specified that the receptacle outlet to be located within 50 ft of the electrical service equipment.
As well, instead of the receptacle being required within 50 ft., it is now required within 25 ft. of the service. This was to accommodate the typical 25 ft. cord used by many service electricians.
The rules apply to indoor service locations other than one-and two-family dwellings. For these locations, at least one 125-volt, single-phase, 15- or 20-ampere receptacle outlet must be installed in an accessible location within 25 ft. of the indoor electrical service equipment. The receptacle must be within the same room or area as the actual service equipment.
Having a maintenance receptacle near the electrical service allows for testing, servicing and connection of portable electrical data acquisition equipment for analyzing the electrical system.
A fireworks rocket explodes at a height of 120 m above the ground. An observer on the ground directly under the explosion experiences an average sound intensity of 7.20 10⁻² W/m² for 0.205 s.
(a) What is the total amount of energy transferred away from the explosion by sound?
(b) What is the sound level in decibels heard by the observer?
Answer:
2670.90667586 J
108.573324964 dB
Explanation:
r = Distance = 120 m
A = Area = [tex]4\pi r^2[/tex]
I = Intensity of sound = [tex]7.2\times 10^{-2}\ W/m^2[/tex]
t = Time taken = 0.205 s
[tex]I_0[/tex] = Threshold intensity = [tex]10^{-12}\ W/m^2[/tex]
Power is given by
[tex]P=IA\\\Rightarrow E=7.2\times 10^{-2}\times 4\pi 120^2[/tex]
Energy is given by
[tex]E=Pt\\\Rightarrow E=7.2\times 10^{-2}\times 4\pi 120^2\times 0.205\\\Rightarrow E=2670.90667586\ J[/tex]
The total amount of energy is 2670.90667586 J
Sound intensity level is given by
[tex]\beta=10log\dfrac{I}{I_0}\\\Rightarrow \beta=10log\dfrac{7.2\times 10^{-2}}{10^{-12}}\\\Rightarrow \beta=108.573324964\ dB[/tex]
The sound level is 108.573324964 dB
The total amount of energy transferred away from the explosion by sound is 1.476 x 10^-2 Joules. The sound level in decibels heard by the observer is 97.50 dB.
Explanation:(a) To find the total amount of energy transferred away from the explosion by sound, we can use the formula:
Energy = Intensity x Time
Substituting the given values:
Energy = (7.20 x 10-2 W/m2) x (0.205 s) = 1.476 x 10-2 Joules
(b) The sound level in decibels heard by the observer can be calculated using the formula:
Sound Level (dB) = 10 x log10(Intensity / Threshold Intensity)
Substituting the given values:
Sound Level (dB) = 10 x log10((7.20 x 10-2 W/m2) / (10-12 W/m2)) = 97.50 dB
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A physical change
A) occurs when iron rusts.
B) occurs when sugar is heated into caramel.
C) occurs when glucose is converted into energy within your cells.
D) occurs when water is evaporated.
E) occurs when propane is burned for heat.
Answer: D) occurs when water is evaporated.
Explanation:
A physical change is defined as a change in which there is alteration in shape, size etc. No new substance gets formed in these reactions.
A chemical change is defined as a change in which a change in chemical composition takes place. A new substance is formed in these reactions.
a). iron rusts :the chemical reaction occurs by combination of iron with oxygen , thus a chemical change.
b) sugar is heated into caramel. the chemical reaction occurs by decomposition , thus a chemical change.
c) glucose is converted into energy within your cells: the chemical reaction occurs by decomposition , thus a chemical change.
d) when water is evaporated : Only the state changes and thus a physical change and cannot be reversed.
e) when propane is burned for heat: the chemical reaction as propane combines with oxygen to form carbon dioxide and water , thus a chemical change.
The kinetic energy of a photoionized electron increases linearly with the wavelength of the ionizing radiation, but is independent of the intensity of the radiation. If False, explain in one sentence what makes the statement False.A. True
B. False
Answer: False
Explanation:
It is dependent on the intensity of the radiation i.e the photo electron should increase with the light amplitude.
A post is wrapped two full turns around with a belt. The tension in the belt is 7500 N by exerting a force of 150 N on its free end. Determine the coefficient of static friction between the belt and the post.
Answer:
Coefficient of friction is
Ū = 0.31
Explanation:
T2 = T1* e^(ūơ)
Where T2 = 7500n = tension in the belt, T1 = 150n = reaction force,
ū = coefficient of friction
Ơ = 2pai * N
Where N = number of turns = 2
Ơ = 4pai
7500 = 150e^(ū*4pai)
50 = e^(ū *4pai)
lin 50 = 4pai * ū
Ū = 3.91/4pai
Ū = 0.31
008 (part 1 of 3) 10.0 points A 0.338 kg particle has a speed of 3.8 m/s at point A and kinetic energy of 10.1 J at point B. What is its kinetic energy at A? Answer in units of J. 009 (part 2 of 3) 10.0 points What is the particle’s speed at B? Answer in units of m/s. 010 (part 3 of 3) 10.0 points What is the total work done on the particle as it moves from point A to B?
Answer:
1) 2.44 joules
2) 7.73 m/s
3) 7.6 joules
Explanation:
Kinetic energy (K) of a particle is:
[tex] K=\frac{mv^{2}}{2} [/tex] (1)
with m the mass, and v the velocity
1) Because we already now velocity on A (va) and the mass of the object we can calculate its kinetic energy:
[tex]K_{a}=\frac{mv_{a}^{2}}{2}=\frac{(0.338kg)(3.8\frac{m}{s})^{2}}{2}=2.44J [/tex]
2) Because on B we know mass and kinetic energy we should solve (1) for v and use our values to find the velocity on B:
[tex]v_{b}=\sqrt{\frac{2K_{b}}{m}}=\sqrt{\frac{2(10.1J)}{(0.338kg)}}=7.73\frac{m}{s} [/tex]
3) Work-energy theorem states that the change of kinetic energy of an object is equal to the total work done on it, so:
[tex]W=K_b-K_a=10.1J-2.44J= 7.6J [/tex]
A rectangular certificate has a perimeter of 32 inches. Its area is 63 square inches. What are the dimensions of the certificate?
The dimensions of the rectangular certificate with a perimeter of 32 inches and an area of 63 square inches can be 7 inches by 9 inches or 9 inches by 7 inches.
Explanation:The subject of this question is Mathematics, specifically an application of algebra to solve for the dimensions of a rectangle when given its perimeter and area. Let’s denote length as L and width as W. The formulas for area and perimeter of a rectangle are given by Area = L * W and Perimeter = 2 * (L + W), respectively. Given that the area is 63 square inches and the perimeter is 32 inches, we can set up two equations.
From the perimeter, we have 2L + 2W = 32, simplifying gives us L = 16 - W.
We can then substitute this into the area equation, so (16 - W) * W = 63.
This simplifies and solved that gives us W = 7 or W = 9.
To find L, we substitute W = 7 / W = 9 into L = 16 - W. We get the pairs (L, W) as (9, 7) or (7, 9).
So the dimensions of the certificate can be 7 inches by 9 inches or 9 inches by 7 inches.
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Alice and Tom dive from an overhang into the lake below. Tom simply drops straight down from the edge, but Alice takes a running start and jumps with an initial horizontal velocity of 25 m/s. Neither person experiences any significant air resistance. Just as they reach the lake below
A) the speed of Alice is larger than that of Tom.
B) the splashdown speed of Alice is larger than that of Tom.
C) they will both have the same speed.
D) the speed of Tom will always be 9.8 m/s larger than that of Alice.
E) the speed of Alice will always be 25 m/s larger than that of Tom.
Final answer:
Alice and Tom will both have the same speed upon reaching the water because gravity accelerates them both at the same rate for their vertical descent. Alice's initial horizontal velocity does not affect her vertical downward acceleration.
Explanation:
The correct answer is C) they will both have the same speed. This is because their vertical descent is solely affected by gravity, which accelerates all objects at the same rate (approximately 9.8 m/s2) regardless of their horizontal velocity component. Alice's initial horizontal velocity of 25 m/s contributes only to her horizontal movement and does not affect the rate at which she falls vertically due to gravity. Consequently, both Alice and Tom will have the same vertical speed when they reach the lake. When you consider both vertical and horizontal components for Alice, her overall splashdown speed will be greater than Tom's; however, the question asks about just the speed. Since only the vertical descent is mentioned, we infer that it's the vertical component being considered. Therefore, their speeds, considering just their vertical motion under gravity, are the same.
Consider a three-phase generator rated 300 MVA, 23 kV, supplying a system load of 240 MVA and 0.9 power factor lagging at 230 kV through a 330 MVA, 23 D/ 230 Y-kV step-up transformer with a leakage reactance of 0.11 per unit a) Neglecting the exciting current and choosing base values at the load of 100 MVA and 230 kV, find the phasor currents IA, IB, and IC supplied to the load in per unit b) By choosing the load terminal voltage VA as reference, specify the proper base for the generator circuit and determine the generator voltage V as well as the phasor currents IA, IB, and IC, from the generator. (Note: Take into account the phase shift of the transformer.) c) Find the generator terminal voltage in kV and the real power supplied by the generator in MW d) By omitting the transformer phase shift altogether, check to see whether you get the same magnitude of generator terminal voltage and real power delivered by the generator.no-load speeds possible with this motor?
The phasor currents for the load are 0.573 per unit. With the transformer phase shift, the generator supplies a voltage of 24.3 KV and 54.6 MW. Ignoring the transformer phase shift, the magnitudes remain the same, but the angles differ.
Explanation:The complex power supplied to the load is P = 240 MVA * 0.9 = 216 MW and reactive power Q = 240 MVA * sqrt (1-0.9^2) = 106.3 MVAR. From this, you can compute the phasor currents IA, IB, and IC using the formula: I = S/V * conjugate of the power factor angle, giving IA = IB = IC = 0.573 per unit.
For b), taking leakage reactance into account, the generator voltage V = 1 + j0.11 * I, giving a voltage of 1.063 per unit. Because the generator is in Delta and the load is in Wye, there is a -30-degree shift in currents from the generator. So, IA, IB, and IC are 0.573 angle -30 degrees per unit.
For c), multiply the generator voltage by base voltage to determine the terminal voltage (1.063*23kV = 24.3 kV). The total real power supplied by the generator PG = VI*cos(theta) = 0.573*1.063*0.9 = 0.546 per unit, or 54.6 MW.
For d), if the transformer phase shift is ignored, we will still obtain the same magnitude of generator terminal voltage and real power, but the angles will be different, i.e., the currents will be in phase with the voltage, and not lagging by 30 degrees, because we didn't take into account the Y-D transformation.
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An oblique rectangular prism with a square base has a volume of 539 cubic units. The edges of the prism measure 7 by 7 by 14 units. How many units longer is the slanted edge length of the prism, 14, compared to its perpendicular height?
Answer:
3 units
Solution:
V=539 cubic units
Square base, with edge a=7 units
Slanted edge length: s=14 units
V=Ab h
Ab=49 square units
539 cubic units = (49 square units) h
h= 11 units
s-h=14 units-11 units
s-h=3 units
Answer:
3 units
Explanation :
Volume of an oblique rectangular prism with a square base = A×B×h
Where h = perpendicular height
From the question, Volume of an oblique rectangular prism with a square base = 539 cubic units
We were asked from the question to find how many units longer the slanted edge length of the prism, 14 is compared to its perpendicular height.
The first step is : Find the perpendicular height
Edges A = 7 units
Edges B = 7 units
Perpendicular height ?
Hence,
539 = 7 × 7 × h
539 = 49h
h = 539 ÷ 49
h = 11 units
Therefore, perpendicular height of the prism = 11 units
To find how many units longer, we would subtract the perpendicular height of the prism from the slanted edge length of the prism
= 14 units - 11 units
= 3 units .
Therefore the slanted edge length of the prism , 14, is 3 units longer compared to its perpendicular height.
The electric force between objects A and B is F. If the charge of object A were twice as large as it is, but everything else was kept the same, what would be the new electric force between objects A and B?
Answer:
[tex]F'=2F[/tex]
Explanation:
According to Coulomb's law and assuming the objects as point charges, the magnitude of the electric force that each object exerts on the other is defined as:
[tex]F=\frac{kq_Aq_B}{d^2}[/tex]
Here k is thee Coulomb constant, [tex]q_A[/tex] and [tex]q_B[/tex] are the charges of the objects and d is the distance of separation between them. We have [tex]q'_A=2q_A[/tex]:
[tex]F'=\frac{kq'_Aq_B}{d^2}\\F'=\frac{k2q_Aq_B}{d^2}\\F'=2\frac{kq_Aq_B}{d^2}\\F'=2F[/tex]
Imagine holding a basketball in both hands, throwing it straight up as high as you can, and then catching it when it falls. At which points in time does a zero net force act on the ball? ignore air resistance
A. When you hold the ball still in your hands after catching it
B. Just after the ball first leaves your hands.
C. At the instant the ball reaches its highest point.
D. At the instant the falling ball hits your hands.
E. When you hold the ball still in your hands before it is thrown.
Answer:
C. At the instant the ball reaches its highest point.
Explanation:
When a body is thrown up, it tends to come down due to the influence of gravitational force acting on the body. The body will be momentarily at rest at its maximum point before falling. At this maximum point, the velocity of the body is zero and since force acting on a body is product of the mass and its acceleration, the force acting on the body at that point will be "zero"
Remember, F = ma = m(v/t)
Since v = 0 at maximum height
F = m(0/t)
F = 0N
This shows that the force acting on the body is zero at the maximum height.
Sam, whose mass is 79 kg, stands at the top of an 11-m-high, 120-m-long snow-covered slope. His skis have a coefficient of kinetic friction on the snow of 0.07. If he uses his poles to get started, then glides down, what is his speed at the bottom?
Answer:
The speed of Sam at the bottom is 7.19 m/s.
Explanation:
Given that,
Mass of Sam = 79 kg
Height = 11 m
Length = 120 m
Coefficient of kinetic friction = 0.07
Suppose, an object of mass m is at rest at the top of a smooth slope of height h and length L. The coefficient of kinetic friction between the object and the surface, micro-kilometer , is small enough that the object will slide down the slope if given a very small push to get it started.
We need to calculate the speed at the bottom
Using conservation of energy
[tex]P.E=K.E+\text{energy lost of friction}[/tex]
[tex]mgh=\dfrac{1}{2}mv^2+\mu mg(\sqrt{L^2-h^2})[/tex]
[tex]v^2=2gh-2\mu g(\sqrt{L^2-h^2}[/tex]
[tex]v=\sqrt{2gh-2\mu g(\sqrt{L^2-h^2}}[/tex]
Where, m = mass
h = height
L= length
v = speed
g = acceleration due to gravity
Put the value into the formula
[tex]v=\sqrt{2\times9.8\times11-2\times0.07\times9.8(\sqrt{120^2-11^2})}[/tex]
[tex]v=7.19\ m/s[/tex]
Hence, The speed of Sam at the bottom is 7.19 m/s.
Velocity is the rate of change of position. Sam's velocity at the bottom of the slope with a height of 11 m is 13.8844 m/s.
What is velocity?Velocity is the rate of change of position of an object with respect to time.
[tex]v = \dfrac{ds}{dt}[/tex]
We know that at the topmost height, the weight of Sam will act as potential energy, while during skiing down this potential energy will be partially converted to kinetic energy and part will be converted to heat or can say will be lost due to the friction, therefore,
Potential Energy = Kinetic energy + Energy loss due to the friction,
[tex]mgh = \frac{1}{2}mv^2 + \mu gh\sqrt{L^2+H^2}\\\\mgh - \mu gh\sqrt{L^2+H^2} = \frac{1}{2}mv^2\\\\2mgh - 2\mu gh\sqrt{L^2+H^2} = mv^2\\\\2gh(m - \mu \sqrt{L^2+H^2}) = mv^2\\\\v^2 = \dfrac{2gh}{m}(m - \mu \sqrt{L^2+H^2})[/tex]
Substitute the values,
Mass, m = 79 kg
Height, H = 11 m
Length, L = 120 m
Coefficient of kinetic friction, μ = 0.07
Acceleration due to gravity, g = 9.81 m/s²
[tex]v^2 = \dfrac{2 \times 9.81 \times 11}{79}[79-0.07\sqrt{120^2+11^2}]\\\\v^2 = 215.82 - 23.0442\\\\v = 13.8844\rm\ m/s[/tex]
Hence, Sam's velocity at the bottom of the slope with a height of 11 m is 13.8844 m/s.
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A dog running in an open field has components of velocity vx = 3.1 m/s and vy = -1.3 m/s at time t1 = 11.3 s . For the time interval from t1 = 11.3 s to t2 = 22.5 s , the average acceleration of the dog has magnitude 0.52 m/s2 and direction 27.5 ∘ measured from the +x−axis toward the +y−axis. At time t2 = 23.6s , what are the x-component of the dog's velocity?
Answer:
x -component of dog's velocity = 8.265m/s
y- component of dog's velocity = 1.389m/s
Explanation:
The detailed and step by step calculation is as shown in the attached file.
A parallel-plate capacitor in air has a plate separation of 1.25 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 244 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume that distilled water is an insulator.
(a) What is the charge on the plates before immersion?
What is the charge on the plates after immersion?
(b) What is the capacitance after immersion?
What is the potential difference after immersion?
(c) What is the change in energy of the capacitor?
?U = Uwater - Uair = 5 nJ
Answer:
(a) Since net charge remains same,after immersion Q is same
(b) I. 14.56pF ii. 3.05V
(c) ΔU = 5.204nJ
Explanation:
a)
C = kεA/d
k=1 for air
ε is 8.85x10-12F/m
A = .0025m2
d = .125m
C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF
Q = CV = .177pF * 244V = 43.188pC
Since net charge remains same,after immersion Q is same
b)
C = kεA/d, for distilled water k is approx. 80
Cwater = Cair x k
=0.177pF x 80 = 14.16pF
Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V
c) Change in energy: ΔU = Uwater - Uair
Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ
Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ
ΔU = 5.204nJ
From whom should a departing VFR aircraft request radar traffic information during ground operations?
Answer:From whom should a departing VFR aircraft request radar traffic information during ground operations? ... Answer: Sequencing to the primary Class C airport, traffic advisories, conflict resolution, and safety alerts.
Explanation:
Final answer:
During ground operations, a departing VFR aircraft should request radar traffic information from ground control. Once airborne, the responsibility switches to tower control for traffic updates.
Explanation:
A departing VFR (Visual Flight Rules) aircraft should request radar traffic information from the ground control (also known as ground movement control) before taxiing. Ground control is responsible for the safety on the taxiways and runways, except the active runway. Once airborne or when holding short of the runway, the aircraft switches to tower frequency, at which point the aircraft may receive radar traffic updates relevant to their departure.
Technology utilizing a network of radio frequency waveguides is part of managing air traffic, which is crucial for maintaining a safe separation between aircraft. Furthermore, radar technology can provide directional information, and as noted in research such as Hasselmann, 1991, traffic information can be directly calculated from radar data even during the ground operations of VFR aircraft.
A 30kg rock is swung in a circular path and in a vertical plane on a .25m length string at the top of the path, the angular speed 12.0 rad/s. what is the tension in the string at that point?
A. 7.9 N.
B. 16 N.
C. 18 N.
D. 83 N.
Answer
given,
mass of the rock, m = 30 Kg
Length of the string, r = 0.25 m
angular speed of the rock, ω = 12 rad/s
Tension in the string at the top of the vertical circle = ?
At the top of the string the tension is
[tex]T_{top}=\dfrac{mv^2}{r} - m g[/tex]
we know v = r ω
[tex]T_{top}= mr\omega^2 - m g[/tex]
[tex]T_{top}= 30\times 0.25\times 12^2-30\times 9.8[/tex]
[tex] T_{top} = 786 N[/tex]
if the given mass is 0.3 Kg then tension in the rope
[tex]T_{top}= 0.30\times 0.25\times 12^2-0.30\times 9.8[/tex]
[tex]T_{top} = 7.86 N[/tex]
If the mass is 0.3 then the correct answer is option A.
Which provides evidence that humans are causing an increase in the carbon dioxide concentration in the atmosphere?
A. The net flux of carbon into the atmosphere without including the contribution from burning fossil fuels is positive.
B. The net flux of carbon out of the atmosphere without including the contribution from burning fossil fuels is zero.
C. The net flux of carbon into the atmosphere without including the contribution from burning fossil fuels is negative.
D. The net flux of carbon out of the atmosphere without including the contribution from burning fossil fuels is negative.
Answer: The net flux of carbon into the atmosphere without including the contribution from burning fossil fuels is positive.
Explanation:
The human activity of burning fossil fuels inevitably contributes to the increase in carbon dioxide and other gases in the Earth's atmosphere. As a result of this occurrence, a natural greenhouse appears. The greenhouse effect is the process of warming the Earth's surface and the lower layers of the atmosphere, resulting from the leakage of heat radiation.
Answer:
C
Explanation:
earth science climate change unit test
Beginning about 55 seconds into the video, you'll see an animation of a photographer looking through her camera at a man, a set of trees, and distant mountains. Notice that, as viewed through the camera, the positions of the man and the trees change (relative to distant mountains) as the photographer moves. Which of the following statements correctly describes what is really happening in this situation?
Answer and Explanation
The concept of parallax and relative motion is responsible for this.
That is, the photographer's motion prompts her to see parallax for the man and the trees, because their positions appear to keep shifting even though they are not really moving.
The apparent movement of the man and trees relative to distant mountains as the photographer moves is a result of a physics concept called parallax. Parallax causes nearer objects to appear to move more than farther ones as the observer's position changes.
Explanation:The phenomenon you're describing occurred as a result of a concept in physics known as parallax. Parallax refers to the apparent movement of objects when the observer's position changes. So in this case, as the photographer moves, the position of the man and the trees appear to change relative to the distant mountains. This is because nearer objects appear to move more than farther objects as the observer moves. So, the man and trees (which are closer) seem to shift their positions more than the mountains (which are farther away).
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Suppose you had one circuit with a 9V battery connected to two 100 ohm light bulbs in series. and a second circuit with a 9V battery connected to two 100 ohm light bulbs in parallel. How does the brightness of the two bulbs in series compare to two bulbs connected in parallel? Which set is dimmer? Explain why.
Answer:
Bulbs in series will be dim
Explanation:
The voltage remain constant in a series circuit, thus bulbs in series circuit will get equally distributed voltage which is not equal to the voltage across the circuit or the voltage supplied
In case of parallel circuit, the voltage across the bulb is the same as the voltage across the circuit due to which bulbs will be brighter in case of parallel circuit
A certain copper wire has a resistance of 13.0 Ω . At some point along its length the wire was cut so that the resistance of one piece is 7.0 times the resistance of the other
Determine the length of the short piece.(% of length of the wire)?
Determine the resistance of the short piece?
Determine the resistance of the long piece?
Answer with Explanation:
Let r be the resistance of short piece of copper wire.
Resistance of copper wire=R=[tex]13\Omega[/tex]
Resistance is directly proportional to length.
If a wire has greater resistance then,the wire will be greater in length.
Therefore,resistance of long piece of wire=7r
Total resistance of copper wire=Sum of resistance of two piece of wires
[tex]r+7r=13[/tex]
[tex]8r=13[/tex]
[tex]r=\frac{13}{8}[/tex]ohm
Resistance of long piece of wire=[tex]7\times\frac{13}{8}=\frac{91}{8}\Omega[/tex]
Resistance of short piece of wire =[tex]\frac{13}{8}\Omega[/tex]
Resistivity of wire and cross section area of wire remains same .
Let L be the total length of wire and L' be the length of short piece of wire.
We know that
[tex]R=\frac{\rho L}{A}[/tex]=[tex]\frac{\rho}{A}L=KL[/tex]
[tex]\frac{R}{L}=K[/tex]
Where K=[tex]\frac{\rho}{A}[/tex]=Constant
Using the formula
[tex]\frac{13}{L}=\frac{\frac{13}{8}}{L'}[/tex]
[tex]\frac{L'}{L}=\frac{13}{8}\times \frac{1}{13}=\frac{1}{8}[/tex]
[tex]L'=\frac{L}{8}[/tex]
Length of short piece of wire=L'=[tex]\frac{L}{8}[/tex]
Length of long piece of wire=[tex]L-L'=L-\frac{L}{8}=\frac{8L-L}{8}=\frac{7}{8}L[/tex]
% of length of short piece of wire=[tex]\frac{\frac{L}{8}}{L}\times 100=12.5%[/tex]%
The resistance of the short piece=[tex]\frac{13}{8}\Omega[/tex]
The resistance of the long piece=[tex]\frac{91}{8}\Omega[/tex]
The length of short piece wire is [tex]12.5[/tex] % of total wire length.
The resistance of the short piece wire is [tex]\frac{13}{8}ohms[/tex]
The resistance of the long piece wire is, [tex]7*\frac{13}{8}=\frac{91}{8}ohms[/tex]
Resistance :Let us consider that resistance of short piece wire is r.
So that, resistance of long piece wire is [tex]7r[/tex].
Total resistance of wire is [tex]13[/tex] ohms.
[tex]7r+r=13\\\\8r=13\\\\r=\frac{13}{8}Ohm[/tex]
The resistance of the short piece wire is [tex]\frac{13}{8}ohms[/tex] The resistance of the long piece wire is, [tex]7*\frac{13}{8}=\frac{91}{8}ohms[/tex]The resistance of wire is directly proportional to length of wire.Let us consider that total length of wire is 13L.
So that, length of short piece wire [tex]=\frac{13}{8} L[/tex]
length of short piece wire is,
[tex]=\frac{\frac{13}{8}L }{13L}*100 =\frac{1}{8} *100=12.5[/tex]
The length of short piece wire is 12.5 % of total wire length.
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A small object moves along the xx-axis with acceleration ax(t)ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At tt = 0 the object is at xx = -14.0 mm and has velocity v0xv0x = 8.70 m/sm/s.
What is the xx-coordinate of the object when tt = 10.0 ss?
Answer:
x = 54.3m (on the +ve x axis)
Explanation:
This is an Initial Value Problem. That means the initial values of certain parameters have been given and that can help solve the problem.
Given that acceleration, a, is:
ax(t) = - 0.032(15.0 - t)
And the initial values are:
x(t = 0) = - 14.0m
v(t = 0) = 8.7m/s
Hence,
a = - 0.032(15 - t)
a = - 0.48 + 0.032t
a = dv/dt = -0.48 + 0.032t
To obtain the velocity, v, integrate the acceleration and apply the initial values of v and t:
v = ∫dv/dt = ∫(-0.48 + 0.032t)
∫dv = ∫(-0.48 + 0.032t)dt
(v - v₀) = -0.48(t - t₀) + 0.032(t²/2 - t₀²/2)
Inputting the initial values t₀ = 0s, v₀ = 8.7m/s:
=> v - 8.7 = -0.48t + 0.032t²/2
v = 8.7 - 0.48t + 0.016t²
To obtain distance, x, integrate the velocity and apply the initial values:
v = dx/dt = 8.7 - 0.48t + 0.016t²
=> ∫dx/dt = ∫(8.7 - 0.48t + 0.016t²)
∫dx= ∫(8.7 - 0.48t + 0.016t²)dt
(x - x₀) = 8.7(t - t₀) - 0.48(t²/2 - t₀²/2) + 0.016(t³/3 - t₀³/3)
Inputting the initial values t₀ = 0s, x₀ = - 14.0m:
(x + 14.0) = 8.7t - 0.48t²/2 + 0.016t³/3
x = 8.7t - 0.48t²/2 + 0.016t³/3 - 14.0
Now that the distance, x, has been obtained, when t = 10s:
x = 8.7*10 - 0.48*10²/2 + 0.016*10³/3 - 14.0
x = 87 - 24 + 5.3 - 14.0
x = 54.3m
Therefore, at time, t = 10s, x = +54.3m. (i.e. 54.3 on the +ve x axis).
Acceleration of a body is the change in velocity with respect to time.
The x-coordinate of the object when value of time is 10 second is 54.3 m.
What is acceleration?Acceleration of a body is the change in velocity with respect to time.
Given information-
A small object moves along the x-axis with acceleration,
[tex]ax(t)=-0.0320(15-t)[/tex]
Initial position and initial velocity of the object is,
[tex]x_0=-14\rm m\\v_0=8.7\rm m/s[/tex]
For the velocity integrate the given equation as,
[tex]ax(t)=-0.0320(15-t)\\\dfrac{dv}{dt} =\int ({-0.48+0.032t} )\, dt\\v-v_0=-0.48t+0.032\times\dfrac{t^2}{2} \\v-8.7=-0.48t+0.016t^2\\v=0.016t^2-0.48t+8.7[/tex]
Integrate it again to find the distance of the object.
[tex]v=0.016t^2-0.48t+8.7\\\int\dfrac{dx}{dt}=\int(0.016t^2-0.48t+8.7)dt\\x-x_0=0.016\times\dfrac{t^3}{3}-0.48\times\dfrac{t^2}{2}+8.7t\\x-14=0.016\times\dfrac{t^3}{3}-0.48\times\dfrac{t^2}{2}+8.7t\\\\x=0.016\times\dfrac{t^3}{3}-0.48\times\dfrac{t^2}{2}+8.7t-14\\[/tex]
Put the value of [tex]t[/tex] as 10 seconds as,
[tex]x=0.016\times\dfrac{10^3}{3}-0.48\times\dfrac{10^2}{2}+8.7\times10-14\\[/tex]
[tex]x=54.3\rm m[/tex]
Hence the x-coordinate of the object when value of time is 10 second is 54.3 m.
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The speed of a boat in still water is v. The boat is to make a round trip in a river whose current travels at speed u.
A) Derive a formula for the time needed to make a round trip of total distance D if the boat makes the round trip by moving upstream and back downstream.
Express your answer in terms of the variables D, v, and u.
B) Derive a formula for the time needed to make a round trip of total distance D if the boat makes the round trip by moving directly across the river and back.
Express your answer in terms of the variables D, v, and u.
Answer:
a) time needed to make a round trip = DV/v2 - u2
b) time needed to make a round trip = D/root ( v2 -u2)
Explanation:
The detailed explanation is as shown in the attachment
For a round trip moving upstream and downstream, the formula for the total time needed is D / (v - u) + D / (v + u). But, if the boat moves directly across the river and back, the total time needed is D / sqrt(v² - u²). This involves the concept of relative velocity.
Explanation:First, we need to understand the concept of relative velocity here. When the boat travels upstream, it moves against the current, so its net speed is (v - u). Downstream, it moves with the current at a net speed of (v + u).
A) The total time for a round trip moving upstream and downstream is the total distance divided by the speed in each direction. The time to travel a distance D upstream is D / (v - u), and the time to travel the same distance downstream is D / (v + u). So, the formula for the total time needed is D / (v - u) + D / (v + u).
B) If the boat is going directly across the river and back, it is moving perpendicular to the current. The boat makes the round trip at a constant speed of sqrt(v² - u²). Hence, the time needed for a round trip of total distance D is D / sqrt(v² - u²).
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A Ford Taurus driven 12,000 miles a year will use about 650 gal of gasoline compared to a Ford Explorer that would use 850 gal. About 19.8 lbm of CO2, which causes global warming, is released to the atmosphere when a gallon of gasoline is burned. Determine the extra amount of CO2 production a man is responsible for during a 5-year period if he trades in his Taurus for an Explorer.
Answer:
19 800 lbm of carbon dioxide more.
Explanation:
Taurus
Amount of gasoline used in 5 years = 650 ga/year * 5 years = 3250 ga
amount of carbon dioxide released = 19.8 lbm/ga * 3250 ga = 64 350 lbm
Explorer
Amount of gasoline used in 5 years = 850 ga/year * 5 years = 4250 ga
amount of carbon dioxide released = 19.8 lbm/ga * 4250 ga = 84 150 lbm
Extra amount of CO2 released = 84 150 lbm - 64 350 lbm = 19 800 lbm
The extra amount of carbon dioxide emission by a man during his 5 years of trading in Taurus and Explorer is 19,800 lbm.
Taurus
Amount of gasoline used in 5 years =
[tex]\bold {650\ gallon/year \times 5 = 3250\ gallon}[/tex]
Amount of carbon dioxide released =
[tex]\bold {19.8\ lbm/gallon \times 3250 = 64 350 lbm}[/tex]
Explorer
Amount of gasoline used in 5 years =
[tex]\bold {850\ gallons/year \times 5 = 4250\ gallons }[/tex]
Amount of carbon dioxide released =[tex]\bold { 19.8\ lbm/gallons \times 4250 gallons = 84 150\ lbm}[/tex]
Extra amount of [tex]\bold {CO_2}[/tex] released =
[tex]\bold {= 84 150 - 64 350 = 19 800\ lbm}[/tex]
Therefore, the extra amount of carbon dioxide emission by a man during his 5 years of trading in Taurus and Explorer is 19,800 lbm.
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Unlike a phonograph record that has a constant angular speed, a CD scans information at a constant linear speed (130 cm/s). Does a CD therefore rotate at a constant angular speed, or varying angular speed?
Answer: A CD rotate at a varying angular speed
A CD rotates at varying angular speed despite having a constant linear speed. This variation in rotation speed is necessary for the CD to operate efficiently.
Unlike a phonograph record, a CD rotates at varying angular speed even though it scans information at a constant linear speed due to the layout of data on the disc. The rotation speed decreases as the laser moves from the inside to the outside of the CD, maintaining a constant linear speed and adjusting the angular speed accordingly. This change in angular speed is essential for the CD to function effectively.