To convert the equatorial diameter of the Moon from kilometers to miles, you can use the given conversion factor:
Equatorial diameter in miles = Equatorial diameter in kilometers × Conversion factor
Given:
Equatorial diameter of the Moon = 3476 kilometers
Conversion factor = 0.6214 miles/kilometer
Now, plug in the values:
Equatorial diameter in miles = 3476 km × 0.6214 miles/km
[tex]\[ \text{Equatorial diameter in miles} \approx 2160.9264 \, \text{miles} \][/tex]
So, the Moon's equatorial diameter is approximately 2160.93 miles.
Final answer:
To convert the Moon's diameter from 3476 kilometers to miles, multiply by the conversion factor of 0.6214 miles per kilometer, resulting in approximately 2160.6344 miles.
Explanation:
The question asks for the conversion of the Moon's diameter from kilometers to miles. The given diameter of the Moon is 3476 kilometers. To convert kilometers to miles, we use the provided conversion factor where 1 kilometer equals 0.6214 miles.
Here's how you can calculate the Moon's diameter in miles:
1. Multiply the Moon's diameter in kilometers by the conversion factor:
3476 kilometers × 0.6214 miles/kilometer
2. Calculating this gives:
2160.6344 miles
So, the Moon's diameter in miles is approximately 2160.6344.
A transverse wave on a string of amplitude 0.12 m and wavelength 12.16 m propagates with speed 112 m/s. What is the maximum speed a point on the medium moves as this wave passes?
To develop this problem we will start using the concept of maximum speed for this type of systems. The maximum velocity can be described as the product between the Amplitude and the Angular velocity. At the same time, said angular velocity can be found through the relationship between linear and "angular wavenumber" velocity. The Angular wavenumber is a wave number defined as the number of radians per unit distance. Finally with the value of the angular velocity found we will proceed to find the maximum speed.
The maximum speed is given by
[tex]v_{max} = A\omega[/tex]
Here,
A = Amplitude
[tex]\omega[/tex]= Angular velocity
The angular velocity can be described as the number of radians per unit distance
[tex]\omega = vk[/tex]
[tex]\omega = v (\frac{2\pi}{\lambda})[/tex]
[tex]\omega = 112(\frac{2\pi}{12.16})[/tex]
[tex]\omega =57.8714rad/s[/tex]
Then,
[tex]v_{max} = 0.12 *57.8714[/tex]
[tex]v_{max} = 6.94m/s[/tex]
Therefore the maximum speed a point on the medium moves as this wave passes is 6.94m/s
A free electron and a free proton are released in identical electric fields.
(i) How do the magnitudes of the electric force exerted on the two particles compare?
It is millions of times greater for the electron.
It is thousands of times greater for the electron.
They are equal.
It is thousands of times smaller for the electron.
It is millions of times smaller for the electron.
(ii) Compare the magnitudes of their accelerations.
It is millions of times greater for the electron.
It is thousands of times greater for the electron.
They are equal.It is thousands of times smaller for the electron.
It is millions of times smaller for the electron.
Answer:
i) They are equal . ii) It is thousands of times greater for the electron.
Explanation:
i) By definition, the electric field is the electric force per unit charge. If the field is the same, the force will depend on the value of the charge under the influence of the field.
As the magnitude of the charge of the electron and the proton are the same, we conclude that the electric force on both must be equal in magnitude.
ii) The acceleration on both particles must meet the Newton´s 2nd Law, so, if the forces are equal in magnitude (neglecting any other external interaction), the acceleration will only depend on the mass of both particles, according this general expression:
a = F/m
As the mass of the electron is approximately two thousands times smaller than the proton´s, it concludes that the acceleration on the electron must be thousands of times greater for the electron.
The magnitude of the electric force exerted on the electron is millions of times greater than that exerted on the proton. The magnitudes of their accelerations are equal.
Explanation:(i) The magnitude of the electric force exerted on the electron is millions of times greater than that exerted on the proton. This is because the electron has a much smaller mass compared to the proton, and the electric force depends on the charge and mass of the particle.
(ii) The magnitudes of their accelerations are equal. The acceleration of a particle in an electric field depends only on the magnitude of the electric field and the mass of the particle. Since both the electron and proton experience the same electric field, their accelerations will be the same.
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In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components. Let vectors A rightarrow = (1,0, -3), rightarrow = (-2,5,1), and C rightarrow = (3,1,1). Calculate the following, and express your answers as ordered triplets of values separated by commas.
Answer:
The calculated vectors are:
[tex]\vec{A}-\vec{B}=(3,-5,-4)[/tex]
[tex]\vec{B}-\vec{C}=(-5,4,0)[/tex]
[tex]-\vec{A}+\vec{B}-\vec{C}=(-6,4,3)[/tex]
[tex]3\vec{A}-2\vec{C}=(-3,-2,-11)[/tex]
Explanation:
To operate with vectors, you sum or rest component to component. To multiply scalars with vectors, you distribute the scalar with each component of the vector. These are the following rules you must apply in these cases:
[tex]\vec{V}+\vec{W}=(V_1,V_2,V_3)+(W_1,W_2,W_3)=(V_1+W_1,V_2+W_2,V_3+W_3)[/tex] (1)
[tex]\vec{V}-\vec{W}=(V_1,V_2,V_3)-(W_1,W_2,W_3)=(V_1-W_1,V_2-W_2,V_3-W_3)[/tex] (2)
[tex]\alpha\cdot\vec{V}=\alpha\cdot(V_1,V_2,V_3)=(\alpha\cdot V_1,\alpha\cdot V_2,\alpha\cdot V_3)[/tex] (3)
The operations in these cases are:
[tex]\vec{A}-\vec{B}=(1,0, -3)-(-2,5,1)=(3,-5,-4)[/tex]
[tex]\vec{B}-\vec{C}=(-2,5,1)-(3,1,1)=(-5,4,0)[/tex]
[tex]-\vec{A}+\vec{B}-\vec{C}=-(1,0, -3)+(-2,5,1)-(3,1,1)=(-6,4,3)[/tex]
[tex]3\vec{A}-2\vec{C}=3(1,0, -3)-2(3,1,1)=(3,0, -9)-(6,2,2)=(-3,-2,-11)[/tex]
Answer:
Part A: (3, -5, -4)
Part B: (-5, 4, 0)
Part C: (-6, 5, 3)
Part D: (-3, -2, -11)
Part E: (17, -12, -6)
Explanation:
This problem involves addition and subtraction of vectors. This can be done by adding and subtracting the respective components of each vector as the case may be.
The full descriptive solution can be found in the attachment below.
If A is the amplitude of a mass on an oscillating spring, then in one period the mass travels a distance of ___________.
a. 0
b. A
c. 2A
d. 4A
Answer:
d. 4A
Explanation:
Given that
Amplitude = A
We know that the distance cover by a particle before coming the the rest from the mean point in the oscillation motion is known as amplitude.
The distance cover by particle from 1 - 2 = A
The distance cover by particle from 2 - 1 = A
The distance cover by particle from 1 - 3 = A
The distance cover by particle from 3-4 = A
Therefore the total distance cover by particle = A+A+A+A = 4 A
Therefore the total distance = 4 A
That is why the answer will be 4 A.
d. 4A
You slide a coin across the floor, and observe that it slows down and eventually stops. A sensitive thermometer shows that its temperature increased. What can we conclude? (Select all that apply.)
When you slide a coin across the floor, it slows down and eventually stops due to the force of friction. The friction converts the coin's kinetic energy into thermal energy, resulting in an increase in temperature.
Explanation:When you slide a coin across the floor, it eventually slows down and stops due to the force of friction acting on it. Friction is a force that opposes the motion of objects in contact, and it causes the coin to lose kinetic energy, slowing it down. As the coin slows down, its kinetic energy is converted into thermal energy, increasing the temperature of the coin and the surface it slides on. This is why a sensitive thermometer shows an increase in temperature when the coin slides.
If you wanted to move an electron from the positive to the negative terminal of the battery, how much work W would you need to do on the electron
Answer:
W = qV
Explanation:
Let V be the potential difference of the battery, and q be the charge on the electron.
The work done in moving a charge through a potential difference V =?
Work = force * distance
Work = F.r
F = qE
But E = V / r
F = q. V / r
Work ( W ) = (qV/ r ) * r
Work = qV
Therefore the force required to move a charge through a potential difference V = qV
The amount of work needed to move an electron from the positive to the negative terminal of a battery using the equation W = qV, where q is the charge of the electron and V is the voltage of the battery is 1.92 x 10-18 joules.
Explanation:In order to move an electron from the positive to the negative terminal of a battery, you would need to do work on the electron.
The amount of work, W, can be calculated using the equation W = qV, where q is the charge of the electron and V is the voltage of the battery.
The charge of an electron is approximately 1.6 x 10-19 coulombs.
For example, if the voltage of the battery is 12 volts, the work done on the electron would be:
W = ([tex]1.6[/tex]×[tex]10^{-19}[/tex][tex]C[/tex]) × ([tex]12[/tex] [tex]v[/tex]) = [tex]1.92[/tex]×[tex]10^{-18}[/tex] joules.
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Unless otherwise specified in the question, responses to numerical question must be within 5% of the correct answer. Try out different responses to the following question, which has an answer of 20, where a correct response is between 19 and 21.
What is the average velocity (in kilometers per hour) of a cart that travels 100 km in 5 hours?
2.An exception to the 5% tolerance rule arises for simple calculations that can be done in your head that come out as integers or or simple decimals. So if the calculation involves dividing 5 by 2 with the result of 2.5, an answer of 2.6 may not be graded as correct.
Give the value of 6.5 times 2.
Try it several times giving the answer (13) in different forms (say, 13 or 1.3e1) and with different accuracies (say, 13.001 or 13.2). Since this is a simple calculation, two of these four will be graded as incorrect.
Part of the problem involves not only answering the question with the correct degree of accuracy, but also making approximations to the correct answer. For example, for the first case we know that the speed is equivalent to the distance traveled in a given time. Therefore it would be defined as
]1) [tex]V = \frac{x}{t}[/tex]
x = Displacement
t = Time
The displacement value is 100km and the time value is 5 hours. Therefore the speed value would be
[tex]V = \frac{100km}{5h} = 20km/h[/tex]
We know that the answer margin is within 5% of the value, then 5% of 20 would be 1. That is, we have a margin of error of '1km / h' to answer the question. Any value that falls within that range can be added or subtracted from the response and the response will be valid. Values included within this value would be
[tex]V_1 = 20km/h +1km/h = 21km/h[/tex]
[tex]V_2 = 20km/h - 1km/h = 19km/h[/tex]
[tex]V_3 = 20km/h + 0.5km/h = 20.5km/h[/tex]
[tex]V_4 = 20km/h -0.5km/h = 19.5km/h[/tex]
[tex]V_5 = 20km/h +0.01km/h = 20.01km/h[/tex]
2) For the second case the margin of tolerance for the response is 5%, so if we multiply the given value we would have a response of.
[tex]x = 6.5*2 = 13[/tex]
5% of 13 is 0.65. Therefore, any value that falls within that range will be a correct answer. The value could then be
[tex]x_1 = 13+0.65 = 13.65[/tex]
[tex]x_2 = 13-0.65 =12.35[/tex]
Incorrect values will be
[tex]x_3 = 13+1 = 14[/tex]
[tex]x_4 = 13-1=12[/tex]
A car starts from rest at a stop sign. It accelerates at 4.6 m/s^2 for 6.2 s , coasts for 2.1s , and then slows down at a rate of 3.3 m/s^2 for the next stop sign. How far apart the are the stop signs?
Answer:
D = 271.54 m
Explanation:
given,
1. car accelerates at 4.6 m/s² for 6.2 s
2. constant speed for 2.1 s
3. slows down at 3.3 m/s²
distance travel for case 1
using equation of motion
[tex]d_1 = u t +\dfrac{1}{2}at^2[/tex]
[tex]d_1 =\dfrac{1}{2}\times 4.6\times 6.2^2[/tex]
d₁ = 88.41 m
case 2
constant speed for 2.1 s now, we have to find velocity
v = u + at
v = 0 + 4.6 x 6.2
v = 28.52 m/s
distance travel in case 2
d₂ = v x t
d₂ = 28.52 x 2.1 = 59.89 m
for case 3
distance travel by the car
v² = u² + 2 a s
final velocity if the car is zero
0² = 28.52² + 2 x (-3.3) x d₃
6.6 d₃ = 813.39
d₃ = 123.24 m
total distance travel by the car
D = d₁ + d₂ + d₃
D = 88.41 + 59.89 + 123.24
D = 271.54 m
The acceleration due to gravity on the moon is about 5.4 ft/s2 . If your weight is 150 lbf on the earth:
What is your mass on the moon, in slugs.
Answer:
4.662 slugs
Explanation:
Your mass on the moon should always be the same as any planet you are on (due to law of mass conservation), only your weight be different as gravitational acceleration is different on each planet.
If you weight 150 lbf on Earth, and gravitational acceleration on Earth is 32.174 ft/s2. The your mass on Earth is
m = W / g = 150 / 32.174 = 4.662 slugs
which is also your mass on the moon.
In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The front end crumples so much that the truck is 0.62 m shorter than before.
(a) What is the average speed of the truck during the collision (that is, during the interval between first contact with the wall and coming to a stop)?
vavg = ______ m/s
(b) About how long does the collision last? (That is, how long is the interval between first contact with the wall and coming to a stop?)
Δt = ______ s
(c) What is the magnitude of the average force exerted by the wall on the truck during the collision?
Fwall, avg = ______ N
(d) It is interesting to compare this force to the weight of the truck. Calculate the ratio of the force of the wall to the gravitational force mg on the truck. This large ratio shows why a collision is so damaging.
Fwall, avg / mg =
(e) What approximations were necessary in making this analysis? (Select all that apply.)
Neglect the horizontal component of the force of the road on the truck tires.
Assume a nearly constant force exerted by the wall, so the speed changes at a nearly constant rate.
The deceleration of the truck is approximately equal to g.
Answer:
a) 11 m/s
b) 0.0564 s
Explanation:
Given:
m = 2100 kg
vi = 22 ..... m/s before collision
vf = 0 ......after collision to stop
Δs = 0.62 distance traveled after collision .. crumpling of truck
Part a
[tex]V_{avg} = \frac{vi- vf}{2}\\\\V_{avg} = \frac{22- 0}{2}\\\\ V_{avg} = 11 m/s[/tex]
Part b
[tex]vf = vi + a*t\\vf^2 = vi^2 + 2*a*s\\\\0 = 22 + a*t\\t = -22 /a\\\\0 = 484 +2*a*(0.62)\\a = - 390.3232 m/s^2\\\\t = -22/(-390.3232)\\\\t = 0.0564 s[/tex]
(a) The required average speed of the truck during collision is 11 m/s.
(b) The required time interval for the collision is 0.058 s.
(c) The required magnitude of the average force exerted by the wall on the truck is [tex]7.96 \times 10^{5} \;\rm N[/tex].
(d) The required ratio of the force on the truck and the gravitational force is 38.67 : 1.
(e) The required approximations are:
Neglect the horizontal component of the force of the road on the truck tires.Assume a nearly constant force exerted by the wall, so the speed changes at a nearly constant rate.The section of analysis that deals with the motion of any object in one dimension are known as linear kinematics. The terms such as speed, velocity, and acceleration are the variables under kinematics.
Given data:
The mass of the truck is, m = 2100 kg.
The speed of the truck is, v = 22 m/s.
The distance crumpled by the truck is, d = 0.62 m.
(a)
Since the truck is going to stop finally (v' = 0) therefore the average speed is calculated as,
[tex]v_{av.}=\dfrac{v - v'}{2}\\\\ v_{av.}=\dfrac{22 - 0}{2}\\\\ v_{av.}=11 \;\rm m/s[/tex]
Thus, the required average speed of the truck during collision is 11 m/s.
(b)
Now, apply the first kinematic equation of motion as,
[tex]v' = v + at[/tex]
Here, a is the linear acceleration and t is the time interval for the collision.
Solving as,
[tex]0 = 22 + at\\\\ a = -22/t[/tex]
Now, apply the second kinematic equation as,
[tex]v'^{2}=u^{2}+2ad\\\\ 0^{2}=22^{2}+2 \times \dfrac{-22}{t} \times 0.62\\\\ \dfrac{27.28}{t}=484\\\\ t = 0.058 \;\rm s[/tex]
Thus, we can conclude that the required time interval for the collision is 0.058 s.
(c)
The expression for the magnitude of average force exerted by wall on truck is,
[tex]F_{av.} = \dfrac{mv}{t}[/tex]
Solving as,
[tex]F_{av.}=\dfrac{2100 \times 22}{0.058}\\\\ F_{av.}=7.96 \times 10^{5} \;\rm N[/tex]
Thus, the required magnitude of the average force exerted by the wall on truck is [tex]7.96 \times 10^{5} \;\rm N[/tex].
(d)
The ratio of force of the truck and the gravitation force on the truck is,
[tex]= \dfrac{F_{av}}{mg}[/tex]
Here, g is the gravitational acceleration.
Solving as,
[tex]=\dfrac{7.96 \times 10^{5} \;\rm N}{2100 \times 9.8}\\\\ =38.67[/tex]
Thus, the required ratio of the force on the truck and the gravitational force is 38.67 : 1.
(e)
The approximations that were necessary to make these approximations include the negligence of the horizontal component of a force and the assumption of a constant force exerted by the wall.
This is because:
The horizontal force may act as an unbalanced force.And the constant force by wall results in the variation of the speed of the truck at the constant rate.Learn more about the linear kinematics here:
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A thin cylindrical shell and a solid cylinderhave the same mass and radius. The two arereleased side by side and roll down, withoutslipping, from the top of an inclined planethat is 1.3 m above the ground.Find the final linear velocity of the thincylindrical shell.The acceleration of gravity is9.8 m/s2
Answer:
v = 5.05m/s
Explanation:
H = 1.3m
initial velocity = 0
final velocity = v = ?
g =9.8 m/s^2
we apply the conservation of energy; all potential energy is comletely converting to kinetic energy
[tex]mgh = \frac{mv^{2}}{2}[/tex]
mass is same; it cancels out
[tex]v =\sqrt{2gh} = \sqrt{2(9.81)(1.3)}[/tex]
v = 5.05m/s
Answer:
the final linear velocity is 3.56931 m/s
Explanation:
the solution is in the attached Word file
A simple harmonic oscillator completes 1550 cycles in 30 min. (a) Calculate the period. s (b) Calculate the frequency of the motion. Hz
Answer:
(a) 1.16 s
(b)0.861 Hz
Explanation:
(a) Period : The period of a simple harmonic motion is the time in seconds, required for a object undergoing oscillation to complete one cycle.
From the question,
If 1550 cycles is completed in (30×60) seconds,
1 cycle is completed in x seconds
x = 30×60/1550
x = 1.16 s
Hence the period is 1.16 seconds.
(b) Frequency : This can be defined as the number of cycles that is completed in one seconds, by an oscillating body. The S.I unit of frequency is Hertz (Hz).
Mathematically, Frequency is given as
F = 1/T ........................... Equation 1
Where F = frequency, T = period.
Given: T = 1.16 s.
Substitute into equation 1
F = 1/1.16
F = 0.862 Hz
Hence thee frequency = 0.862 Hz
A parallel-plate capacitor is constructed of two square plates, size L x L, separated by distance d. The plates are given charge ±Q.
a. What is the ratio Ef/Ei of the final to initial electric field strengths if L is doubled?
Answer:
Explanation:
Given
Area of capacitor Plates [tex]A=L\times L[/tex]
distance between plates is d
capacitance C is given by
[tex]C=\frac{\epsilon A}{d}[/tex]
[tex]C=\frac{\epsilon \cdot L^2}{d}[/tex]
Provided V is Voltage
[tex]Charge(Q)=capacitance(C)\times Voltage(V)[/tex]
If L is doubled
Capacitance [tex]C'=\frac{\epsilon \cdot (2L)^2}{d}[/tex]
[tex]C'=4\times \frac{\epsilon \cdot L^2}{d}[/tex]
Electric field is given by
[tex]E=\frac{Q}{\epsilon _0A}[/tex]
[tex]E_i=\frac{Q}{\epsilon _0L^2}---1[/tex]
[tex]E_f=\frac{Q}{\epsilon _0(2L)^2}---2[/tex]
divide 1 and 2 we get
[tex]\frac{E_i}{E_f}=\frac{(2L)^2}{L^2}[/tex]
[tex]\frac{E_f}{E_i}=\frac}{1}{4}[/tex]
A block of wood is floating in water; it is depressed slightly and then released to oscillate up and down. Assume that the top and bottom of the block are parallel planes which remain horizontal during the oscillations and that the sides of the block are vertical. Show that the period of the motion (neglecting friction) is 2π (sqrt(h/g), where h is the vertical height of the part of the block under water when it is floating at rest. Hint: Recall that the buoyant force is equal to the weight of displaced water.Solve using 2nd ODEs was the instructions given by the professor.
Answer:
Please refer to the attachment below since we need to prove that the period of motion is 2π*(sqrt(h/g))
Explanation:
Please refer to the attachment for explanation.Answer:
The Proof for T=2π (sqrt(h/g) for a floating block exhibiting SHM is shown in the pictures attached below
Explanation:
Which of the following is true about free energy? A. It is all the energy that a molecule contains B. The free energy change of a reaction is dependent on the reaction path C. The free energy change of a reaction is dependent on the enzyme D. The free energy change of a reaction can determine the reaction direction
Free energy is a thermodynamic potential that can be used to calculate the maximum reversible work that can be performed by a thermodynamic system at a constant temperature and pressure. It is fulfilled that if the energy change is less than zero it will mean that the relationship will proceed towards the product, while if the relationship is greater than zero the reaction will proceed towards the reactant. Therefore the correct option is D.
The free energy change of a reaction can determine the reaction direction
An individual is nearsighted; his near point is 18.0 cm and his far point is 40.0 cm.
(a) What lens power is needed to correct his nearsightedness?
-2.5diopters
(b) When the lenses are in use, what is this person's near point?...?
Answer
given,
near point = 18 cm
far point = 40 cm
a) The lens should form an upright, virtual image at far point from the distant object.
therefore, f = q = -40 cm = -0.4 m
where f is the focal length.
the required power
[tex]P =\dfrac{1}{f}[/tex]
[tex]P =\dfrac{1}{-0.40}[/tex]
P = -2.5 D
b) If the lens is used the Person's near point
The lens should form an upright, virtual image at near point from the distant object should be q = - 18 cm = -0.18 m
[tex]p = \dfrac{qf}{q-f}[/tex]
[tex]p = \dfrac{(-0.18)(-0.4)}{-0.18-(-0.4)}[/tex]
p = 32.72 cm
The person's near point is 32.72 cm
The voltage across a membrane forming a cell wall is 74.0 mV and the membrane is 8.40 nm thick. What is the electric field strength in volts per meter?
Answer:
Electric field strength=8.81×10⁶V/m
Explanation:
Given Data
voltage v= 74.0 mV
Membrane thickness d=8.40 nm
To find
Electric field strength E=?
Solution
Electric field strength =voltage/Membrane thickness
[tex]E=v/d\\E=\frac{74.0*10^{-3} }{8.40*10^{-9} }\\ E=8.81*10^{6}V/m[/tex]
Calculate the wavelengths of the first five emission lines of the Balmer series for hydrogen
Answer:
Explanation:
The Balmer series in a hydrogen atom relates the possible electron transitions down to the n = 2 position to the wavelength of the emission that scientists observe. In quantum physics, when electrons transition between different energy levels around the atom (described by the principal quantum number, n) they either release or absorb a photon. The Balmer series describes the transitions from higher energy levels to the second energy level and the wavelengths of the emitted photons. You can calculate this using the Rydberg formula.
Consider n equal positive charged particles each of magnitude Q/n placed symmetrically around a circle of radius a. Calculate the magnitude of the electric field at a point a distance x from the center of the circle and on the line passing through the center and perpendicular to the plane of the circle. (Use any variable or symbol stated above along with the following as necessary: ke.) (b) Now consider a ring of radius a that carries a uniformly distributed positive total charge Q. Recall the calculation of the electric field at point a point a distance x from the center of the ring and on the line passing through the center and perpendicular to the plane of the ring. Explain why the result in part (a) is identical to the result for the ring.
Answer:
To make it easier to Understand, consider the circle to be in the usual 3 dimensional x-z plane (the "horizontal plane") and the point of measurement C to be at a distance p (instead of x to avoid confusion with the x-axis) on the y-axis (the "vertical direction").
The answer for a and b is the same. This is because a the horizontal component of a charged portion of a uniform ring is canceled by the opposite portion of the ring since they are placed at an equal distance from the position in which the electric field is being measured or applied, just as are the opposing point charges described in part a.
Explanation:
A.) Using Coulomb's law of point charges, each charge on the circle would exert a field Ec at C given by:
(1) Ec = Ke * (Q / n) / d²
where:
Ke is Coulomb's constant,
Q / n is the magnitude of the charge, and
d = the distance between the charge and the point of measurement C, with d² = a² + c²
Since the charges are in a circle in the x-z plane, all force components in both the x- and the z-directions are canceled by symmetry; the vertical force (that in the y-direction) is the only component that does not cancel.
Therefore the resultant vector Ecy points up (+y-direction) and has a magnitude of:
(2) Ecy = Eq * sin(theta)
= (Ke * (Q / n) / d²) * (c / d)
Then, summing the forces from all the charges, the magnitude of the total electric field is given by:
(3) Ey = n * Ecy
= n * [ (Ke * (Q / n) / d²) * (c / d) ]
= c * Ke * Q / d^3
B.) The equation is the same. This is because both the x- and z-components (the two planar components) of a charged portion of a uniform ring are canceled by the opposite portion of the ring, since they lie at an equal distance but opposite direction from c.
This is the same way the opposing point charges described in part A behave.
In both scenarios described, because of symmetry, the electric fields generated by opposing charge pairs cancel each other out, resulting in a net electric field of zero at the specified point.
Explanation:The magnitude of the electric field in the scenario described is given by Coulomb's law, which states that the electric field E due to a charge Q at a distance r from a point is given by E = ke*(Q/r^2), where ke is Coulomb's constant.
In this case, the magnitude of each charge is Q/n and we have n charges symmetrically distributed along the circle. Because we are calculating the electric field at a point on the line passing through the center of the circle and perpendicular to the plane of the circle, each pair of charges on the circle will produce an electric field that is symmetric and thus its contribution will be canceled out by another pair's contribution. The final resultant electric field will hence be zero.
(b) Now let's consider a ring of radius a that carries a uniformly distributed positive total charge Q. Because the charge is symmetrically distributed, the electric field at any point on the line passing through the center and perpendicular to the plane of the ring will be sum of electric fields due to all small charged particles which make up this ring, and due to symmetry of the distribution, it will add up to zero.
Therefore, the results in part (a) and (b) are identical because in both cases, due to symmetry, the electric fields generated by opposing charge pairs cancel each other out, resulting in a net electric field of zero at the specified point.
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The column has a mass of 600 lb/ft and a force of 50 kips applied at the top. The diagonal brace has a 10 kip axial force that can be either tension or compression (note that in seismic and wind loading on buildings, braces experience both tension and compression). Assume all elements are pin-connected at their ends, including the two supports.
Answer: the maximum compression force = 59.47 kip
the minimum compression force = 43.33 kip
Explanation:
In this we required to do force balance in order to get maximum and minimum compression force in the column.
the picture below explains the steps to solve the question with diagrams to ease understanding.
The charge entering the positive terminal of an element is q = 5 sin 4πt mC while the voltage across the element (plus to minus) is v = 3 cos 4πt V.
(a) Find the power delivered to the element at t = 0.3 s.
(b) Calculate the energy delivered to the element between 0 and 0.6 s.
Answer:
(a). The power delivered to the element is 187.68 mW
(b). The energy delivered to the element is 57.52 mJ.
Explanation:
Given that,
Charge [tex]q=5\sin4\pi t\ mC[/tex]
Voltage [tex]v=3\cos4\pi t\ V[/tex]
Time t = 0.3 sec
We need to calculate the current
Using formula of current
[tex]i(t)=\dfrac{dq}{dt}[/tex]
Put the value of charge
[tex]i(t)=\dfrac{d}{dt}(5\sin4\pi t)[/tex]
[tex]i(t)=5\times4\pi\cos4\pi t[/tex]
[tex]i(t)=20\pi\cos4\pi t[/tex]
(a).We need to calculate the power delivered to the element
Using formula of power
[tex]p(t)=v(t)\times i(t)[/tex]
Put the value into the formula
[tex]p(t)=3\cos4\pi t\times20\pi\cos4\pi t[/tex]
[tex]p(t)=60\pi\times10^{-3}\cos^2(4\pi t)[/tex]
[tex]p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi t}{2})[/tex]
Put the value of t
[tex]p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi\times0.3}{2})[/tex]
[tex]p(t)=30\pi\times10^{-3}(1+\cos8\pi \times0.3)[/tex]
[tex]p(t)=187.68\ mW[/tex]
(b). We need to calculate the energy delivered to the element between 0 and 0.6 s
Using formula of energy
[tex]E(t)=\int_{0}^{t}{p(t)dt}[/tex]
Put the value into the formula
[tex]E(t)=\int_{0}^{0.6}{30\pi\times10^{-3}(1+\cos8\pi \times t)}[/tex]
[tex]E(t)=30\pi\times10^{-3}\int_{0}^{0.6}{1+\cos8\pi \times t}[/tex]
[tex]E(t)=30\pi\times10^{-3}(t+\dfrac{\sin8\pi t}{8\pi})_{0}^{0.6}[/tex]
[tex]E(t)=30\pi\times10^{-3}(0.6+\dfrac{\sin8\pi\times0.6}{8\pi}-0-0)[/tex]
[tex]E(t)=57.52\ mJ[/tex]
Hence, (a). The power delivered to the element is 187.68 mW
(b). The energy delivered to the element is 57.52 mJ.
To solve the problem, the power at a certain moment (t=0.3s) is calculated by substitifying the values of voltage and charge into the power formula. The energy delivered to the element can be calculated by integrating the power function over the period from 0 to 0.6 s.
Explanation:The subject of this question is related to electrical power and energy in a circuit, and it requires a knowledge of trigonometry. The instantaneous power in an electrical circuit is given by the product of charge (q) and voltage (v). Hence we can calculate the power at t = 0.3s by substituting the given values into the power formula.
Power, p = qv = (5 sin 4πt) * (3 cos 4πt) = 15 sin 4πt cos 4πt. At t = 0.3s, power p = 15 sin 4π * 0.3 cos 4π * 0.3 = 15 sin 1.2π cos 1.2π.
For part (b), the energy delivered to the element between 0 and 0.6 s can be obtained from the integral of the power function over the given time interval, which requires integration skills in calculus.
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A U.S. Department of Energy report estimates that over 100 billion kWh/year can be saved in the United States by various energy-conservation techniques applied to the pump-driven systems. Calculate (a) how many 1000-MW generating plants running constantly supply this wasted energy and (b) the annual savings in dollars if the cost of electricity is 0.10 $/kWh.
Answer:
(a). 12 plants
(b). 3171 $
Explanation:
(a)first convert units of 100 billion kWh/year into Watts(W)
also convert the units of 1000 MW into Watts(W)
1 billion = 10^9
1 year = 365*24 = 8760 hrs
so
100 billion kWh/year = 1[tex]\frac{100*(10^9)*(10^3)}{8760}[/tex]
= [tex]1.142*10^{10}[/tex]W
1000 MW = [tex]1000*10^{6} = 10^{9}W[/tex]
no. of plants = [tex]\frac{1.14155*10^{10} }{10^9}[/tex] = 11.4
So 12 plants required
(b)
savings = unit price*total units
= [tex]0.1 * 1.142*10^{10}( \frac{1}{1000*3600} )[/tex]
= 3170.9 =3171 $
Answer:
a) Number of generating plants N = 11.42
That means N > 11
N = 12
b) annual savings S = $1×10^10
S = $10 billion
Explanation:
Given;
Amount of energy to be saved A=100 billion kWh/year
Capacity of each generating plant C= 1000 MW
Rate in dollars of cost of electricity R= $0.10/kWh
The number N of generating plants with capacity C that can supply the Amount A of of energy cam be given as;
N = A/C ......1
And also the Annual savings S in dollars if the rate of electricity cost R is used and amount of energy A is saved is:
S = AR .....2
But we need to derive the value of A in Watts
A = 100 billion kWh/year
There are 8760hours in a year,
A = 1×10^14 ÷ 8760 W
A = 11415525114.1W or 11415525.1141kW
C = 1000MW = 1× 10^9 W
a) Using equation 1,
N = 11415525114.1/(1×10^9)
N = 11.42
That means N > 11
N = 12
b) using equation 2
S = 1×10^11 kWh × $0.10/kWh
S = $1×10^10
S = $10 billion
You are standing in a moving bus, facing forward, and you suddenly fall forward as the bus comes to an immediate stop. The force acting on you that causes you to fall forward is A) the normal force due to your contact with the floor of the bus. B) the force due to static friction between you and the floor of the bus. C) the force of gravity. D) the force due to kinetic friction between you and the floor of the bus. E) No forces were acting on you to cause you to fall.
Answer:
D
Explanation:
Static friction is a force that keeps an object at rest. It must be overcome to start moving the object. Once an object is in motion, it experiences kinetic friction.
When a bus suddenly stops, a passenger tends to keep moving forward due to inertia (Newton's first law). The force causing you to fall forward in this instance is the insufficient static friction between you and the floor of the bus, which isn't enough to counteract your forward momentum.
Explanation:The force acting on you that causes you to fall forward when the bus comes to an immediate stop is B) the force due to static friction between you and the floor of the bus. This is because of Newton's first law, also known as the law of inertia, which states that an object at rest tends to stay at rest and an object in motion tends to stay in motion, unless acted upon by an external force.
While you're standing in the moving bus, both you and the bus are moving forward. When the bus suddenly stops, your body tends to keep moving forward due to inertia, causing you to fall forward. Here, the static friction between you and the floor of the bus isn't enough to counteract your momentum, leading to your forward fall.
It's important to note that other forces such as gravity and normal force are in effect too. Gravity pulls you downward, and the bus floor exerts an upward normal force to counterbalance it. These two forces are balanced and don't contribute to your forward movement. The force causing the unbalanced motion (falling forward) is the lack of sufficient friction to oppose your inertia.
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It is possible to charge a balloon to several thousand volts using electrostatics, for example, by rubbing it on one’s hair or on different types of clothing. Does this mean that the balloon is also storing several Joules of energy? Why or why not?
Answer:
Yes it is possible to charge balloon to several thousand of volts and the balloon will also store several Joules of energy.
Explanation:
By rubbing a balloon on one's hair or on different types of clothing, the balloon either gain or loss electron.
If the balloon gains electron it becomes negatively charged, it contains more electron and subsequently charged to several thousand volts.
Also, if the balloon losses electron, it becomes positively charged. In this case it contains more proton, which makes the balloon positively charged to several thousand volts.
However, amount of joules depends on volts produced in the balloon.
Volt = Joules/coulomb,
Joules = volts*coulomb
Because the charge of the particles (electron and proton) are small, amount of joules will always be small than volts.
So, it is possible to charge balloon to several thousand of volts and the balloon will store several Joules of energy
A sandbag is dropped from a balloon which is ascending vertically at a constant speed of 5 m/s. The bag is released with the same upward velocity of 5 m/s when t = 0 and hits the ground when t = 8 s.
(a) Determine the speed of the bag as it hits the ground.
(b) Determine the altitude of the balloon when the bag hits the ground
Answer:
a) 73.48 m/s
b) 313.92 m
Explanation:
Data provided in the question:
ascending velocity = - 5 m/s [ negative sign depicts upward movement]
Time taken by bag to hit the ground, t = 8 s
a) from the Newton's equation of motion
we have
[tex]s=ut+\frac{1}{2}at^2[/tex]
where,
u is the initial speed
a is the acceleration = 9.81 m/s² (since it is a case of free fall )
s is the distance
thus,
[tex]s=(-5)(8)+\frac{1}{2}(9.81)(8)^2[/tex]
s = - 40 + 313.92
s = 273.92 m
from
v = u + at
v is the final speed
v = -5 + (9.81)(8)
or
v = 73.48 m/s
b) Distance traveled by balloon = Speed × Time
= 5 × 8
= 40 m
Therefore,
Altitude of the balloon
= Distance traveled by bag + Distance traveled by balloon
= 273.92 m + 40 m
= 313.92 m
The speed of the bag as it hits the ground is 5 m/s. The altitude of the balloon when the bag hits the ground is 40 meters.
Explanation:To determine the speed of the bag as it hits the ground, we can use the equation for velocity:
v = u + at
Since the bag is dropped with an initial velocity of 5 m/s and there is no acceleration in the vertical direction, the final velocity of the bag as it hits the ground is also 5 m/s.
To determine the altitude of the balloon when the bag hits the ground, we can calculate the distance traveled by the bag using the equation for distance:
s = ut + (1/2)at^2
Plugging in the values, we get:
s = 5(8) + (1/2)(0)(8^2) = 40 meters
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A charge Q is located inside a rectangular box. Theelectric flux through each of the six surfaces of the box is: electric flux 1 = +1500 N*m^2/C, electric flux 2 = +2200N*m^2/C, electric flux 3 = +4600 N*m^2/C, electric flux 4 = -1800N*m^2/C, electric flux 5 = -3500 N*m^2/C, and electric flux 6 =-5400 N*m^2/C. What is Q?
To solve this problem we will apply the laws of gaus that relate the Electric Flow as the charge of the object on the permittivity constant of free space. Mathematically this is
[tex]\phi = \frac{Q}{\epsilon_0}[/tex]
Rearranging to find the charge,
[tex]Q = \phi \epsilon_0[/tex]
Here
Q = Charge
[tex]\phi =[/tex] Electric Flux
[tex]\epsilon_0 =[/tex] Permittivity of free space
The total flux would be
[tex]\phi_T = \phi_1+\phi_2+\phi_3+\phi_4+...+\phi_{\infty}[/tex]
[tex]\phi = ( 1500+2200+4600-1800-3500-5400 ) N\cdot m^2 / C[/tex]
[tex]\phi = - 2400 N\cdot m^2 / C[/tex]
Replacing we have that,
[tex]Q = (-2400 N\cdot m^2/C)( 8.85*10^{-12} C^2 / N \cdot m^2)[/tex]
[tex]Q = -21240 * 10^{-12} C[/tex]
[tex]Q = - 21.24 nC[/tex]
Therefore the charge Q inside a rectangular box is -21.24nC
Final answer:
Using Gauss's Law, the charge Q inside the box can be calculated by summing the electric flux values across all six surfaces and then multiplying by the permittivity of free space, resulting in a charge of approximately -1.24 x 10^-8 C.
Explanation:
To determine the charge Q inside the rectangular box, we need to use Gauss's Law, which relates the electric flux through a closed surface to the charge enclosed within that surface. The net electric flux (Φnet) through a closed surface is equal to the charge inside (Q) divided by the permittivity of free space (ε0). Mathematically, this is expressed as Φnet = Q/ε0.
The net flux is the algebraic sum of the fluxes through each surface, so we have:
Φnet = (electric flux 1) + (electric flux 2) + (electric flux 3) + (electric flux 4) + (electric flux 5) + (electric flux 6)
Substituting the given values, we get:
Φnet = (+1500 N·m2/C) + (+2200 N·m2/C) + (+4600 N·m2/C) + (-1800 N·m2/C) + (-3500 N·m2/C) + (-5400 N·m2/C)
Summing these yields:
Φnet = -1400 N·m2/C
Assuming ε0 is the permittivity of free space (approximately 8.854 x 10-12 C2/N·m2), we can find Q:
Q = Φnet ε0
Q = (-1400 N·m2/C)(8.854 x 10-12 C2/N·m2)
Q ≈ -1.24 x 10-8 C
Thus, the charge Q inside the box is approximately -1.24 x 10-8 C.
A stream of air flowing at 20 liters/min with P = 0.20 MPa and T = 400 K is mixed with a stream ofmethane flowing at 5 liters/min with P = 0.20 MPa and T = 300 K. The combined gas stream exitingthe mixer is at P = 0.10 MPa and T = 370 K. What is the volumetric flow rate and composition of theexiting gas stream?
Answer:
a) the mole fraction of air in the exiting stream is Xa=0.25 (25%) and for methane Xm=0.75 (75%)
b) the volumetric flow rate is 49.33 L/s
Explanation:
Assuming ideal gas behaviour, then
for air
Pa*Va=Na*R*Ta
for methane
Pm*Vm=Nm*R*Tm
dividing both equations
(Pa/Pm)*(Va/Vm)= (Na/Nm)*(Ta/Tm)
Na/Nm = (Pa/Pm)*(Va/Vm) * (Tm/Ta) = (0.2/0.2)*(20/5)*(300/400) = 1*4*3/4 = 3
Na=3*Nm
therefore the moles of gas of the outflowing stream are (assuming that the methane does not react with the air):
Ng= Na+Nm = 4*Na
the mole fraction of A is
Xa= Na/Ng= Na/(4*Na) = 1/4 (25%)
and
Xm= 1-Xa = 3/4 (75%)
also for the exiting gas
Pg*Vg=Ng*R*Tg = Na*R*Tg + Nm*R*Tg = Pa*Va * (Tg/Ta) + Pm*Vm * (Tg/Tm)
Vg = Va * (Pa/Pg)*(Tg/Ta) + Vm *(Pm/Pg)* (Tg/Tm)
Vg = 20 L/min * (0.2/0.1)*(370/400) + 5 L/min * (0.2/0.1)*(370/300) = 49.33 L/s
A listener increases his distance from a sound source by a factor of 4.49.
Assuming that the source emits sound uniformly in all directions, what is the change in the sound intensity level in dB?
Answer: Δβ (dB) = -13.1dB
Explanation:
The intensity of sound is inversely proportional to the square of the distance between them.
I ∝ 1/r²
I₁/I₂= r₂²/r₁² .....1
When the listener increases his distance from the source by a factor of 4.49.
Then,
r₂/r₁= 4.49
From equation 1
I₁/I₂ = (4.49)²
I₁/I₂ = 20.16
I₂/I₁ = 1/20.16
The change in sound intensity in dB can be given as
Δβ (dB) = 10 log(I₂/l₁) = 10log(1/20.6) = -13.1dB
The change in the sound intensity level in dB is -13.1 dB.
The given parameters;
increase in distance = 4.49The relationship between intensity of sound and distance is calculated as follows;
[tex]I = \frac{k}{r^2} \\\\I_1r_1^2 = I_2r_2^2\\\\I_2 = \frac{I_1 r_1^2 }{r_2^2} \\\\I_2 = \frac{I_1 r_1^2}{(4.49r_1)^2} \\\\I_2 = \frac{I_1r_1^2}{20.16r_1^2} \\\\I_2 = \frac{I_1}{20.16} \\\\\frac{I_2}{I_1} = \frac{1}{20.16}[/tex]
The change in sound intensity in dB is calculated as follows;
[tex]\Delta \beta = 10 \ log[\frac{I_2}{I_1} ]\\\\\Delta \beta = 10 \times log [\frac{1}{20.16} ]\\\\\Delta \beta = -13.1 \ dB[/tex]
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A wheel with a 0.10-m radius is rotating at 35 rev/s. It then slows uniformly to 15 rev/s over a 3.0-s interval.
a. What is the angular acceleration of a point on the wheel?
b. how to do and explain will give lifesaver!
Answer:
The angular acceleration of a point on the wheel is [tex]41.89\ rad/s^2[/tex] and it is decelerating.
Explanation:
It is given that,
Radius of the wheel, r = 0.1 m
Initial angular velocity of the wheel, [tex]\omega_i=35\ rev/s=219.91\ rad/s[/tex]
Final angular velocity of the wheel, [tex]\omega_f=15\ rev/s=94.24\ rad/s[/tex]
Time, t = 3 s
We need to find the angular acceleration of a point on the wheel. It is given by the rate of change of angular velocity divided by time taken. It is given by :
[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}[/tex]
[tex]\alpha =\dfrac{(94.24-219.91)\ rad/s}{3\ s}[/tex]
[tex]\alpha =-41.89\ rad/s^2[/tex]
So, the angular acceleration of a point on the wheel is [tex]41.89\ rad/s^2[/tex] and it is decelerating. Hence, this is the required solution.
Final answer:
The angular acceleration of a point on the wheel can be found using the formula: angular acceleration = change in angular velocity / time interval. Given the initial and final angular velocities and the time interval, we can calculate the angular acceleration.
Explanation:
The angular acceleration of a point on the wheel can be found using the formula:
angular acceleration = change in angular velocity / time interval
Given that the initial angular velocity is 35 rev/s, the final angular velocity is 15 rev/s, and the time interval is 3.0 s, we can substitute these values into the formula:
angular acceleration = (15 rev/s - 35 rev/s) / 3.0 s
Simplifying the equation, we get:
angular acceleration = -20 rev/s / 3.0 s
Converting rev/s to rad/s, we have:
angular acceleration = -20 rev/s ×(2π rad/rev) / 3.0 s
angular acceleration ≈ -41.89 rad/s²
Therefore, the angular acceleration of a point on the wheel is approximately -41.89 rad/s².
A metal sphere has a charge of +12C. What is the net charge after 5.9 x 1013 electrons have been placed on it?
Answer:
Explanation:
Given
Charge on metal sphere [tex]Q=12\mu C[/tex]
no of electrons [tex]n=5.9\times 10^{13}[/tex]
Charge on each electron [tex]q=-1.6\times 10^{19}\ mu C[/tex]
Charge by Possessed by Electrons [tex]Q_2=-1.6\times 10^{19}\times 5.9\times 10^{13}[/tex]
[tex]Q_2=-9.44\mu C[/tex]
Net Charge on Sphere [tex]Q_{net}=Q+Q_2[/tex]
[tex]Q_{net}=12-9.44=2.56\mu C[/tex]