Answer:
The aanswers to the question are
(a) 5.33 m³
(b) 1.83 kg/m³
Explanation:
Volume of leak = 40 L, density of propane = 0.24g/cm³
Mass of leak = Volume × Density = 40000 cm³×0.24 g/cm³ = 9600 gram
Molar mass of propane = 44.1 g/mol Number of moles = 9600/44.1 = 217.69 moles
at 1 atmosphare and 298.15 K we have
PV = nRT therefore V = nRT/P = (217.69×8.3145×298.15)/101325 = 5.33 m³
The volume of the vapour = 5.33 m³
(b) Density = mass/volume
Recalculating the above for T = 293 K we have V = 5.33×293÷298.15 = 5.23 m³
Therefore density of propane vapor = 9600/5.23 = 1834.22 g/m³ or 1.83 kg/m³
One end of a horizontal rope is attached to a prong of an electrically driven tuning fork that vibrates at a frequency 129 Hz. The other end passes over a pulley and supports a mass of 1.50 kg. The linear mass density of the rope is 0.0590 kg/m. What is the speed of a transverse wave on the rope? What is the wavelength? How would your answers to parts (a) and (b) be changed if the mass were increased to 2.80 kg?
Answer:
(a). The speed of transverse wave on the rope is 15.78 m/s.
(b). The wavelength is 0.122 m.
(c). The changed speed of transverse wave on the rope is 21.56 m/s.
The changed wavelength is 0.167 m.
Explanation:
Given that,
Frequency = 129 Hz
mass = 1.50 kg
Linear mass density of the rope = 0.0590 kg/m
(a). We need to calculate the speed of a transverse wave on the rope
Using formula of speed
[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]
Put the value into the formula
[tex]v=\sqrt{\dfrac{1.50\times9.8}{0.0590}}[/tex]
[tex]v=15.78\ m/s[/tex]
(b). We need to calculate the wavelength
Using formula of wavelength
[tex]\lambda =\dfrac{v}{f}[/tex]
Put the value into the formula
[tex]\lambda=\dfrac{15.78}{129}[/tex]
[tex]\lambda=0.122\ m[/tex]
(c). If the mass were increased to 2.80 kg.
We need to calculate the speed of a transverse wave on the rope
Using formula of speed
[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]
Put the value into the formula
[tex]v=\sqrt{\dfrac{2.80\times9.8}{0.0590}}[/tex]
[tex]v=21.56\ m/s[/tex]
We need to calculate the wavelength
Using formula of wavelength
[tex]\lambda =\dfrac{v}{f}[/tex]
Put the value into the formula
[tex]\lambda=\dfrac{21.56}{129}[/tex]
[tex]\lambda=0.167\ m[/tex]
Hence, (a). The speed of transverse wave on the rope is 15.78 m/s.
(b). The wavelength is 0.122 m.
(c). The changed speed of transverse wave on the rope is 21.56 m/s.
The changed wavelength is 0.167 m.
The speed of the transverse wave on the rope is 2.48 m/s and the wavelength is 0.019 m. If the mass were increased to 2.80 kg, the speed of the wave would stay the same but the wavelength would be different.
Explanation:To find the speed of a transverse wave on the rope, we can use the equation:
v = √(T/μ)
where v is the speed of the wave, T is the tension in the rope, and μ is the linear mass density of the rope. Plugging in the values given in the question, we get:
v = √(1.50 kg * 9.8 m/s^2 / 0.0590 kg/m) = 2.48 m/s
To find the wavelength of the wave, we can use the equation:
λ = v/f
where λ is the wavelength, v is the speed of the wave, and f is the frequency of the tuning fork. Plugging in the values given in the question, we get:
λ = 2.48 m/s / 129 Hz = 0.019 m
If the mass were increased to 2.80 kg, the speed of the wave on the rope would remain the same. However, the wavelength would be different because it is determined by the frequency of the tuning fork and the speed of the wave, not the mass of the hanging weight.
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A football player runs the pattern given in the drawing by the three displacement vectors , , and . The magnitudes of these vectors are A = 4.00 m, B = 13.0 m, and C = 19.0 m. Using the component method, find the (a) magnitude and (b)direction of the resultant vector + + . Take to be a positive angle.
Answer:
Explanation:
check the attached for the solution
A springboard diver leaps upward from the springboard, rises dramatically to a peak height, and than drops impressively into the water below the board. Neglect any influences of air or the atmosphere. During this trip, the diver experiences ________.
Answer:
a constant downward net force.
Explanation:
The diver experiences a constant downward net force because while on his was up he decelerates until he gets to his maximum height when the acceleration is zero, during this phase force is not constant. While on his way down neglecting influences of air or atmosphere he falls with a constant downward net acceleration hence his net downward force will be constant.
In order to ensure that a cable is not affected by electromagnetic interference, how far away should the cable be from fluorescent lighting?
Answer:
the answer is at least 3 feet
Explanation:
In order to ensure that a cable is not affected by electromagnetic interference, it should be at least 3 feet away from fluorescent lighting.
This is because, cables can be adversely affected by electromagnetic interference - which is a disturbance that affects an electrical circuit due to either electromagnetic induction or radiation emitted from an external source - and insulation alone cannot provide adequate protection for these cables.
Therefore, the cables should be kept a few feet away from flourescent lighting in order to prevent this interference.
To minimize the EMI on a cable from fluorescent lighting, it should be kept at a distance of at least 2 feet or 0.6 meters. The distance can vary depending on the cable type, the electromagnetic field size and the data's sensitivity.
Explanation:To minimize the electromagnetic interference (EMI) effect on a cable from fluorescent lighting, it is recommended to maintain a distance of at least 2 feet or 0.6 meters. This is because the fluorescent light produces a magnetic field that can interact with the cable and cause electromagnetic interference. The above mentioned distance is considered a safe threshold, yet it can vary depending on the type of cable, the strength of the electromagnetic field produced by the light and the sensitivity of the data being transmitted.
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A closed Gaussian surface in the shape of a cube of edge length 1.9 m with one corner at x = 1 4.8 m, y = 1 3.9 m.
The cube lies in a region where the electric field vector is given by [tex]E = 3.4 \hat{i} + 4.4 y^2 \hat{j} + 3.0 \hat{k}[/tex] NC with y in meters.
What is the net charge contained by the cube?
Answer:
Net charge = 2.59nC
Explanation:
Gauss' Law states that the net electric flux is given by:
∮→E⋅→d/A = q enc/ϵ0
At this point, we have to solve the net electric flux through each side.
One corner is at (4.8, 3.9, 0).
The other corners are each 1.9m apart, so the other corners are at
(4.8, 5.8, 0), (6.7, 3.9, 0) and (6.7, 5.8, 0).
The other four corners are 1.9m away in the z-axis:
(4.8, 5.8, 2), (6.7, 3.9, 2) and (6.7, 5.8, 2).
Therefore, there are two planes (parallel to the x-z plane), one at
y = 3.9
and the other at
y = 5.8 which have a constant y-coordinate and are facing in the −^j and +^j respectively.
Area = 1.9m * 1.9m = 3.61m²
The flux through the first plane (area of 3.61m²) is given by:
E.A = (3.4i + 4.4 * 3.9²j + 3.0k) * (-3.61m²j) = -241.59564
The flux through the other plane is
E.A = (3.4i + 4.4 * 5.8²j + 3.0k)* (3.61m²j) = 534.33776
Now, for the other planes. There are no ^j components for the unit vectors for the area.
Therefore, even though they change in y-coordinate, those terms cancel out.
Therefore, for the planes with unit vector in the x-direction, we get:
E.A = (3.4i + 4.4y²j + 3.0k) * (1.9m * 1.9m) = ±12.274
And in the z-direction:
E.A = (3.4i + 4.4y²j + 3.0k) * (1.9m * 1.9m) = ±10.83
Now, when we sum all these fluxes together, the contribution from the x- and z-directions cancel out. Therefore, our net flux is:
-241.59564 + 534.33776 = 292.74212
Therefore, the enclosed charge is given by
q enc = ϵ0* (292.74212)
= 2.5856871136363E−9C
= 2.59E-9 nC--_- Approximated
= 2.59nC
Answer:
Q_enclosed = 1.576 nC
Explanation:
Given:
- The edge length of the cube L = 1.9 m
- One corner of the cube P_1 = ( 4.8 , 13.9 ) m
- The Electric Field vector is given by:
E = 3.4 i + 4.4*y^2 j + 3.0 k N/C
Find:
What is the net charge contained by the cube?
Solution:
- The flux net Ф through faces parallel to y-z plane is:
net Ф_yz = E_x . A . cos (θ)
Where, E_x is the component of E with unit vector i.
θ is the angle between normal vector dA and E.
Hence,
net Ф_yz = 3.4 . 1.9^2 . cos (0) + 3.4 . 1.9^2 . cos (180)
net Ф_yz = 3.4 . 1.9^2 - 3.4 . 1.9^2 . cos (180)
net Ф_yz = 0.
- Similarly, The flux net Ф through faces parallel to x-y plane is:
net Ф_xy = E_z . A . cos (θ)
Where, E_z is the component of E with unit vector k.
net Ф_xy = 3 . 1.9^2 . cos (0) + 3 . 1.9^2 . cos (180)
net Ф_xy = 3 . 1.9^2 - 3 . 1.9^2 . cos (180)
net Ф_xy = 0
-The flux net Ф through faces parallel to x-z plane is:
net Ф_xz = E_y . A . cos (θ)
Where, E_y is the component of E with unit vector j.
net Ф_xz = 4.4y_1^2 * 1.9^2 . cos (0) + 4.4y_2^2. 1.9^2 . cos (180)
Where, The y coordinate for face 1 y_1 = 3.9 - 1.9 = 2, & face 2 y_2 = 3.9
net Ф_xz = - 4.4*2^2*1.9^2 . cos (0) - 4.4*3.9^2. 1.9^2 . cos (180)
net Ф_xz = -63.536 + 241.59564 = 178.0596 Nm^2/C
- From gauss Law we have:
Total net Ф_x,y,z = Q_enclosed / ∈_o
Where,
Q_enclosed is the charge contained in the cube
∈_o is the permittivity of free space = 8.85*10^-12
Hence,
Total net Ф_x,y,z = net Ф_xz + net Ф_yz + net Ф_xy
Total net Ф_x,y,z = 178.0596 + 0 + 0 = 178.0596 Nm^2/C
We have,
Q_enclosed = Total net Ф_x,y,z * ∈_o
Q_enclosed = 178.0596 * 8.85*10^-12
Q_enclosed = 1.576 nC
A horse pulls on an object with a force of 300 newtons and does 12,000 joules of work. How far was the object moved?
1. 40
2. 0.03
3. 5 ×10↑5
4. 10 ×10↑4
Answer:
1.
Explanation:
because 12,000 divided by 300 is 40 so its 40m
A cannonball is shot out of a long cannon and then a short cannon with the same amount of force. The short cannon cannonball has Group of answer choices
more mometum, velocity, and impulse.
more momentum only.
more impulse only.
less mometum.
more velocity only.
Answer:
The correct answer is less momentum.
Explanation:
The momentum of an object is directly linked to the force exerted on that object. In a longer barrel, the shot cannonball will have a greater impulse changing to occur a larger momentum and therefore a higher speed will be applied to the bullet, contrary to what would happen in a shorter barrel.
Final answer:
The cannonball shot from a short cannon does not have more momentum, velocity, or impulse by virtue of the cannon's length alone, since momentum and impulse depend on both the force applied and the time the force acts on the cannonball.
Explanation:
If a cannonball is shot from both a long and a short cannon with the same amount of force, we need to understand that force, momentum, and impulse are interrelated but distinct concepts in physics. The product of an object's mass and velocity is its momentum.
Impulse is the change in momentum, often calculated as force multiplied by the time the force is applied. Because the question states that the force is the same, the impulse given to the cannonball would be identical regardless of the cannon length, assuming the force is applied over the same time period.
However, a longer cannon might allow the cannonball more time to accelerate, potentially resulting in greater final velocity and therefore more momentum, since momentum is mass times velocity. Conversely, a shorter cannon may allow less time for the force to act, possibly resulting in a lower velocity and, by extension, less momentum.
Thus, the cannonball from the short cannon does not necessarily have more velocity only, as it depends on the time the force acts on the cannonball.
In summary, the answer choice would be that the short cannon cannonball does not categorically have more momentum, velocity, or impulse based solely on the length of the cannon.
What sentence(s) is/are true when we talk about equipotential lines?
a. Electric potential is the same along an equipotential line
b. Work is necessary to move a charged particle along these lines.
c. They are always perpendicular to electric field lines
d. They are always parallel to electric field lines
Answer:
a. True
Explanation:
Equipotential lines are the imaginary lines in the space where actually the electric potential is same at each and every point.
Work is not required to move along such points of the equipotential line because the movement is always perpendicular to the electric field lines because these lines are always perpendicular to the electric field lines.
The electric potential for a point charge is given mathematically as:
[tex]V=\frac{1}{4\pi.\epsilon_0}\times \frac{Q}{r}[/tex]
where:
[tex]Q=[/tex] magnitude of the point charge
[tex]r=[/tex] radial distance form the charge
[tex]\epsilon_0=[/tex] permittivity of free space
Equipotential lines in physics are lines where the electric potential remains constant, perpendicular to electric field lines, and require no work to move a charge along them.
Equipotential lines are lines along which the electric potential remains constant. These lines are perpendicular to the electric field lines. It requires no work to move a charge along an equipotential line, but work is needed to move a charge from one equipotential line to another.
Two charges that are separated by one meter exert 1-N forces on each other. If the charges are pushed together so the separation is 25 centimeters, the force on each charge will be______
Final answer:
If the charges are pushed together so the separation is 25 centimeters, the force on each charge will be 16 N.
Explanation:
In electrostatics, the force between two charged particles is given by Coulomb's law. The formula to calculate the force is F = k * (q1 * q2) / r^2, where F is the force, k is the electrostatic constant, q1 and q2 are the charges of the particles, and r is the distance between them.
Using this formula, if the distance between the charges is decreased from 1 meter to 25 centimeters, the force on each charge will increase. The force is inversely proportional to the square of the distance, so when the distance is reduced to 0.25 meters (25 centimeters), the force will be 16 times stronger. Therefore, the force on each charge will be 16 N.
A tennis ball has a mass of 0.059 kg. A professional tennis player hits the ball hard enough to give it a speed of 41 m/s (about 92 miles per hour.) The ball moves toward the left, hits a wall and bounces straight back to the right with almost the same speed (41 m/s).
A) What is the average speed of the ball during the period from first contact with the wall to the moment the ball's speed is momentarily zero?
B) How much time elapses between first contact with the wall, and coming to a stop?
C) What is the magnitude of the average force exerted by the wall on the bal dring contact?
D) In contrast, what is the magnitude of the gravitational force of the Earth on the ball?
There is an omission of some sentences in the question which affects the answering of question B and C, so we will based the omission of the sentences on assumption in order to solve the question that falls under it.
NOTE: The omitted sentences are written in bold format
A tennis ball has a mass of 0.059 kg. A professional tennis player hits the ball hard enough to give it a speed of 41 m/s (about 92 miles per hour.) The ball moves toward the left, hits a wall and bounces straight back to the right with almost the same speed (41 m/s).
As indicated in the diagram below, high-speed photography shows that the ball is crushed about d = 2.0 cm at the instant when its speed is momentarily zero, before rebounding.
A) What is the average speed of the ball during the period from first contact with the wall to the moment the ball's speed is momentarily zero?
B) How much time elapses between first contact with the wall, and coming to a stop?
C) What is the magnitude of the average force exerted by the wall on the bal dring contact?
D) In contrast, what is the magnitude of the gravitational force of the Earth on the ball?
Answer:
a) [tex]V_{avg} = 20.5m/s[/tex]
b) 9.76 × 10⁻⁴s
c) 247.9 N
d) 5.8 N
Explanation:
Given that;
Initial speed [tex](V_i)[/tex] = 0
Final speed [tex](V_f)[/tex] = 41 m/s
Distance (d) = 0.002
mass (m) = 0.059 kg
g = 9.8 m/s²
a)
The average speed of the ball can be calculated as;
[tex]V_{avg} = \frac{V_i+V_f}{2}[/tex]
[tex]V_{avg} = \frac{0+41}{2}[/tex]
[tex]V_{avg} = 20.5m/s[/tex]
b)
The time elapsed can be calculated by using the second equation of motion which is given as:
[tex]S=(\frac{V_i+V_f}{2})t[/tex]
If we make time (t) the subject of the formula; we have:
[tex](V_i+V_f)t=2S[/tex]
[tex]t= (\frac{2S}{V_I+V_f})[/tex]
[tex]=\frac{2(0.02)}{41+0}[/tex]
[tex]=\frac{0.04}{41}[/tex]
= 0.000976
=9.76 × 10⁻⁴s
c)
the magnitude of the average force (F) exerted by the wall on the bal dring contact can be determined using;
Force (F) = mass × acceleration
where acceleration [tex](a)= \frac{Vo}{t}[/tex]
[tex]\frac{41}{0.00976}[/tex]
acceleration (a) = 4200.82 m/s²
F = m × a
= 0.059 × 4200.82
= 247.85
≅ 247.9 N
d)
the magnitude of the gravitational force of the Earth on the ball
Force (F) = mass (m) × gravity (g)
= 0.059kg × 9.8 m/s²
= 5.782 N
≅ 5.8 N
The average speed of the tennis ball during contact with the wall is zero, and without the time of contact, we cannot determine the time elapsed or the average force exerted by the wall. However, the gravitational force on the ball is 0.5782 N.
Explanation:The question relates to the change in momentum and the forces involved when a tennis ball bounces off a wall. Specifically, a tennis ball with a mass of 0.059 kg is hit at a speed of 41 m/s, bounces off a wall, and comes back at the same speed. To tackle the posed questions, it is essential to apply concepts from Newton's laws of motion and the conservation of momentum.
Part AThe average speed of the ball during contact is zero since the speed decreases uniformly from 41 m/s to zero.
Part BWithout the time of contact with the wall, this cannot be determined. Previous examples of collisions show time of contact can vary, so it must be provided to answer this part of the question.
Part CTo calculate the magnitude of the average force exerted by the wall on the ball, we would need the time of contact with the wall. Since it is not given, this cannot be calculated accurately.
Part DThe magnitude of the gravitational force of the Earth on the ball is calculated as the product of the mass of the ball and the acceleration due to gravity (9.8 m/s²), which is 0.059 kg * 9.8 m/s² = 0.5782 N.
A roofing tile slides down a roof and falls off the roof edge 10 m above the ground at a speed of 6 m/s. The roof makes an angle of 30 degrees to the horizontal. How far from the exit point on the roof does the tile land?
The roofing tile will land around 3.1 meters away from the roof edge. This is found by using equations of motion and breaking the problem into vertical and horizontal components.
Explanation:To solve this, we can break the problem down into two dimensions (horizontal and vertical) and use equations of motion. On the vertical, the roofing tile starts off 10 m above the ground, and falls under acceleration due to gravity. On the horizontal, the tile leaves the roof edge with a horizontal speed of 6 m/s * cos(30), and continues with this speed (since there's no horizontal acceleration).
The time it takes the tile to hit the ground can be found using the equation y = v(initial)*t - 0.5*g*t^2. Assuming upward is positive and taking the initial velocity on the vertical as 6 m/s * sin(30), y = -10 m, v(initial) = 3 m/s and g = 9.8 m/s^2, we can solve for t to get approximately 0.598 seconds.
The distance from the exit point on the roof can be found by multiplying the horizontal speed 5.2 m/s (6 m/s * cos(30)) by the time (0.598 sec) to get approximately 3.1 m.
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An object moves in a circle of radius R at constant speed with a period T. If you want to change only the period in order to cut the object's acceleration in half, the new period should be A) Th/2 B) 7V2. C) T/4. D) 4T E) TI2
Answer: Option (B) is the correct answer.
Explanation:
Expression for centripetal acceleration is as follows.
a = [tex]r \omega^{2}[/tex] .......... (1)
Also, we know that
[tex]\omega = \frac{2 \pi}{T}[/tex] ........... (2)
Putting the value from equation (2) into equation (1) as follows.
a = [tex]r (\frac{2 \pi}{T})^{2}[/tex]
= [tex]r \frac{2 (\pi)^{2}}{4}[/tex]
As, [tex]a \propto \frac{1}{T^{2}}[/tex]
a = [tex]\frac{k}{T^{2}}[/tex]
or, [tex]aT^{2} = k[/tex]
Now, we will reduce a to [tex]\frac{a}{2}[/tex]. So, new value of [tex]T^{2}[/tex] will be equal to [tex]2T^{2}[/tex].
Therefore, value of new period will be as follows.
[tex]T' = \sqrt{2T^{2}}[/tex]
= [tex]\sqrt{2}T[/tex]
Thus, we can conclude that the new period is equal to [tex]T \sqrt{2}[/tex].
The new period should be [tex]T\sqrt{2}[/tex]
Calculation of new period:The expression for centripetal acceleration should be [tex]a = r\omega^2[/tex]
Now
we know that [tex]w = 2\pi \div T[/tex]
Now here we put the values
[tex]a = r(2\pi\div T)^2\\\\= r (2(\pi)^2\div 4\\\\Since\ a \alpha \frac{1}{T^2}\\\\ aT^2 = K[/tex]
Now here we have to decrease to a by 2. So the new T value should be [tex]2T^2[/tex]
So, the new period should be
[tex]T = \sqrt{2T^2}\\\\ = \sqrt{2T}[/tex]
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What are constellations?
1. Apparent groupings of stars and planets visible on a given evening
2. Groups of galaxies gravitationally bound and close together in the sky
3. Groups of stars gravitationally bound and appearing close together in the sky
4. Groups of stars making an apparent pattern in the celestial sphere
5. Ancient story boards, useless to modern astronomers
Answer:
Option 4
Explanation:
A constellation can be defined as that region formed by the stars in such a way that the formation by the group of stars in that area appear to seem an imaginary pattern of some mythological creature, animal, god or some inanimate object formed apparently.
Thus in accordance with the above definition a constellation is a group of stars that forms some apparent pattern in the celestial sphere.
What possible source of errors would be in this experiment besides human error and why?
Explanation:
Besides human error, other sources of error include friction on the cart (we assumed there is no friction, but in reality, that's never the case), and instrument errors (imprecision in the spring scale).
Besides human error, experiments can also encounter sampling errors, nonsampling errors, uncertainty factors, and spontaneous errors. These can be caused by an inaccurate sampling process, equipment malfunctions, variability in samples, and limitations in the measuring device.
Explanation:Possible sources of error in an experiment, besides human error, could include sampling errors and non-sampling errors. Sampling errors occur when the sample utilized may not be large enough or not accurately representative of the larger population.
This can lead to inaccurate assumptions and conclusions. Non-sampling errors, on the other hand, are unrelated to the sampling process and could be a result of equipment or instrument malfunction (like a faulty counting device). An example for this would be an uncertainty in measurement induced by the fact that smallest division on a measurement ruler is 0.1 inches.
Further, variability in samples can also lead to errors called spontaneous errors. These can be due to natural phenomena such as exposure to ultraviolet or gamma radiation, or to intercalating agents. Lastly, uncertainty factors can contribute to errors as well. These factors include limitations of the measuring device, inaccuracies caused by the instrument itself (e.g., a paper cutting machine that causes one side of the paper to be longer than the other).
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Charles is having a lot of problems with errors in a very complicated spreadsheet that he inherited from a colleague, and he turns to another co-worker, Seymour, for tips on how to trace errors in the sheet. If Charles sees which of the following, Seymour explains, there is a mistyped function name in the sheet.
a.#FORM?
b.#NAME?
c.#####
d.#FNCT?
Answer:
b.#NAME?
Explanation:
Remember, in Spreadsheet programs like Ms Excel several types of errors can occur such as value error.
However, since Seymour explains that there is a mistyped function name in the sheet it is more likely to display on the affected cell as #NAME?.
For example the function =SUM is wrongly spelled =SOM.
Therefore it is important to make sure the function name is spelled correctly.
Two manned satellites approaching one another at a relative speed of 0.150 m/s intend to dock. The first has a mass of 4.50 ✕ 103 kg, and the second a mass of 7.50 ✕ 103 kg. Assume that the positive direction is directed from the second satellite towards the first satellite. (a) Calculate the final velocity after docking, in the frame of reference in which the first satellite was originally at rest. Incorrect: Your answer is incorrect. m/s (b) What is the loss of kinetic energy in this inelastic collision? Incorrect: Your answer is incorrect. J (c) Repeat both parts, in the frame of reference in which the second satellite was originally at rest. final velocity Incorrect: Your answer is incorrect. Check the sign of your answer for velocity in part (c). m/s loss of kinetic energy Correct: Your answer is correct. J
a) Final velocity after docking: +0.094 m/s
b) Kinetic energy loss: 31.6 J
c) Final velocity after docking: -0.056 m/s
d) Kinetic energy loss: 31.6 J
Explanation:
a)
Since the system of two satellites is an isolated system, the total momentum is conserved. So we can write:
[tex]p_i = p_f[/tex]
Or
[tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2)v[/tex]
where, in the reference frame in which the first satellite was originally at rest, we have:
[tex]m_1 = 4.50\cdot 10^3 kg[/tex] is the mass of the 1st satellite
[tex]m_2 = 7.50\cdot 10^3 kg[/tex] is the mass of the 2nd satellite
[tex]u_1 = 0[/tex] is the initial velocity of the 1st satellite
[tex]u_2 = +0.150 m/s[/tex] is the initial velocity of the 2nd satellite
v is their final velocity after docking
Solving for v,
[tex]v=\frac{m_2 u_2}{m_1 +m_2}=\frac{(7.50\cdot 10^3)(0.150)}{4.50\cdot 10^3 + 7.50\cdot 10^3}=0.0938 m/s[/tex]
b)
The initial kinetic energy of the system is just the kinetic energy of the 2nd satellite:
[tex]K_i = \frac{1}{2}m_2 u_2^2 = \frac{1}{2}(7.50\cdot 10^3)(0.150)^2=84.4 J[/tex]
The final kinetic energy of the two combined satellites is:
[tex]K_f = \frac{1}{2}(m_1 +m_2)v^2=\frac{1}{2}(4.50\cdot 10^3+7.50\cdot 10^3)(0.0938)^2=52.8 J[/tex]
Threfore, the loss in kinetic energy during the collision is:
[tex]\Delta K = K_f - K_i = 52.8 - 84.4=-31.6 J[/tex]
c)
In this case, we are in the reference frame in which the second satellite is at rest. So, we have
[tex]u_2 = 0[/tex] (initial velocity of satellite 2 is zero)
[tex]u_1 = -0.150 m/s[/tex] (initial velocity of a satellite 1)
Therefore, by applying the equation of conservation of momentum,
[tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2)v[/tex]
And solving for v,
[tex]v=\frac{m_1 u_1}{m_1 +m_2}=\frac{(4.50\cdot 10^3)(-0.150)}{4.50\cdot 10^3 + 7.50\cdot 10^3}=-0.0563 m/s[/tex]
d)
The initial kinetic energy of the system is just the kinetic energy of satellite 1, since satellite 2 is at rest:
[tex]K_i = \frac{1}{2}m_1 u_1^2 = \frac{1}{2}(4.50\cdot 10^3)(-0.150)^2=50.6 J[/tex]
The final kinetic energy of the system is the kinetic energy of the two combined satellites after docking:
[tex]K_f = \frac{1}{2}(m_1 + m_2)v^2=\frac{1}{2}(4.50\cdot 10^3+ 7.50\cdot 10^3)(-0.0563)^2=19.0 J[/tex]
Therefore, the kinetic energy lost in the collision is
[tex]\Delta K = K_f - K_i = 19.0 -50.6 = -31.6 J[/tex]
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Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a 100 W incandescent bulb uses only 23.0 W of power. The compact bulb lasts 1.00×104 hours, on the average, and costs $ 12.0 , whereas the incandescent bulb costs only 76.0 ¢, but lasts just 750 hours. The study assumed that electricity cost 9.00 ¢ per kWh and that the bulbs were on for 4.0 h per day.
Answer:
The resistance is 626.0 Ω.
Explanation:
Given that,
Power of compact bulb= 100 W
Power of incandescent bulb = 23.0 W
Time [tex]t= 1.00\times10^{4}\ hours[/tex]
What is the resistance of a “100 W”fluorescent bulb? (Remember the actual rating is only 23W of powerfor a 120V circuit)
We need to calculate the resistance of the bulb
Using formula of power
[tex]P=\dfrac{V^2}{R}[/tex]
[tex]R=\dfrac{V^2}{P}[/tex]
Where, P = power
R = resistance
V = voltage
Put the value into the formula
[tex]R=\dfrac{120^2}{23}[/tex]
[tex]R=626.0\ \Omega[/tex]
Hence, The resistance is 626.0 Ω.
What kind of line does Edward Hopper use in New York Movie to divide theatre space from lobby space?
Answer:
hard line, soft line
Explanation:
If an aircraft is loaded 90 pounds over maximum certificated gross weight and fuel (gasoline) is drained to bring the aircraft weight within limits, how much fuel should be drained?
Answer:
15 gallons.
Gallons = (Pounds) ÷ (Pounds ÷ Gallons)
Explanation:
To bring an aircraft's weight under the maximum certificated gross weight by 90 pounds, 15 gallons of gasoline should be drained.
If an aircraft is loaded with 90 pounds over the maximum certificated gross weight and needs to be brought within the weight limit by draining fuel, we need to calculate how much fuel, in pounds, should be drained. Gasoline's weight can vary slightly with temperature, but for general aviation purposes, it's usually taken to be about 6 pounds per gallon.
Hence, to remove 90 pounds of weight, the amount of fuel that should be drained would be 90 divided by 6.
Performing the calculation, 15 gallons of fuel should be drained to remove 90 pounds of weight from the aircraft and bring it back to or below the maximum certificated gross weight.
A pair of oppositely charged parallel plates is separated by 5.38 mm. A potential difference of 623 V exists between the plates. What is the strength of the electric field between the plates? The fundamental charge is 1.602 × 10−19 . Answer in units of V/m.
Answer:
115799.256V/m
Explanation:
Given V = 623v, d = 5.38mm = 0.00538m
Strength of electric field E = V/d
E = 623/0.00538
= 115799.256V/m
A baseball is hit at an initial speed of 40 m/s at an angle of 60° above the horizontal and reaches a maximum height of h meters. What would be the maximum height reached if it were hit at 80 m/s? a.) 2h b.) 4h c.) 6h d.) 8h
Answer:
b.) 4h
Explanation:
A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The coefficient of static friction between the board and the box is μs. The coefficient of kinetic friction between the board and the box is, as usual, less than μs.
Throughout the problem, use g for the magnitude of the acceleration due to gravity. In the hints, use Ff for the magnitude of the friction force between the board and the box.
uploaded image
Find Fmin, the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box (which will then fall off of the opposite end of the board).
Express your answer in terms of some or all of the variables mu_s, m_1, m_2, g, and L. Do not include F_f in your answer.
Answer: Fmin = (m₁ + m₂) μsg
Explanation:
To begin, we would first define the parameters given in the question.
Mass of the box = m₁
Mass of the board = m₂
We have a Frictionless surface given that Fr is acting as the frictional force between the box and the board.
from our definition of force, i.e. the the frictional force against friction experienced by the box, we have
Fr = m₁a ...................(1)
Also considering the force between the box and the board gives;
Fr = μsm₁g .................(2)
therefore equating both (1) and (2) we get
m₁a = μsm₁g
eliminating like terms we get
a = μsg
To solve for the minimum force Fmin that must be applied to the board in order to pull the board out from under the box, we have
Fmin = (m₁ + m₂) a ...........(3)
where a = μsg, substituting gives
Fmin = (m₁ + m₂) μsg
cheers i hope this helps
The constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box is [tex]\mu_{s} g({m_{1}+m_{2}})[/tex].
Given data:
The mass of small box is, [tex]m_{1}[/tex].
The mass of board is, [tex]m_{2}[/tex].
The length of board is, L.
The coefficient of static friction between the board and box is, [tex]\mu_{s}[/tex].
The linear force acting between the box and the board provides the necessary friction to box. Therefore,
[tex]F=F_{f}\\F=\mu_{s}m_{1}g\\m_{1} \times a = \mu_{s} \times m_{1}g\\a= \mu_{s} \times g[/tex]
a is the linear acceleration of board.
Then, the minimum force applied on the board is,
[tex]F_{min}=({m_{1}+m_{2}})a\\F_{min}=({m_{1}+m_{2}})(\mu_{s} \times g)\\F_{min}=\mu_{s} g({m_{1}+m_{2}})[/tex]
Thus, the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box is
[tex]\mu_{s} g({m_{1}+m_{2}})[/tex].
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A 3.42 kg mass hanging vertically from a spring on the Earth (where g = 9.8 m/s2) undergoes simple oscillatory motion. If the spring constant is 12 N/m, find the period of oscillation of this setup on the moon, where g = 1.6 m/s2.
Answer:
Time period of oscillation on moon will be equal to 3.347 sec
Explanation:
We have given mass which is attached to the spring m = 3.42 kg
Spring constant K = 12 N/m
We have to find the period of oscillation
Period of oscillation is equal to [tex]T=2\pi \sqrt{\frac{m}{K}}[/tex], here m is mass and K is spring constant
So period of oscillation [tex]T=2\times 3.14\times \sqrt{\frac{3.42}{12}}[/tex]
[tex]T=2\times 3.14\times 0.533=3.347sec[/tex]
So time period of oscillation will be equal to 3.347 sec
As it is a spring mass system and from the relation we can see that time period is independent of g
So time period will be same on earth and moon
An electron beam in an oscilloscope is deflected by the electric field produced by oppositely charged metal plates. If the electric field between the plates is 2.07 x 105 N/C directed downward, what is the force on each electron when it passes between the plates?
Answer:
Explanation:
Given
Electric Field Strength [tex]E_0=2.07\times 10^5\ N/C[/tex]
Charge on electron [tex]q=1.6\times 10^{-19}\ C[/tex]
Force on any charged particle in an Electric field is given by
[tex]Force=charge\ on\ electron\times Electric\ Field\ strength [/tex]
[tex]F=1.6\times 10^{-19}\times 2.07\times 10^5[/tex]
[tex]F=3.312\times 10^{-14}\ N[/tex]
as the electric field is pointing downward therefore force will be acting downward on a positively charged particle but opposite for an electron i.e. in upward direction
A friend of yours is loudly singing a single note at 412 Hz while racing toward you at 25.8 m/s on a day when the speed of sound is 347 m/s . What frequency do you hear?
Answer:
5541Hz
Explanation:
If the frequency of a wave is directly proportional to the velocity we have;
F = kV where;
F is the frequency
K is the constant of proportionality
V is the velocity
Since f = kV
K = f/v
K = F1/V1 = F2/V2
Given f1 = 412Hz v1 = 25.8m/s f2 = ? V2 = 347m/s
Substituting in the formula we have;
412/25.8=f2/347
Cross multiplying
25.8f2 = 412×347
F2 = 412×347/25.8
F2 = 5541Hz
The frequency heard will be 5541Hz
A riverboat took 2 h to travel 24km down a river with the current and 3 h to make the return trip against the current. Find the speed of the boat in still water and the speed of the current.
Explanation:
Let the speed of river be r and speed of boat be b.
Distance traveled = 24 km
Time taken to travel 24 km with current = 2 hr
We have
Distance = Speed x Time
24 = (b + r) x 2
b + r = 12 --------------------eqn 1
Time taken to travel 24 km against current = 3 hr
We have
Distance = Speed x Time
24 = (b - r) x 3
b - r = 8 --------------------eqn 2
eqn 1 + eqn 2
2b = 20
b = 10 km/hr
Substituting in eqn 1
10 + r = 12
r = 2 km/hr
Speed of boat in still water = 10 km/hr
Speed of current of river = 2 km/hr
The speed of the boat in still water is 10 km/h and the speed of the current is 2 km/h. This is a solved by using a system of linear equations with speed of the boat and the current as variables.
Explanation:This problem is a case for a system of linear equations. Let's denote the speed of the boat in still water as b km/h and the speed of the current as c km/h. When the boat travels down the river, the boat and the current speeds are added, because they move in the same direction. When the boat travels up the river, the speeds are subtracted, because they move in opposite directions.
So we have the following equations:
b + c = 24km/2h = 12 km/h (down the river)b - c = 24km/3h = 8 km/h (up the river)The solution to this system of equations will give you the speed of the boat in still water and the speed of the current. Adding these two equations together, we get:
2b = 20 km/hTherefore, b = 10 km/h, which is the speed of the boat in still water. Substitute b = 10 km/h into the first equation to find the speed of the current, c = 12 km/h - 10 km/h = 2 km/h.
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A racing car travels on a circular track of radius 158 m, moving with a constant linear speed of 19.1 m/s. Find its angular speed. Answer in units of rad/s.
Answer:
[tex]\omega=0.12\frac{rad}{s}[/tex]
Explanation:
In a uniform circular motion, since a complete revolution represents 2π radians, the angular velocity, which is defined as the angle rotated by a unit of time, is given by:
[tex]\omega=\frac{2\pi}{T}(1)[/tex]
Here T is the period, that is, the time taken to complete onee revolution:
[tex]T=\frac{2\pi r}{v}(2)[/tex]
Replacing (2) in (1):
[tex]\omega=\frac{2\pi}{\frac{2\pi r}{v}}=\frac{v}{r}\\\omega=\frac{19.1\frac{m}{s}}{158m}\\\omega=0.12\frac{rad}{s}[/tex]
The angular speed of the racing car is approximately 0.1208861 rad/s.
The angular speed of the racing car can be calculated using the formula that relates linear speed (v) to angular speed and the radius (r) of the circular path, which is:
[tex]\[ v = r \cdot \omega \][/tex]
Given that the linear speed (v) of the car is 19.1 m/s and the radius (r) of the circular track is 158 m, we can solve for the angular speed (Ï) as follows:
[tex]\[ \omega = \frac{v}{r} \][/tex]
[tex]\[ \omega = \frac{19.1 \text{ m/s}}{158 \text{ m}} \][/tex]
[tex]\[ \omega =\frac{19.1}{158} \text{ rad/s} \][/tex]
[tex]\[ \omega \ = 0.1208861 \text{ rad/s} \][/tex]
Therefore, the angular speed of the racing car is approximately 0.1208861 rad/s.
To express this in a more simplified fraction, we can write:
[tex]\[ \omega \ = \frac{19.1}{158} \text{ rad/s} \][/tex]
[tex]\[ \omega \ = \frac{191}{1580} \text{ rad/s} \][/tex]
[tex]\[ \omega \ = \frac{191}{1580} \times \frac{10}{10} \text{ rad/s} \][/tex]
[tex]\[ \omega \ = \frac{1910}{15800} \text{ rad/s} \][/tex]
[tex]\[ \omega \ = \frac{191}{1580} \text{ rad/s} \][/tex]
Since 191 and 1580 do not have any common factors other than 1, this fraction is already in its simplest form. Thus, the final answer for the angular speed of the racing car is:
[tex]\[ \ {0.1208861 \text{ rad/s}} \][/tex]
A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/s by pulling on the rope tied to the crate with a force of 50 N. The rope makes an angle of 25° with the horizontal.
a. What are the horizontal and vertical components of the applied force?
b. What is the magnitude of each of the forces? i) Applied Force ii) Weight ii) Normal Force iii) Frictional Force
c. How much work is done by each of the forces? i) WApplied Force ii) WWeight iii) WNormal Force iv) WFrictional Force
d. What is the total amount of work done on the object?
e. What is the coefficient of friction of the crate on the floor?
Answer:
A) Fx = 45.3N Fy = 21.1N
B) Fm = 45.3N R = 196N Ff = 35.4N
C) i) 0KJ ii) 11.5KJ iii) 11.5KJ iv) 0KJ
D) 23KJ
E) 0.047
Explanation:
a) Horizontal component of the force Fx = 50cos 25° = 45.3N
Vertical component = 50sin25° = 21.1N
b) Magnitude of the applied force = moving force Fm = Fx = 50cos25° =45.3N
Magnitude of the weight = mg = 20×9.8 = 196N
Magnitude of normal force which is the reaction is equal to the weight = mg = 20×9.8 = 196N
Frictional force = moving force = 50cos45° = 35.4N
c) since workdone = Force done × perpendicular distance in the direction of the force
- workdone on the moving force is 0Joules since it has no perpendicular distance
- workdone on weight is the weight × distance = (20×9.8)×12 = 11.5KJ
= work done on normal force = workdone by the weight = 11.5KJ
Workdone on the frictional force is is 0Joules since the force is along the horizontal (no perpendicular distance)
d) the total work done = work done by Applied Force +Weight + Normal Force + Frictional Force= 0+11.5+11.5 = 23KJ
e) the coefficient of friction = moving force / normal reaction = 45.3/196 = 0.047
Explanation:
A.
Horizontal component, Fy = F * cos(a)
= 50cos25
= 50 * 0.91
= 45.32 N
Vertical component,Fx = F * sin(a)
= 50sin25
= 50 * 0.42
= 21.13 N
B.
Applied force = 50 N
Weight,W = m * g
= 20 * 9.81
= 196.2 N
Normal force, N = W - Fx
= 200 - 21.13
= 179 N
Frictional force = Fy
= 45.32 N.
C.
Workdone by Normal and Weight forces are = 0, because they both act perpendicular to the movement.
Workdone by friction = Workdone by applied forces
= force * distance
= 12 * 45.32
= 543.84 J
D.
Total amount of work done on the crate = 0 (the movement with a constant speed).
E.
The coefficient of friction, u = Fy/(W - Fx)
= 45.5/(196.2 - 21.13)
= 0.26
What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?
Answer : The energy of one photon of hydrogen atom is, [tex]3.03\times 10^{-19}J[/tex]
Explanation :
First we have to calculate the wavelength of hydrogen atom.
Using Rydberg's Equation:
[tex]\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )[/tex]
Where,
[tex]\lambda[/tex] = Wavelength of radiation
[tex]R_H[/tex] = Rydberg's Constant = 10973731.6 m⁻¹
[tex]n_f[/tex] = Higher energy level = 3
[tex]n_i[/tex]= Lower energy level = 2
Putting the values, in above equation, we get:
[tex]\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )[/tex]
[tex]\lambda=6.56\times 10^{-7}m[/tex]
Now we have to calculate the energy.
[tex]E=\frac{hc}{\lambda}[/tex]
where,
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\lambda[/tex] = wavelength = [tex]6.56\times 10^{-7}m[/tex]
Putting the values, in this formula, we get:
[tex]E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}[/tex]
[tex]E=3.03\times 10^{-19}J[/tex]
Therefore, the energy of one photon of hydrogen atom is, [tex]3.03\times 10^{-19}J[/tex]
A 2,000 kg car starts from rest and coasts down from the top of a 5.00 m long driveway that is sloped at an angel of 20o with the horizontal. If an average friction force of 4,000 N impedes the motion of the car, find the speed of the car at the bottom of the driveway.
Answer:
The speed at the bottom of the driveway is3.67m/s.
Explanation:
Height,h= 5sin20°= 1.71m
Potential energy PE=mgh= 2000×9.8×1.71
PE= 33516J
KE= PE- Fk ×d
0.5mv^2= 33516 - (4000×5)
0.5×2000v^2= 33516 - 20000
1000v^2= 13516
v^2= 13516/1000
v =sqrt 13.516
v =3.67m/s