Answer:
1000 J
Explanation:
Work = force × distance
W = (50 N) (20 m)
W = 1000 J
If a pickup is placed 16.25 cm from one of the fixed ends of a 65.00-cm-long string, which of the harmonics from n=1 to n=12 will not be "picked up" by this pickup?
Answer:
The answer to this question can be defined as follows:
Explanation:
Therefore the 4th harmonicas its node is right and over the pickup so, can not be captured from 16.25, which is 1:4 out of 65. Normally, it's only conceptual for the certain harmonic, this will be low, would still be heard by the catcher.
Instead, every harmonic node has maximum fractions along its string; the very first node is the complete string length and the second node is half a mile to the third node, which is one-third up and so on.
Answer:
b
Explanation:
because:/
A particle is released as part of an experiment. Its speed t seconds after release is given by v (t )equalsnegative 0.4 t squared plus 2 t, where v (t )is in meters per second. a) How far does the particle travel during the first 2 sec? b) How far does it travel during the second 2 sec?
Answer:
a) 2.933 m
b) 4.534 m
Explanation:
We're given the equation
v(t) = -0.4t² + 2t
If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.
See attachment for the calculations
The conclusion of the attachment will be
7.467 - 2.933 and that is 4.534 m
Thus, The distance it travels in the second 2 sec is 4.534 m
Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of the bridge is a height h above the water. The bungee cord, which has length L when unstretched, will first straighten and then stretch as Kate falls.
Assume the following:
A) The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant .
B) Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward.
C) Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle.
Use g for the magnitude of the acceleration due to gravity.
1) How far below the bridge will Kate eventually be hanging, once she stops oscillating and comes finally to rest? Assume that she doesn't touch the water.
Express the distance in terms of quantities given in the problem introduction.
Update:
I already found the answer to 1) to be:
d = L + (mg / k)
2) If Kate just touches the surface of the river on her first downward trip (i.e., before the first bounce), what is the spring constant k? Ignore all dissipative forces.
Express k in terms of L, h, m, and g.
Answer:
1) d = h- L - mg / k , 2) k = 2 mg/h (-1 +2 L / h)
Explanation:
1) Let's use the translational equilibrium equation
[tex]F_{e}[/tex] -W = 0
F_{e} = W
k x = mg
x = mg / k
from the statement of the exercise the height of the bridge, for the reference system in the river, is
h = L + x + d
x = h - L - d
we substitute
h - L -d = mg / k
d = h- L - mg / k
2) They ask us for the spring constant. For this part we can use energy conservation
Starting point. At the point before jumping
Em₀ = U = m g h
Final Point. When it's hanging
Em_f = [tex]K_{e}[/tex] + U = ½ k x² + mg d
Em₀ = Em_f
mg h = 1 / k x² + m g d
in the exercise they indicate that Kate touches the surface of the river, so the distance d = 0
mg h = ½ k x²
k = 2 mg h / x²
we substitute the value of x
k = 2mg h / (h -L)²
k = 2mg h / (- L + h)²
we simplify the expression
k = 2mg h / [h² (1- L / h)²]
k = 2m /h (1- L / h)⁻²
In these jumps the bridge height is always greater than the length of the rope L / h <1, so we can expand the last expression
(1- L / h)⁻² = 1 - 2 (1 -L / h) + 2 3/2! (1 -L / h)² + ...
for simplicity let's keep up to the linear term, we substitute in the solution
k = 2 mg/h [1 - 2 (1- L / h)]
k = 2 mg/h (-1 +2 L / h)
What force is required so that a particle of mass m has the position function r(t) = t3 i + 7t2 j + t3 k?
Answer:
[tex]F(t)=m\,\,a(t)=6\,m\,t\,\hat i+14\,m\,\hat j+6\,m\,t\,\hat k\\F(t)=\,(6\,m\,t,14\,m,6\,m\,t)[/tex]
Explanation:
Recall that force is defined as mass times acceleration, and acceleration is the second derivative with respect to time of the position. Since the position comes in terms of time, and with separate functions for each component in the three dimensional space, we first calculate the velocity (with the first derivative, and then the acceleration as the second derivative:
[tex]r(t)=t^3\,\hat i+7\,t^2\,\hat j+t^3\,\hat k\\v(t)=3\,t^2\,\hat i+14\,t\,\hat j+3\,t^2\,\hat k\\a(t)=6\,t\,\hat i+14\,\hat j+6\,t\,\hat k[/tex]
Therefore, the force will be given by the product of this acceleration times the mass "m":
[tex]F(t)=m\,\,a(t)=6\,m\,t\,\hat i+14\,m\,\hat j+6\,m\,t\,\hat k[/tex]
A resistor, inductor, and capacitor are connected in series, each with effective (rms) voltage of 65 V, 140 V, and 80 V respectively. What is the value of the effective (rms) voltage of the applied source in the circuit
Answer:
The value of the effective (rms) voltage of the applied source in the circuit is 132 V
Explanation:
Given;
effective (rms) voltage of the resistor, [tex]V_R[/tex] = 65 V
effective (rms) voltage of the inductor, [tex]V_L[/tex] = 140 V
effective (rms) voltage of the capacitor, [tex]V_C[/tex] = 80 V
Determine the value of the effective (rms) voltage of the applied source in the circuit;
[tex]V= \sqrt{V_R^2 + (V_L^2-V_C^2} )\\\\V= \sqrt{65^2 + (140^2-80^2} )\\\\V = \sqrt{4225+ 13200} \\\\V = \sqrt{17425} \\\\V = 132 \ V[/tex]
Therefore, the value of the effective (rms) voltage of the applied source in the circuit is 132 V.
A tightly wound toroid of inner radius 1.2 cm and outer radius 2.4 cm has 960 turns of wire and carries a current of 2.5 A.
Requried:
a. What is the magnetic field at a distance of 0.9 cm from the center?
b. What is the field 1.2 cm from the center?
Answer:
a
[tex]B = 0.0533 \ T[/tex]
b
[tex]B = 0.04 \ T[/tex]
Explanation:
From the question we are told that
The inner radius is [tex]r = 1.2 \ cm = 0.012 \ m[/tex]
The outer radius is [tex]r_o = 2.4 \ cm = \frac{2.4}{100} = 0.024 \ m[/tex]
The nu umber of turns is [tex]N = 960[/tex]
The current it is carrying is [tex]I = 2. 5 A[/tex]
Generally the magnetic field is mathematically represented as
[tex]B = \frac{\mu_o * N* I }{2 * \pi * r }[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with a constant value
[tex]\mu = 4\pi * 10^{-7} N/A^2[/tex]
And the given distance where the magnetic field is felt is r = 0.9 cm = 0.009 m
Now substituting values
[tex]B = \frac{ 4\pi * 10^{-7} * 960* 2.5 }{2 * 3.142 * 0.009 }[/tex]
[tex]B = 0.0533 \ T[/tex]
Fro the second question the distance of the position considered from the center is r = 1.2 cm = 0.012 m
So the magnetic field is
[tex]B = \frac{ 4\pi * 10^{-7} * 960* 2.5 }{2 * 3.142 * 0.012 }[/tex]
[tex]B = 0.04 \ T[/tex]
The magnetic field at a distance of 0.9 cm from the center of the toroid is 0.053 T.
The magnetic field at a distance of 1.2 cm from the center of the toroid is 0.04 T.
The given parameters;
radius of the toroid, r = 1.2 cm = 0.012 mouter radius of the toroid, R = 2.4 cm = 0.024 mnumber of turns, N = 960 turnscurrent in wire, I = 2.5 AThe magnetic field at a distance of 0.9 cm from the center of the toroid is calculated as follows;
[tex]B = \frac{\mu_o NI}{2\pi r} \\\\B = \frac{(4\pi \times 10^{-7})\times (960) \times (2.5)}{2\pi \times 0.009} \\\\B = 0.053 \ T[/tex]
The magnetic field at a distance of 1.2 cm from the center of the toroid is calculated as follows;
[tex]B = \frac{\mu_o NI}{2\pi r} \\\\B = \frac{(4\pi \times 10^{-7})\times (960) \times (2.5)}{2\pi \times 0.012} \\\\B = 0.04 \ T[/tex]
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Water flowing through a garden hose of diameter 2.76 cm fills a 20.0-L bucket in 1.45 min. (a) What is the speed of the water leaving the end of the hose
Answer:
v = 31.84 cm/s or 0.318 m/s
the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s
Explanation:
Given;
Diameter of hose d = 2.76 cm
Volume filled V = 20.0 L = 20,000 cm^3
Time t = 1.45 min = 105 seconds
The volumetric flow rate of water is;
F = V/t = 20,000cm^3 ÷ 105 seconds
F = 190.48 cm^3/s
The volumetric flow rate is equal the cross sectional area of pipe multiply by the speed of flow.
F = Av
v = F/A
Area A = πd^2/4
Speed v = F/(πd^2/4)
v = 4F/πd^2 ......1
Substituting the given values;
v = (4×190.48)/(π×2.76^2)
v = 31.83767439628 cm/s
v = 31.84 cm/s or 0.318 m/s
the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s
How much energy does the light bulb dissipate in 1 minmin when half that voltage is supplied? Express your answer in kilojoules to two significant figures.
Answer:
E = 15 P₀
Explanation:
The power dissipated in a light bulb is
P = V I
V = I R
P = V² / R
the power is defined by
P = W / t
work equals energy
P = E / t
we substitute
V² / R = E / t
E = V² t / R
let's reduce the time to SI units
t = 1 min = 60 s
let's calculate the dissipated energy
In the exercise it does not indicate the nominal voltage of the bulb, but in general this voltage is V₀= 120 V
The applied voltage is half the nominal voltage
V = V₀ / 2
V = 120/2 = 60 V
E = (V₀ / 2)² t / R
E = ¼ t V₀² / R
E = ¼ 60 P₀
E = 15 P₀
Many times the nominal power (P₀) is written on the box of the bulb
A building is located on earth's equator. As the earth rotates about its axis, which floor of the building has the greatest angular speed?
Answer:
The angular speed of the earth rotation is equal. Therefore
Our angular speed due to Earth’s rotation is same at every point on the earth irrespective of the elevation. So your angular speed due to earth’s rotation on the top floor of the building will be same as it is on the ground floor.
Explanation:
An amusement park ride has a vertical cylinder with an inner radius of 3.4 m, which rotates about its vertical axis. Riders stand inside against the carpeted surface and rotate with the cylinder while it accelerates to its full angular velocity. At that point the floor drops away and friction between the riders and the cylinder prevents them from sliding downward. The coefficient of static friction between the riders and the cylinder is 0.87. What minimum angular velocity in radians/second is necessary to assure that the riders will not slide down the wall?
Answer:
The minimum angular velocity necessary to assure that the riders will not slide down the wall is 1.58 rad/second.
Explanation:
The riders will experience a centripetal force from the cylinder
[tex]F_{C}[/tex] = mrω^2 .... equ 1
where
m is the mass of the rider
r is the inner radius of the cylinder = 3.4 m
ω is the angular speed of of the rider
For the riders not to slide downwards, this centripetal force is balanced by the friction between the riders and the cylinder. The frictional force is given as
[tex]F_{f}[/tex] = μR ....equ 2
where
μ = coefficient of friction = 0.87
R is the normal force from the rider = mg
where
m is the rider's mass
g is the acceleration due to gravity = 9.81 m/s
substitute mg for R in equ 2, we'll have
[tex]F_{f}[/tex] = μmg ....equ 3
Equating centripetal force of equ 1 and frictional force of equ 3, we'll get
mrω^2 = μmg
the mass of the rider cancels out, and we are left with
rω^2 = μg
ω^2 = μg/r
ω = [tex]\sqrt{\frac{ug}{r} }[/tex]
ω = [tex]\sqrt{\frac{0.87*9.81}{3.4} }[/tex]
ω = 1.58 rad/second
The minimum angular velocity necessary so that the riders will not slide down the wall is 1.58 rad/s
The riders will experience a centripetal force from the cylinder
[tex]F = mrw^2[/tex]
where m is the mass of the rider
r is the inner radius of the cylinder = 3.4 m
ω is the angular speed of the rider
For the riders not to slide downwards, this centripetal force must be balanced by friction. The frictional force is given as
f = μN
where
μ = coefficient of friction = 0.87
N is the normal force = mg
f = μmg
Equating centripetal force of and frictional force of we'll get
[tex]mrw^2 = umg[/tex]
[tex]rw^2 = ug[/tex]
[tex]w^2 = ug/r[/tex]
[tex]w= \sqrt{ug/r}[/tex]
[tex]w= \sqrt{0.87*9.8/3.4}[/tex]
ω = 1.58 rad/s is the minimum angular velocity needed to prevent the rider from sliding.
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what is drift speed ? {electricity}
Answer: In physics a drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.
Explanation:
A particle with kinetic energy equal to 282 J has a momentum of magnitude 26.4 kg · m/s. Calculate the speed (in m/s) and the mass (in kg) of the particle.
Answer:
[tex]v=21.36\,\,\frac{m}{s}\\[/tex]
[tex]m=1.2357\,\,kg[/tex]
Explanation:
Recall the formula for linear momentum (p):
[tex]p = m\,v[/tex] which in our case equals 26.4 kg m/s
and notice that the kinetic energy can be written in terms of the linear momentum (p) as shown below:
[tex]K=\frac{1}{2} m\,v^2=\frac{1}{2} \frac{m^2\,v^2}{m} =\frac{1}{2}\frac{(m\,v)^2}{m} =\frac{p^2}{2\,m}[/tex]
Then, we can solve for the mass (m) given the information we have on the kinetic energy and momentum of the particle:
[tex]K=\frac{p^2}{2\,m}\\282=\frac{26.4^2}{2\,m}\\m=\frac{26.4^2}{2\,(282)}\,kg\\m=1.2357\,\,kg[/tex]
Now by knowing the particle's mass, we use the momentum formula to find its speed:
[tex]p=m\,v\\26.4=1.2357\,v\\v=\frac{26.4}{1.2357} \,\frac{m}{s} \\v=21.36\,\,\frac{m}{s}[/tex]
What is the change in internal energy of an engine if you put 15 gallon of gasoline into its tank? The energy content of gasoline is 1.5 x 106 J/gallon. All other factors, such as the engine’s temperature, are constant. How many hours the engine can work if the power of the engine’s motor is 600 W? (8 marks)
Answer:
ΔU = 2.25 x 10⁸ J
t = 104.17 s
Explanation:
The change in internal energy of the engine can be given by the following formula:
ΔU = (Mass of Gasoline)(Energy Content of Gasoline)
ΔU = (1.5 x 10⁶ J/gallon)(15 gallon)
ΔU = 2.25 x 10⁸ J
Now, for the time of operation, we use the following formula of power.
P = W/t = ΔU/t
t = ΔU/P
where,
t = time of operation = ?
ΔU = Change in internal energy = 2.25 x 10⁸ J
P = Power of motor = 600 W
Therefore,
t = (2.25 x 10⁸ J)/(600 W)
t = (375000 s)(1 h/3600 s)
t = 104.17 s
The radius of he Earth orbit around the sun (assumed circular) is 1.50 X 10^8km, with T=365d. What is the radial acceleration of Earth towards the sun?
Answer:
ar = 5.86*10^-3 m/s^2
Explanation:
In order to calculate the radial acceleration of the Earth, you first take into account the linear speed of the Earth in its orbit.
You use the following formula:
[tex]v=\sqrt{\frac{GM_s}{r}}[/tex] (1)
G: Cavendish's constant = 6.67*10^-11 m^3 kg^-1 s^-2
Ms: Sun's mass = 1.98*10^30 kg
r: distance between Sun ad Earth = 1.50*10^8 km = 1.50*10^11 m
Furthermore, you take into account that the radial acceleration is given by:
[tex]a_r=\frac{v^2}{r}[/tex] (2)
You replace the equation (1) into the equation (2) and replace the values of all parameters:
[tex]a_r=\frac{1}{r}\frac{GM_s}{r}=\frac{GM_s}{r^2}\\\\a_r=\frac{(6.67*10^{-11}m^3kg^{-1}s^{-2})(1.98*10^{30}kg)}{(1.50*10^{11}m)^2}\\\\a_r=5.86*10^{-3}\frac{m}{s^2}[/tex]
The radial acceleration of the Earth, towards the sun is 5.86*10^-3 m/s^2
In an experiment different wavelengths of light, all able to eject photoelectrons, shine on a freshly prepared (oxide-free) zinc surface. Which statement is true
Answer:
the energy of the photons is greater than the work function of the zinc oxide.
h f> = Ф
Explanation:
In this experiment on the photoelectric effect, it is explained by the Einstein relation that considers the light beam formed by discrete energy packages.
K_max = h f - Ф
in the exercise phase, they indicate that different wavelengths can inject electrons, so the energy of the photons is greater than the work function of the zinc oxide.
h f > = Ф
What accurately depicts the change in average kinetic energy of the particles undergoes in matter as the temperature of the sample is decreased?
Answer:
As a sample of matter is continually cooled, the average kinetic energy of its particles decreases. Eventually, one would expect the particles to stop moving completely. Absolute zero is the temperature at which the motion of particles theoretically ceases.
Explanation:
A 5000 kg railcar hits a bumper (a spring) at 1 m/s, and the spring compresses 0.1 meters. Assume no damping. a) Find the spring constant k.
Answer:
k = 0.5 MN/m
Explanation:
Mass of the railcar, m = 5000 kg
Speed of the rail car, v = 1 m/s
The Kinetic energy(KE) of the railcar is given by the equation:
KE = 0.5 mv²
KE = 0.5 * 5000 * 1²
KE = 2500 J
The spring's compression, x = 0.1 m
The potential energy(PE) stored in the spring is given by the equation:
PE = 0.5kx²
PE = 0.5 * k * 0.1²
PE = 0.005k
According to the principle of energy conservation, Kinetic energy of the railcar equals the potential energy stored in the spring
KE = PE
2500 = 0.005k
k = 2500/0.005
k = 500000 N/m
k = 0.5 MN/m
What is Dark Matter?
Explanation:
dark matter is a form of matter thought to account for approximately 85% of the matter in the universe and about a quarter of its total mass–energy density or about 2.241×10⁻²⁷ kg/m³. Its presence is implied in a variety of astrophysical observations, including gravitational effects that cannot be explained by accepted theories of gravity unless more matter is present than can be seen. For this reason, most experts think that dark matter is abundant in the universe and that it has had a strong influence on its structure and evolution. Dark matter is called dark because it does not appear to interact with the electromagnetic field, which means it doesn't absorb, reflect or emit electromagnetic radiation, and is therefore difficult to detect.
Which observation have scientists used to support Einstein's general theory of relativity?
The orbital path of Mercury around the Sun has changed.
O GPS clocks function at the same rate on both Earth and in space.
O The Sun has gotten more massive over time.
Objects act differently in a gravity field than in an accelerating reference frame.
Answer:
Objects act differently in a gravity field than in an accelerating reference frame.
Explanation:
The main thrust of the theory general relativity as proposed by Albert Einstein boarders on space and time as the two fundamental aspects of spacetime. Spacetime is curved in the presence of gravity, matter, energy, and momentum. The theory of general relativity explains gravity based on the way space can 'curve', that is, it seeks to relate gravitational force to the changing geometry of space-time.
The Einstein general theory of relativity has replaced Newton's ideas proposed in earlier centuries as a means of predicting gravitational interactions. This concept is quite helpful but cannot be fitted into the context of quantum mechanics due to obvious incompatibilities.
Answer:
A - The orbital path of mercury around the sun has changed.
Explanation:
got right on edg.
Why can a magnetic monopole not exist, assuming Maxwell's Equations are currently correct and complete?
Answer:
Because closed magnetic field loops have to be formed between both ends of the magnet, a magnet will always have two poles.
Explanation:
Magnetic Monopoles do not exist in nature because a magnetic field always forms a loop that runs from one end of the magnet to the other.
Since this loop of the magnetic field has an origination and termination point which are at the two ends of the magnet (North and South poles). A magnet will always be bipolar which is in this case, North and South; even at an atomic level.
Consider a heat engine that inputs 10 kJ of heat and outputs 5 kJ of work. What are the signs on the total heat transfer and total work transfer
Answer:
Total heat transfer is positive
Total work transfer is positive
Explanation:
The first law of thermodynamics states that when a system interacts with its surrounding, the amount of energy gained by the system must be equal to the amount of energy lost by the surrounding. In a closed system, exchange of energy with the surrounding can be done through heat and work transfer.
Heat transfer to a system is positive and that transferred from the system is negative.
Also, work done by a system is positive while the work done on the system is negative.
Therefore, from the question, since the heat engine inputs 10kJ of heat, then heat is being transferred to the system. Hence, the sign of the total heat transfer is positive (+ve)
Also, since the heat engine outputs 5kJ of work, it implies that work is being done by the system. Hence the sign of the total work transfer is also positive (+ve).
which statement did Ernest Rutherford make about atoms?
Answer:
Option A
Explanation:
Ernest Rutherford concluded that the atom has a small, dense center which constitutes the mass of the whole atom. He called it a "Nucleus". He also said that most of the space in the atom is empty.
a certain plane parallel capacitor stores energy E when the plates have a charge Q on each plate. Then distance between the plates is double. In order to store triply as much energy, how much charge should it have in its plates
Answer:
[tex]Q'=\sqrt{6}Q[/tex]
Explanation:
You have that a parallel plate capacitor has a total energy of E when the distance between the plates is d and the charge on each plate is Q.
You take into account the following formula for the stored energy in the capacitor:
[tex]E=\frac{1}{2}\frac{Q^2}{C}[/tex] (1)
The capacitance C of the parallel plate capacitor is given by the following formula is:
[tex]C=\epsilon_o\frac{A}{d}[/tex] (2)
A: area of the plates
ε0: dielectric permittivity of vacuum
You replace the expression (2) into the equation (1):
[tex]E=\frac{1}{2}\frac{Q^2A}{\epsilon_o d}[/tex] (3)
the previous formula is the expression for the total energy stored for the given parameters A, d and Q.
If the distance between the plates is twice and it is required that the energy is three times the initial energy, to find the value of the charge you use the equation (3):
[tex]E'=\frac{1}{2}\frac{Q'^2A}{\epsilon_o d'}[/tex] (4)
d' = 2d
E' = 3E
Q': required charge
You replace the values of d' and E' in the equation (4) and then divide the result with the equation (3):
[tex]3E=\frac{1}{2}\frac{Q'^2A}{\epsilon_o(2d)}=\frac{1}{4}\frac{Q'^2A}{\epsilon_od}\\\\\frac{3E}{E}=\frac{1/4\frac{Q'^2A}{\epsilon_od}}{1/2\frac{Q^2A}{\epsilon_o d}}\\\\3=\frac{1}{2}\frac{Q'^2}{Q^2}[/tex]
Finally, you solve for Q':
[tex]3=\frac{1}{2}\frac{Q'^2}{Q^2}\\\\Q'=\sqrt{6}Q[/tex]
Then, the required charge is √6Q , to obtain three times the initial energy E, when the distance between plates is doubled.
How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the orbit of Mars is 12.6 light-minutes.
Question:
A spaceship enters the solar system moving toward the Sun at a constant speed relative to the Sun. By its own clock, the time elapsed between the time it crosses the orbit of Jupiter and the time it crosses the orbit of Mars is 35.0 minutes
How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the orbit of Mars is 12.6 light-minutes.
Answer:
S = 5.508 × 10¹¹m
V = 2.62 × 10⁸ m/s
Explanation:
The radius of the orbit of Jupiter, Rj is 43.2 light-minutes
radius of the orbit of Mars, Rm is 12.6 light-minutes
Distance travelled S = (Rj - Rm)
= 43.2 - 12.6 = 30.6 light- minutes
= 30.6 × (3 ×10⁸m/s) × 60 s
= 5.508 × 10¹¹m
time = 35mins = (35 × 60 secs)
= 2100 secs
speed = distance/time
V = 5.508 × 10¹¹m / 2100 s
V = 2.62 × 10⁸ m/s
A fluid moves through a tube of length 1 meter and radius r=0.002±0.0002 meters under a pressure p=4⋅105±1750 pascals, at a rate v=0.5⋅10−9 m3 per unit time. Use differentials to estimate the maximum error in the viscosity η given by
Answer:
The maximum error is [tex]\Delta \eta = 2032.9[/tex]
Explanation:
From the question we are told that
The length is [tex]l = 1\ m[/tex]
The radius is [tex]r = 0.002 \pm 0.0002 \ m[/tex]
The pressure is [tex]P = 4 *10^{5} \ \pm 1750[/tex]
The rate is [tex]v = 0.5*10^{-9} \ m^3 /t[/tex]
The viscosity is [tex]\eta = \frac{\pi}{8} * \frac{P * r^4}{v}[/tex]
The error in the viscosity is mathematically represented as
[tex]\Delta \eta = | \frac{\delta \eta}{\delta P}| * \Delta P + |\frac{\delta \eta}{\delta r} |* \Delta r + |\frac{\delta \eta}{\delta v} |* \Delta v[/tex]
Where [tex]\frac{\delta \eta }{\delta P} = \frac{\pi}{8} * \frac{r^4}{v}[/tex]
and [tex]\frac{\delta \eta }{\delta r} = \frac{\pi}{8} * \frac{4* Pr^3}{v}[/tex]
and [tex]\frac{\delta \eta }{\delta v} = - \frac{\pi}{8} * \frac{Pr^4}{v^2}[/tex]
So
[tex]\Delta \eta = \frac{\pi}{8} [ |\frac{r^4}{v} | * \Delta P + | \frac{4 * P * r^3}{v} |* \Delta r + |-\frac{P* r^4}{v^2} |* \Delta v][/tex]
substituting values
[tex]\Delta \eta = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}} | * 1750 + | \frac{4 * 4 *10^{5} * (0.002)^3}{0.5*10^{-9}} |* 0.0002 + |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2} |* 0 ][/tex]
[tex]\Delta \eta = \frac{\pi}{8} [56 + 5120 ][/tex]
[tex]\Delta \eta = 647 \pi[/tex]
[tex]\Delta \eta = 2032.9[/tex]
A railroad boxcar rolls on a track at 2.90 m/s toward two identical coupled boxcars, which are rolling in the same direction as the first, but at a speed of 1.20 m/s. The first reaches the second two and all couple together. The mass of each is 3.05 ✕ 104 kg.(a)What is the speed (in m/s) of the three coupled cars after the first couples with the other two? (Round your answer to at least two decimal places.)Incorrect: Your answer is incorrect.What is the momentum of the two coupled cars? What is the momentum of the first car in terms of its mass and initial speed? Note all cars are initially traveling in the same direction. Apply conservation of momentum to find the final speed. m/s(b)Find the (absolute value of the) amount of kinetic energy (in J) converted to other forms during the collision.J
Answer:
momentum of the coupled cars V = 1.77 m/s
kinetic energy coverted to other forms during the collision ΔK.E = -2.892×10⁴J
Explanation:
given
m₁ =3.05 × 10⁴kg
u₁ =2.90m/s
m₂=6.10× 10⁴kg
u₂=1.20m/s
using law of conservation of momentum
m₁u₁ + m₂u₂ = (m₁ + m₂) V
3.05 × 10⁴ ×2.90 + 6.10× 10⁴× 1.20 = (9.15×10⁴)V
V = 1.617×10⁵/9.15×10⁴
V = 1.77m/s
K.E =1/2mV²
ΔK.E = K.E(final) - K.E(initial)
ΔK.E = ¹/₂ × 9.15×10⁴ ×(1.77)² - ¹/₂ ×3.05 × 10⁴ × (2.90)² -¹/₂ × 6.10× 10⁴× (1.20)²
ΔK.E = ¹/₂ × (28.67-25.65-8.784) ×10⁴
ΔK.E = -2.892×10⁴J
The final speed is 1.77 m/s
The initial momentum is 8.84 × 10⁴ kgm/s [first car] and 7.3 × 10⁴ kgm/s [coupled car]
2.892×10⁴J of energy is converted.
Inelastic collision:Since the first boxcar collides and couples with the two coupled boxcars, the collision is inelastic. In an inelastic collision, the momentum of the system is conserved but there is a loss in the total kinetic energy of the system.
Let the mass of the railroad boxcar be m₁ =3.05 × 10⁴kg
The initial speed of the railroad boxcar is u₁ = 2.90m/s
Mass of the two coupled boxcars m₂ = 2 × 3.05 × 10⁴kg = 6.10× 10⁴kg
And the initial speed be u₂ = 1.20m/s
The initial momentum of the first car is:
m₁u₁ = 3.05 × 10⁴ × 2.90 = 8.84 × 10⁴ kgm/s
The initial momentum of the coupled car is:
m₁u₁ = 6.10 × 10⁴ × 1.20 = 7.3 × 10⁴ kgm/s
Let the final speed after all the boxcars are coupled be v
From the law of conservation of momentum, we get:
m₁u₁ + m₂u₂ = (m₁ + m₂)v
3.05 × 10⁴ ×2.90 + 6.10× 10⁴× 1.20 = (9.15×10⁴)Vv
v = 1.617×10⁵/9.15×10⁴
v = 1.77m/s
The difference between initial and final kinetic energies is the amount of energy converted into other forms, which is given as follows:
ΔKE = K.E(final) - K.E(initial)
ΔKE = ¹/₂ × 9.15×10⁴ ×(1.77)² - ¹/₂ ×3.05 × 10⁴ × (2.90)² -¹/₂ × 6.10× 10⁴× (1.20)²
ΔKE = ¹/₂ × (28.67-25.65-8.784) ×10⁴
ΔKE = -2.892×10⁴J
Learn more about inelastic collision:
https://brainly.com/question/13861542?referrer=searchResults
A pulley system is used at a dock to lift shipments of fish off a boat. If you apply a force of 100 N to the pulley, it pulls the shipment with a force of 830 N. a. What is the mechanical advantage of the pulley? b. The pulley has an efficiency of 80%. If you perform 600 J of work, how much useful work does the pulley do?
Explanation:
a. Mechanical advantage = force out / force in
MA = 830 / 100
MA = 8.3
b. Efficiency = work out / work in
0.80 = W / 600 J
W = 480 J
A length of organ pipe is closed at one end. If the speed of sound is 344 m/s, what length of pipe (in cm) is needed to obtain a fundamental frequency of 50 Hz
Answer:
The length = 27.52m
Explanation:
v=f x wavelength
a heat engine with an efficiency of 30.0% performs 2500 j of work. how much heat is discharged to the lower temperature reservoir
Answer:
Q₂ = 5833.33 J
Explanation:
First we need to find the energy supplied to the heat engine. The formula for the efficiency of the heat engine is given as:
η = W/Q₁
where,
η = efficiency of engine = 30% = 0.3
W = Work done by engine = 2500 J
Q₁ = Heat supplied to the engine = ?
Therefore,
0.3 = 2500 J/Q₁
Q₁ = 2500 J/0.3
Q₁ = 8333.33 J
Now, we find the heat discharged to lower temperature reservoir by using the formula of work:
W = Q₁ - Q₂
Q₂ = Q₁ - W
where,
Q₂ = Heat discharged to the lower temperature reservoir = ?
Therefore,
Q₂ = 8333.33 J - 2500 J
Q₂ = 5833.33 J
HELP AGAIN A car going .80m/s accelerates uniformly at .20m/s^2 . What is the distance covered in 1.3 minutes? brainliest goes to person who shows formula/work
Answer:
initial speed (u) = 0.8 m/s
acceleration (a) = 0.2 m/s/s
time (t) = 1.3 min OR 1.3*60 seconds
= 78 seconds
we will use the second equation of motion to find the distance
distance (s) = ut + 1/2 a(t^2)
s = 0.8 * 78 + 1/2 * 0.2 * (6084)
s = 62.4 + 608.4
s = 670.8 m