Answer:
The mole fraction of hexane is 0.67
Explanation:
This problem can be solved using Dalton´s law and Raoult´s law.
The vapor pressure of a mixture of gases is the sum of the partial pressures of each gas (Dalton´s law).
Example:
In a mixture of gases A and B
Pt = PA + PB
where:
Pt = total pressure
PA = partial pressure of A
PB = partial pressure of B
In an ideal solution, the vapor pressure of each component is equal to the vapor pressure of the pure component times the mole fraction of the component in the solution (Raoult´s law).
Example:
In a solution containing A and B, the vapor pressure of A in the solution will be:
PA = P(pure A) * Xa
Where
PA = vapor pressure of A in the solution
P(pure A) = vapor pressure of pure A
Xa = mole fraction of A
In our problem, we have that the vapor pressure of the solution is 241 torr.
Then, using Dalton´s law:
Pt = P(hexane) + P(Pentane)
Using Raoult´s law:
P(hexane) = P(pure hexane) * X(hexane)
P(pentane) = P(pure pentane) * X(pentane)
We also know that the sum of the molar fractions of each component in a solution equals 1:
X(hexane) + X(pentane) = 1
X(pentane) = 1 - X(hexane)
Replacing in the Dalton´s law in terms of X(hexane):
Pt = P(pure hexane) * X(hexane) + P(pure pentane) * (1 - X(hexane))
Solving for X(hexane):
Pt = P(pure hex) * X(hex) + P(pure pent) - P(pure pent) * X(hex)
Replacing with the data:
241 torr = 151 torr * X(hex) + 425 torr - 425 torr*X(hex)
-184 torr = -274 torr * X(hex)
-184torr/-274 torr = X(hex)
X(hex) = 0.67
The mole fraction of hexane in the solution is 1.59.
Explanation:To find the mole fraction of hexane in the solution, we can use Raoult's law, which states that the partial pressure of a component in a solution is equal to the product of the mole fraction of that component and its vapor pressure.
Let's calculate the mole fraction of hexane:
Partial pressure of hexane = mole fraction of hexane * vapor pressure of hexane
241 torr = x * 151 torr
241 torr / 151 torr = x = 1.59
The mole fraction of hexane in the solution is 1.59.
The weight of a body above sea level varies inversely with the square of the distance from the center of Earth. If a woman weighs 131 pounds when she is at sea level, 3960 miles from the center of Earth, how much will she weigh when she is at the top of a mountain, 4.8 miles above sea level?
Answer:
89.16pounds
Explanation:
The equation that defines this problem is as follows
W=k/X^2
where
W=Weight
K= proportionality constant
X=distance from the center of Earth
first we must find the constant of proportionality, with the first part of the problem
k=WX^2=131x3960^2=2054289600pounds x miles^2
then we use the equation to calculate the woman's weight with the new distance
W=2054289600/(4800)^2=89.16pounds
What is the entropy change of a 0.349 g spoonful of water that evaporates completely on a hot plate whose temperature is slightly above the boiling point of water?
Answer:
[tex]\Delta S = 2.11 J/K[/tex]
Explanation:
As we know that entropy change for phase conversion is given as
[tex]\Delta S = \frac{\Delta Q}{T}[/tex]
Here we know that heat required to change the phase of the water is given as
[tex]\Delta Q = mL[/tex]
here we have
[tex]m = 0.349 g = 3.49\times 10^{-4} kg[/tex]
L = 2250000 J/kg
now we have
[tex]\Delta Q = (3.49 \times 10^{-4})\times 2250000[/tex]
[tex]\Delta Q = 788.9 J[/tex]
also we know that temperature is approximately same as boiling temperature
so we have
[tex]T = 373 k[/tex]
so here we have
[tex]\Delta S = \frac{788.9}{373}[/tex]
[tex]\Delta S = 2.11 J/K[/tex]
A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.40 m/s2. Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.80 s, Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off and ignore effects of air resistance. What is the maximum height above ground reached by the helicopter?
Answer:
314.92 m
Explanation:
Acceleration of the helicopter = 5.4 m/s² = a
Time taken by the helicopter to reach maximum height = 10.8 s = t
Initial velocity = 0 = u
Final velocity = v
Displacement = s
Equation of motion
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 10.8+\frac{1}{2}5.4\times 10.8^2\\\Rightarrow s=314.92\ m[/tex]
Maximum height above ground reached by the helicopter is 314.92 m
A particle undergoes two displacements. The first has a magnitude of 11 m and makes an angle of 82 ◦ with the positive x axis. The result after the second displacement is 8.7 m directed at an angle of 135◦ to the positive x axis using counterclockwise as the positive angular direction schoenzeit (ms83473) – Assignment 01-1 – pusch – (19512020) 5 135◦ 82◦ 11 m 8.7 m Find the angle of the second displacement measured counterclockwise from the positive x axis (i.e., a positive angle). Answer in units of ◦ .
Final answer:
To calculate the angle of the second displacement, typically one would use vector addition and trigonometry, but the question, as presented, does not provide sufficient information to determine this angle.
Explanation:
To find the angle of the second displacement for the particle, we need to consider the information given: the first displacement is 11 m at an angle of 82° from the positive x-axis, and the resulting displacement after the second move is 8.7 m at 135° to the positive x-axis, using the counterclockwise direction as positive.
Using vector addition, the resulting displacement's direction can help us determine the second displacement angle. However, without more information about the specifics of the second displacement (like its magnitude), we cannot calculate the exact angle or vector using typical methods such as vector components or the law of cosines.
Given the problem as stated, assuming standard vector addition, a common approach would involve drawing a vector diagram and applying trigonometry; but in this case, we seem to be missing details to perform accurate calculations.
a vector has an x-component of 19.5m and a y-component of 28.4m. Find the magnitude and direction of the vector
Answer:
magnitude=34.45 m
direction=[tex]55.52\°[/tex]
Explanation:
Assuming the initial point P1 of this vector is at the origin:
P1=(X1,Y1)=(0,0)
And knowing the other point is P2=(X2,Y2)=(19.5,28.4)
We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.
For the magnitude we will use the formula to calculate the distance [tex]d[/tex] between two points:
[tex]d=\sqrt{{(Y2-Y1)}^{2} +{(X2-X1)}^{2}}[/tex] (1)
[tex]d=\sqrt{{(28.4 m - 0 m)}^{2} +{(19.5 m - 0m)}^{2}}[/tex] (2)
[tex]d=\sqrt{1186.81 m^{2}}[/tex] (3)
[tex]d=34.45 m[/tex] (4) This is the magnitude of the vector
For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:
[tex]tan \theta=\frac{Y2-Y1}{X2-X1}[/tex] (5)
[tex]tan \theta=\frac{24.8 m - 0m}{19.5 m - 0m}[/tex] (6)
[tex]tan \theta=\frac{24.8}{19.5}[/tex] (7)
Finding [tex]\theta[/tex]:
[tex]\theta= tan^{-1}(\frac{24.8}{19.5})[/tex] (8)
[tex]\theta= 55.52\°[/tex] (9) This is the direction of the vector
A person pushes horizontally on a 50-kg crate, causing it to accelerate from rest and slide across the surface. If the push causes the crate to accelerate at 2.0 m/s2, what is the velocity of the crate after the person has pushed the crate a distance of 6 meters?
Answer:
[tex]v_{f} =4.9\frac{m}{s}[/tex] : velocity of the crate after the person has pushed the crate a distance of 6 meters
Explanation:
Crate kinetics
Crate moves with uniformly accelerated movement
v f²=v₀²+2a*d (formula 1)
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Known data
v₀=0 The speed of the crate is equal to zero because part of the rest.
a= 2m/s²
d= 6m
Distance calculating
We replace data in the Formula (1)
v f²=0+2*2*d
v f²=2*2*6
v f²=24
[tex]v_{f} =\sqrt{24}[/tex]
[tex]v_{f} =4.9\frac{m}{s}[/tex]
To find the final velocity of a 50-kg crate after being pushed for 6 meters with an acceleration of 2.0 m/s², the kinematic equation v² = u² + 2as is used, resulting in a final velocity of 4.9 m/s.
Explanation:To calculate the velocity of a 50-kg crate after being pushed a distance of 6 meters with an acceleration of 2.0 m/s2 from rest, we use the kinematic equation v2 = u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance moved. Since the crate starts from rest, the initial velocity, u, is 0 m/s. Substituting the given values we get:
v2 = 0 m/s + 2(2.0 m/s2)(6 m),
v2 = 24 m2/s2,
v = √(24 m2/s2),
v = 4.9 m/s.
Therefore, the velocity of the crate after being pushed 6 meters will be 4.9 m/s.
A swan on a lake becomes airborne by flapping its wings and running on top of the water. If the swan must reach a velocity of 6.50 m/s to take off and it accelerates from rest at an average rate of 0.350 m/s2 , what distance Δx does it travel before becoming airborne?
Answer:
The distance traveled by the swan is 60.35 meters.
Explanation:
Given that,
A swan accelerate from rest (u = 0) to 6.5 m/s to take off.
Acceleration of the swan, [tex]a=0.35\ m/s^2[/tex]
We need to find the distance Δx it travel before becoming airborne. From the third equation of motion as :
[tex]\Delta x=\dfrac{v^2-u^2}{2a}[/tex]
[tex]\Delta x=\dfrac{(6.5)^2}{2\times 0.35}[/tex]
[tex]\Delta x=60.35\ m[/tex]
So, the distance traveled by the swan is 60.35 meters. Hence, this is the required solution.
During the developing of a TLC plate, it is common to place a cover on the chromatography chamber and have a piece of moist filter paper line the walls of the TLC chamber. Why is the moist filter paper in the TLC chamber of importance during thin-layer chromatography?
Answer:
The moist filter paper is in charge of preventing evaporation and ensuring the proper saturation of the air of the chromatography chamber.
Explanation:
Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures.
A strip of moist filter paper is put into the chromatography chamber so that its bottom touches the solvent and the paper lies on the chamber wall and reaches almost to the top of the container.
The container is closed and left for a few minutes to let the solvent vapors ascend the moist filter paper and saturate the air in the chamber.
The moist filter paper is in charge of preventing evaporation and ensuring the proper saturation of the air of the chamber.
The distance between two adjacent peaks on a wave is called the wavelength. The wavelength of a beam of ultraviolet light is 113 nanometers (nm). What is its wavelength in meters?
Answer:
0.000000113 or 1.13*[tex]10^{-7}[/tex] meters
Explanation:
One nanometer is [tex]10^{-9}[/tex] meters. So 113 nanometers would be 113*[tex]10^{-9}[/tex], or 1.13*[tex]10^{-7}[/tex] meters. That's expessed on "cientific notation." On the "usual" notation, it will be 0.000000113 meters.
For a standard production car, the highest roadtested acceleration ever reported occurred in 1993, when a Ford RS200 Evolution went from zero to 26.8 m/s (60 mph) in 3.275 s. Find the magnitude of the car's acceleration.
Answer:
a=8.1832m/s^2
Explanation:
Vo=initial speed=0m/s
Vf=final speed=26.8m/s
t=time=3.275s
the vehicle moves with a constant acceleration therefore we can use the following equation
A=aceleration=(Vf-Vi)/t
A=(26.8m/s-0m/s)/3.275s=8.1832m/s^2
the magnitude of the car´s aceleration is 8.1832m/s^2
A voltaic cell is constructed with two Zn2+-Zn electrodes, where the half-reaction is Zn2+ + 2e− → Zn (s) E° = -0.763 V The concentrations of zinc ion in the two compartments are 4.50 M and 1.11 ⋅ 10−2 M, respectively. The cell emf is ________ V.
Answer : The cell emf for this cell is 0.077 V
Solution :
The balanced cell reaction will be,
Oxidation half reaction (anode): [tex]Zn(s)\rightarrow Zn^{2+}+2e^-[/tex]
Reduction half reaction (cathode): [tex]Zn^{2+}+2e^-\rightarrow Zn(s)[/tex]
In this case, the cathode and anode both are same. So, [tex]E^o_{cell}[/tex] is equal to zero.
Now we have to calculate the cell emf.
Using Nernest equation :
[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}{diluted}}{[Zn^{2+}{concentrated}]}[/tex]
where,
n = number of electrons in oxidation-reduction reaction = 2
[tex]E_{cell}[/tex] = ?
[tex][Zn^{2+}{diluted}][/tex] = 0.0111 M
[tex][Zn^{2+}{concentrated}][/tex] = 4.50 M
Now put all the given values in the above equation, we get:
[tex]E_{cell}=0-\frac{0.0592}{2}\log \frac{0.0111M}{4.50M}[/tex]
[tex]E_{cell}=0.077V[/tex]
Therefore, the cell emf for this cell is 0.077 V
The cell emf, or electromotive force, for a voltaic cell, like the Zn2+-Zn kind described, can be calculated with the Nernst equation. The emf is determined by the standard electrode potential as well as the concentrations of the redox species in each half-cell.
Explanation:A voltaic cell runs on a spontaneous redox reaction occurring indirectly, with the oxidant and reductant redox couples contained in separate half-cells. In your example, both half-cells are Zn2+-Zn electrodes where the reaction is Zn2+ + 2e− → Zn (s) with a standard electrode potential, E°, of -0.763 V. The cell emf, or electromotive force, is calculated through the Nernst Equation:
Ecell = E° - (RT/nF) * ln(Q)
where Q is the reaction quotient, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the redox reaction, and F is Faraday's constant. Since we're dealing with a concentration cell, Q is given by the ratio of the concentrations of the reduced and oxidized species. Depending on the temperature and assuming that Zn2+ is reduced at the cathode and oxidized at the anode, the cell emf can be calculated accordingly.
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True or False: When using the endpoint mover, you will receive credit for each endpoint that is positioned correctly.
Final answer:
The statement regarding receiving credit for correctly positioned endpoints using an endpoint mover is typically true. Credit is awarded for each endpoint positioned correctly in a digital math exercise or tool, which may involve placing endpoints according to specific criteria.
Explanation:
The statement "When using the endpoint mover, you will receive credit for each endpoint that is positioned correctly" is generally True. In the context of mathematical tools or interactive digital platforms, an endpoint mover likely refers to a feature that allows users to manipulate the endpoints of a line segment or other geometric object. When students are tasked with positioning these endpoints correctly according to given criteria, they would typically receive credit or points for each endpoint that is placed accurately.
For example, in a digital math exercise, if you're asked to position the endpoints of a line segment so that one end is at the point (3,2) and the other end is at the point (-1,5), and you succeed in placing them correctly, you would receive credit for both endpoints. However, details may vary based on the specific platform or educational program, so it's essential to follow the instructions provided.
A carousel is spinning at a speed of 100 cm/s, with the outermost horse at a distance 5 m from the center. This horse experiences a force of 3 x 10-3 or 0.003 N. Consider a horse placed at exactly half the distance from the center as the outermost horse. What will be the centripetal force experienced by this horse, if all the other variables are kept the same?
Answer:
0.006 N
Explanation:
First find the mass of the horse which is the only unknown variable.
Then use it with the new data. Working out is in the attachment.
The densities of cardboard, aluminum, and lead are 0.6 g/cm^3, 2.7 g/cm^3, and 11.4 g/cm^3, respectively. Suppose that you are studying the range of a (nonexistent) elementary particle, the Heidbrinkion, and that it takes 49 cm of cardboard, or 42 cm of aluminum, or 17 cm of lead to stop half of the Heibrinkions emitted from a source. Calculate the absorber thickness for each material.
__________ g/cm^2 for cardboard
__________ g/cm^2 for aluminum
__________ g/cm^2 for lead
Answer:
For cardboard = 29.4 g/cm²
For aluminium = 113.4 g/cm²
For lead = 193.8 g/cm²
Explanation:
Given:
Density of the cardboard, d₁ = 0.6 g/cm³
Density of the aluminium, d₂ = 2.7 g/cm³
Density of the lead, d₃ = 11.4 g/cm³
Length of the cardboard, L₁ = 49 cm
Length of the aluminium, L₂ = 42 cm
Length of the lead, L₃ = 17 cm
Now,
The absorber thickness is calculated as:
= Density × Length
therefore,
For cardboard = d₁ × L₁ = 0.6 × 49 = 29.4 g/cm²
For aluminium = d₂ × L₂ = 2.7 × 42 = 113.4 g/cm²
For lead = d₃ × L₃ = 11.4 × 17 = 193.8 g/cm²
An engineer in a locomotive sees a car stuck
on the track at a railroad crossing in front of
the train. When the engineer first sees the
car, the locomotive is 200 m from the crossing
and its speed is 13 m/s.
If the engineer’s reaction time is 0.46 s,
What should be the magnitude of the minimum deceleration to avoid an accident?
Answer in units of m/s²
Answer:
[tex]-0.44 m/s^2[/tex]
Explanation:
First of all, we need to calculate the distance covered by the locomotive during the reaction time, which is
t = 0.46 s
During this time, the locomotive travels at
v = 13 m/s
And the motion is uniform, so the distance covered is
[tex]d_1 = vt = (13)(0.46)=6.0 m[/tex]
The locomotive was initially 200 m from the crossing, so the distance left to stop is now
[tex]d=200 - 6.0 = 194.0 m[/tex]
And now the locomotive has to slow down to a final velocity of [tex]v=0[/tex] in this distance. We can find the minimum deceleration needed by using the suvat equation:
[tex]v^2 - u^2 = 2ad[/tex]
where
v = 0 is the final velocity
u = 13 m/s is the initial velocity
a is the deceleration
d = 194.0 m is the distance to stop
Solving for a,
[tex]a=\frac{v^2-u^2}{2d}=\frac{0^2-13^2}{2(194)}=-0.44 m/s^2[/tex]
A jet plane lands with a velocity of +107 m/s and can accelerate at a maximum rate of -5.18 m/s2 as it comes to rest. From the instant it touches the runway, what is the minimum time needed before it can come to rest?
Answer:
20.7 s
Explanation:
The equation to calculate the velocity for a uniform acceleration a, time t and initial velocity v₀:
v = a*t + v₀
Solve for t:
t = (v - v₀)/a
A ball is dropped from a tower that is 512 feet high. Use the formula below to find height of the ball 5 seconds after it was dropped? h=512−14t2where h represents the height, in feet, and t represents the time, in seconds, after it was dropped.
Answer:
The height of the ball after 5 seconds is 162 ft.
Explanation:
First, replace the variable t with how many seconds the ball has dropped, which in this case is 5.
[tex]h = 512 - 14 {t}^{2} \\ h = 512 - 14 \times {5}^{2} [/tex]
Solve.
[tex]h = 512 - 14 \times {5}^{2} \\ h = 512 - 14 \times 25 \\ h = 512 - 350 \\ h = 162[/tex]
For a duration of 5 seconds, the ball had managed to drop 350 feet, with 162 feet left to go to touch ground level.
Ship A is located 4.1 km north and 2.3 km east of ship B. Ship A has a velocity of 22 km/h toward the south and Ship B has a velocity of 40 km/h in a direction 38° north of east. What are the (a) x-component and (b) y-component of the velocity of A relative to B?
Answer:
a) x component = -31.25 km/hr
b) y component = 46.64 km/hr
Explanation:
Given data:
A position is 4km north and 2.5 km east to B
Ship A velocity = 22 km/hr
ship B velocity = 40 km/hr
A velocity wrt to velocity of B
[tex]\vec{V_{AB}} =\vec{V_A} - \vec{V_B}[/tex]
[tex]\vec{V_A} = 22 km/hr[/tex]
[tex]\vec{V_B} = 40 cos38\hat{i} + 40sin 38 \hat{j}[/tex]
[tex]= 31.52\hat{i} + 24.62 \hat{j}[/tex]
putting respective value to get velocity of A with respect to B
[tex]\vec{V_{AB}} = -22 \hat {j} - (31.52\hat{i} + 24.62 \hat{j})[/tex]
[tex]\vec{V_{AB}} = -31.52\hat{i} - 46.62\hat{j}[/tex]
a) x component = -31.25 km/hr
b) y component = 46.64 km/hr
A ball is thrown horizontally from the top of a 65 m building and lands 115 m from the base of the building. Ignore air resistance, and use a coordinate system whose origin is at the top of the building, with positive y upwards and positive x in the direction of the throw.
(a) How long is the ball in the air in seconds?
(b) What must have been the initial horizontal component of the velocity in m/s?
(c) What is the vertical component of the velocity just before the ball hits the ground in m/s?
(d) What is the magnitude of the velocity of the ball just before it hits the ground in m/s?
Answer:
The answer to your question is:
a) t = 3.64 s
b) vox = 31.59 m/s
c) vy = 35.71 m/s
d) v = 47.67 m/s
Explanation:
a) To calculate the time, we know that voy = 0 m/s so we this this formula
h = voy + 1/2(gt²)
65 = 0 + 1/2(9.81)(t²)
65 = 4.905t²
t² = 65/4.905 = 13.25
t = 3.64 s
b) To calculate vox we use this formula vox = d/t ; vox is constant
vox = 115/3.64 = 31.59 m/s
c) To calculate voy we use the formula vy = voy + gt
but voy = 0
vy = gt = 9.81 x 3.64 = 35.71 m/s
d) To calculate v we use the pythagorean theorem
c2 = a2 + b2
c2 = 31.59² + 35.71² = 997.92 + 1275.20
c2 = 2273.12
c = 47.67 m/s
Is it possible for two pieces of the same metal to have different recrystallization temperatures? Is it possible for recrystallization to take place in some regions of a part before it does in other regions of the same part? Explain
Recrystallization is a temperature-induced process that can vary among materials. Metal parts may recrystallize at different temperatures, and recrystallization can happen unevenly within a single part.
Recrystallization is a process where a material undergoes structural changes due to increased temperature. Metallic glasses can have different recrystallization temperatures based on the composition of the metals. Recrystallization can occur non-uniformly in a part, with some regions undergoing the process before others due to variations in temperature or cooling rates.
A wave on the ocean surface with wavelength 44 m travels east at a speed of 18 m/s relative to the ocean floor. If, on this stretch of ocean, a powerboat is moving at 14 m/s (relative to the ocean floor), how often does the boat encounter a wave crest, if the boat is traveling (a) west, and (b) east?
Answer:
A)t=1.375s
B)t=11s
Explanation:
for this problem we will assume that the east is positive while the west is negative, what we must do is find the relative speed between the wave and the powerboat, and then with the distance find the time for each case
ecuations
V=Vw-Vp (1)
V= relative speed
Vw= speed of wave
Vp=Speesd
t=X/V(2)
t=time
x=distance=44m
A) the powerboat moves to west
V=18-(-14)=32m/s
t=44/32=1.375s
B)the powerboat moves to east
V=18-14=4
t=44/4=11s
Final answer:
The frequency at which the boat encounters a wave crest is 0.09 Hz regardless of the direction the boat is traveling.
Explanation:
To find the frequency at which the boat encounters a wave crest, we need to determine the time it takes for one wave crest to pass by the boat. The speed of the boat relative to the ocean floor doesn't affect the frequency, only the speed of the wave. The formula to calculate the frequency of a wave is:
Frequency = Wave Speed / Wavelength
When the boat is traveling west:Wave speed = Speed of the wave + Speed of the boat (since the boat is traveling in the opposite direction)
Frequency = (Speed of the wave + Speed of the boat) / Wavelength
Frequency = (18 m/s + (-14 m/s)) / 44 m
Frequency = 4 m/s / 44 m = 0.09 Hz
Wave speed = Speed of the wave - Speed of the boat (since the boat is traveling in the same direction)
Frequency = (Speed of the wave - Speed of the boat) / Wavelength
Frequency = (18 m/s - 14 m/s) / 44 m
Frequency = 4 m/s / 44 m = 0.09 Hz
The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 542 nm. What is the work function W0 of this metal? Express your answer in electron volts.
The work function of the metal is approximately 3.67 eV.
Explanation:The work function, or binding energy, of a metal is the minimum energy required to eject an electron from the metal surface. In the photoelectric effect, the maximum kinetic energy of ejected electrons (photoelectrons) is given by the equation KE = hf - BE, where hf is the photon energy and BE is the work function. To find the work function W0 of the metal, we can use Equation 6.16, which states that the threshold wavelength for observing the photoelectric effect is given by λ = (hc)/(W0), where h is Planck's constant, c is the speed of light, and W0 is the work function.
Given that the maximum wavelength that can still eject electrons is 542 nm, we can rearrange the equation to solve for W0:
W0 = (hc)/λ
Plugging in the values, we have:
W0 = (6.626 × 10^-34 J·s)(3 × 10^8 m/s)/(542 × 10^-9 m)
Simplifying the calculation, we find that the work function W0 of the metal is approximately 3.67 eV.
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Final answer:
The maximum wavelength for ejecting electrons from the metal surface is 542 nm. Using the energy equation for photons, we calculate the work function (W0) to be approximately 2.29 electron volts.
Explanation:
The question relates to the photoelectric effect and the calculation of the work function of a metal when electromagnetic radiation of a certain wavelength is incident upon it. The work function (W0) is the minimum energy needed to eject an electron from the metal surface. According to the question, the maximum wavelength that can eject an electron is 542 nm.
To find the work function in electron volts, we can use the equation:
E(photon) = hc / λ
Where:
E(photon) is the energy of the photon
h is Planck's constant (4.135667696 × [tex]10^-^1^2[/tex] eV·s)
c is the speed of light (3 × 108 m/s)
λ is the wavelength of the photon in meters (542 nm = 542 × [tex]10^-^9[/tex] m)
First, convert the energy into electron volts:
E(photon) = (4.135667696 × [tex]10^-^1^2[/tex] eV·s × 3 × 108 m/s) / (542 × [tex]10^-^9[/tex] m)
After calculation, we find that E(photon) is approximately 2.29 eV. For a photon to eject an electron, its energy must be equal to or greater than the work function of the metal. Therefore, the work function of the metal is also 2.29 eV.
a vector is 253m long and points in a 55.8degree direction. Find the y-component of the vector.
Answer:
209 m
Explanation:
The y-component of a vector is the magnitude times the sine of the angle.
y = 253 sin 55.8°
y = 209
Answer:
The answer to your question is: 209 m
Explanation:
Data
length = 253 m
d = 55.8°
y - component = ?
Formula
To solve this problem we need to use a right triangle a the trigonometric functions (sine, cosine, tangent, etc)
Now we have to choose among the trigonometric functions which one to use that relates the opposite side and the hypotenuse.
And that one is sineФ = os/h
we clear os = sineФ x h
os = sine 55.8 x253
os = 209 m
soh, cah, toa
The acceleration of a bus is given by ax(t)=αt, where α = 1.28 m/s3 is a constant.
The text looks incomplete. Here the complete question found on google:
"The acceleration of a bus is given by ax(t)=αt, where α = 1.28m/s3 is a constant. Part A If the bus's velocity at time t1 = 1.13s is 5.09m/s , what is its velocity at time t2 = 2.02s ? If the bus's position at time t1 = 1.13s is 5.92m , what is its position at time t2 = 2.02s ?"
A) 6.88 m/s
The velocity of the bus can be found by integrating the acceleration. Therefore:
[tex]v(t) = \int a(t) dt = \int (\alpha t)dt=\frac{1}{2}\alpha t^2+C[/tex] (1)
where
[tex]\alpha = 1.28 m/s^3[/tex]
C is a constant term
We know that at [tex]t_1 = 1.13 s[/tex], the velocity is [tex]v=5.09 m/s[/tex]. Substituting these values into (1), we can find the exact value of C:
[tex]C=v(t) - \frac{1}{2}\alpha t^2 = 5.09 - \frac{1}{2}(1.28)(1.13)^2=4.27 m/s[/tex]
So now we can find the velocity at time [tex]t_2 = 2.02 s[/tex]:
[tex]v(2.02)=\frac{1}{2}(1.28)(2.02)^2+4.27=6.88 m/s[/tex]
B) 11.2 m
To find the position, we need to integrate the velocity:
[tex]x(t) = \int v(t) dt = \int (\frac{1}{2}\alpha t^2 + C) dt = \frac{1}{3}\alpha t^3 + Ct +D[/tex] (2)
where D is another constant term.
We know that at [tex]t_1 = 1.13 s[/tex], the position is [tex]v=5.92 m[/tex]. Substituting these values into (2), we can find the exact value of D:
[tex]D = x(t) - \frac{1}{6}\alpha t^3 -Ct = 5.92-\frac{1}{6}(1.28)(1.13)^3 - (4.27)(1.13)=0.79 m[/tex]
And so now we can find the position at time [tex]t_2 = 2.02 s[/tex] using eq.(2):
[tex]x(2.02)=\frac{1}{6}(1.28)(2.02)^3 + (4.27)(2.02) +0.79=11.2 m[/tex]
The question deals with calculating the time-dependent acceleration of a bus using the equation ax(t)=αt, applicable to Physics (specifically mechanics), and is suitable for a high school level.
Explanation:The question pertains to the acceleration of an object, specifically a bus, which makes this a Physics problem. Acceleration is defined as the change in velocity over time and is a vector quantity having both magnitude and direction. Given the equation ax(t)=αt, where α is a constant equal to 1.28 m/s3, we can recognize that the acceleration is linearly increasing over time because for each second, the acceleration increases by 1.28 m/s2. This kind of problem typically appears in high school Physics.
For instance, to calculate the acceleration of a bus at a specific time point, we simply multiply the constant α by the time t at which we want to know the acceleration. So if we wanted to know the acceleration at t=15 s, we would calculate it as ax(15s) = 1.28 m/s3 × 15 s = 19.2 m/s2. This demonstrates that as time progresses, the acceleration value increases.
A flagpole consists of a flexible, 7.14 m tall fiberglass pole planted in concrete. The bottom end of the flagpole is fixed in position, but the top end of the flagpole is free to move. What is the lowest frequency standing wave that can be formed on the flagpole if the wave propagation speed in the fiberglass is 2730 m/s?
Answer:
The lowest frequency is 95.6 Hz
Explanation:
The standing waves that can be formed in this system must meet some conditions, such as until this is fixed at the bottom here there must be a node (point without oscillation) and being free at its top at this point there should be maximum elongation (antinode)
For the lowest frequency we have a node at the bottom point and a maximum at the top point, this corresponds to ¼ of the wavelength, so the full wave has
λ = 4L
As the speed any wave is equal to the product of its frequency by the wavelength
v = f λ
f = v / λ
f = v / 4L
f = 2730 / (4 7.14)
f= 95.6 1 / s = 95.6 Hz
The lowest frequency of the standing wave that can be formed on the flagpole is 95.59 Hz.
Data obtained from the question Length (L) = 7.14 mWavelength (λ) = 4L = 4 × 7.14 = 28.56 mVelocity (v) = 2730 m/sFrequency (f) =? How to determine the frequencyThe velocity, frequency and wavelength of a wave are related according to the following equation:
Velocity (v) = wavelength (λ) × frequency (f)
v = λf
With the above formula, we can obtain the frequency as follow:
v = λf
2730 = 28.56 × f
Divide both side by 28..56
f = 2730 / 28.56
f = 95.59 Hz
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A 5kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant. The mass of the planet is unknown. After 2s, the object has fallen 30m. Air resistance is considered to be negligible. What is the gravitational force exerted on the 5kg object near the planet’s surface?
Answer:
75 N
Explanation:
t = Time taken = 2 seconds
u = Initial velocity
v = Final velocity
s = Displacement = 30 m
a = Acceleration
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 30=0\times 2+\frac{1}{2}\times a\times 2^2\\\Rightarrow a=\frac{30\times 2}{2^2}\\\Rightarrow a=15\ m/s^2[/tex]
The acceleration due to gravity on the planet is 15 m/s²
Force
F = ma
[tex]F=5\times 15\\\Rightarrow F=75\ N[/tex]
The gravitational force exerted on the object near the planet’s surface is 75 N
We have that for the Question it can be said that he gravitational force exerted on the 5kg object near the planet’s surface
F=75N
From the question we are told
A 5kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant. The mass of the planet is unknown. After 2s, the object has fallen 30m.
Air resistance is considered to be negligible. What is the gravitational force exerted on the 5kg object near the planet’s surface?
Generally the equation for the Motion is mathematically given as
[tex]s=ut+1/2at^2\\\\Therefore\\\\30=0+1/2(a)(2)^2\\\\a=15m/s^2[/tex]
Therefore
F=ma
F=5*15
F=75N
Hence, the gravitational force exerted on the 5kg object near the planet’s surface
F=75N
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Solve the following equation.
LaTeX: \left|\frac{x}{7}\right|=1| x 7 | = 1
Group of answer choices
LaTeX: x=7 x = 7
LaTeX: x=-4,\:x=-16 x = − 4 , x = − 16
LaTeX: x=7,\:x=-7 x = 7 , x = − 7
LaTeX: x=-5,\:x=-13
Answer:
x = +7, x = -7
Explanation:
The equation to solve is:
[tex]\left|\frac{x}{7}\right|=1[/tex]
Which means that the absolute value of the fraction [tex]\frac{x}{7}[/tex] must be equal to 1. Since we have an absolute value, we have basically two equations to solve:
1) [tex]\frac{x}{7}=+1[/tex]
Solving this one, we find
[tex]\frac{x}{7}\cdot 7 = 1\cdot 7 \rightarrow x = +7[/tex]
2) [tex]\frac{x}{7}=-1[/tex]
Solving this one, we find
[tex]\frac{x}{7}\cdot 7 = -1\cdot 7 \rightarrow x = -7[/tex]
So the two solutions are +7 and -7.
A man flies a small airplane from Fargo to Bismarck, North Dakota --- a distance of 180 miles. Because he is flying into a head wind, the trip takes him 2 hours. On the way back, the wind is still blowing at the same speed, so the return trip takes only 1 hours. What is the plane's speed in still air, and how fast is the wind blowing?
Answer:
airplane speed 135mph windspeed 45 mph
Explanation:
This information helps us to write down a system of linear equations
When going head wind, the speed of the wind is substracted from that of the airplane and on the return trip it is added, then:
A:=Airlplane speed
W:= Wind speed
(A+W)*1h=180mi (1)
(A-W)*2h=180mi (2)
then from (1) A=180-W (3), replacing this in (2) we get (180-W-W)*2h =180mi, then
360-4W=180, or 180=4W, then W=45 mph. Replacing this in (3) we have that A=180-45=135 mph.
The problem was solved by setting up equations based on the given distances and times. By solving these equations, it was determined that the plane's speed in still air is 105 miles per hour, and the wind's speed is 15 miles per hour.
To solve this problem, we can set up two equations using the relationship between distance, speed, and time. We will denote the plane's speed in still air as P and the wind's speed as W.
When flying against the wind, the plane's effective speed is P - W, and the time taken to cover the 180 miles is 2 hours. We can represent this with the equation:
180 = 2(P - W) ... (1)
On the return trip, with a tailwind, the plane's effective speed is P + W, and the time taken is 1.5 hours. This gives us a
second equation:
180 = 1.5(P + W) ... (2)
By solving these two equations simultaneously, we can find the values of P and W. Multiplying equation (2) by 2 to eliminate the fractions, we get:
360 = 3(P + W) ... (2')
Now, we subtract equation (1) from equation (2'):
360 - 180 = 3(P + W) - 2(P - W)
180 = 3P + 3W - 2P + 2W
180 = P + 5W
Using equation (1), we express P in terms of W:
180 = 2P - 2W
P = 90 + W ... (3)
Substituting (3) into the equation we got after subtracting:
180 = (90 + W) + 5W
180 = 90 + 6W
90 = 6W
W = 15 miles per hour
Now, we substitute the value of W back into equation (3) to find P:
P = 90 + 15
P = 105 miles per hour
The plane's speed in still air is 105 miles per hour and the wind's speed is 15 miles per hour.
A woman at an airport is towing her 20.0-kg suitcase at constant speed by pulling on a strap at an angle θ above the horizontal. She pulls on the strap with a 35.0-N force, and the friction force on the suitcase is 20.0 N. What is the magnitude of the normal force that the ground exerts on the suitcase?
Answer:
Normal force: 167.48 N
Explanation:
First of all it is necessary to draw the free body diagram of the suitcase adding alll the forces stated on the question: the normal force, the friction force and pull force exterted by the woman. Additionally, we need to add the weight, the forces exerted by Earth's gravity. I attached the diagram so you can check it.We need to resolve all the unknown quantities on this exercise, so we need to write down the sum of forces equations on X-Axis and Y-axis. Remember that force exerted by the woman has an angle with respect the horizontal (X-Axis), that is to say it has force compoents on both X and Y axis. The equations will be equal to zero since the suitcase is at constant speed (acceleration is zero).∑[tex]F_{x}[/tex]: [tex]F{x}-20 = 0[/tex]
∑[tex]F_{y}[/tex]: [tex]N -W+F_{y}=0[/tex]
Our objetive is to find the value of the normal force. It means we can solve the sum of Y-axis for N. The solution would be following:[tex]N = W - Fy[/tex]
Keep in mind Weight of the suitcase (W) is equal to the suitcase mass times the acceleration caused by gravity (9.81[tex]\frac{m}{s^{2}}[/tex]. Furthermore, Fy can be replaced using trigonometry as [tex]Fsin(\theta)[/tex] where θ is the angle above the horizontal. So the formula can be written in this way:[tex]N = mg -Fsin(\theta)[/tex]
We need to find the value of θ so we can find the value of N. We can find it out solving the sum of forces on X-axis replacing Fx for Fcos(θ). The equation will be like this:[tex]Fcos(\theta) -20 = 0[/tex] ⇒ [tex]Fcos(\theta) = 20[/tex][tex]\theta
[tex]\theta=cos^{-1}(20/F)[/tex]
Replacing the value of F we will see θ has a value of 55.15°. Now we can use this angle to find the value of N. Replacing mass, the gravity acceleration and the angle by their respective values, we will have the following:[tex]N = 20 x 9.81 - 35sin(55.15)[/tex] ⇒ [tex]N = 167.48 N[/tex]
Consider the three displacement vectors
A=(3i+3j)meters,
B-(i-4j) m
C=(-2i+5j) m
Use the Component method to determine
a) the magnitude and direction of the vector D= A+B+C
b) the magnitude And direction of E=-A-B+C
Answer:
Explanation:
[tex]\overrightarrow{A} = 3\widehat{i}+3\widehat{j}[/tex]
[tex]\overrightarrow{B} = \widehat{i}-4\widehat{j}[/tex]
[tex]\overrightarrow{C} = -2\widehat{i}+5\widehat{j}[/tex]
(a)
[tex]\overrightarrow{D} =\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}[/tex]
[tex]\overrightarrow{D} =\left ( 3+1-2 \right )\widehat{i} +\left ( 3-4+5 \right )\widehat{j}[/tex]
[tex]\overrightarrow{D} =\left 2\widehat{i} +4\widehat{j}[/tex]
Magnitude of [tex]\overrightarrow{D}[/tex] = [tex]\sqrt{2^{2}+4^{2}}[/tex]
= 4.47 m
Let θ be the direction of vector D
[tex]tan\theta =\frac{4}{2}[/tex]
θ = 63.44°
(b)
[tex]\overrightarrow{E} =
- \overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}[/tex]
[tex]\overrightarrow{E} =\left ( - 3- 1 -2 \right )\widehat{i} +\left ( - 3 + 4+5 \right )\widehat{j}[/tex]
[tex]\overrightarrow{E} =- \left 6\widehat{i} +6\widehat{j}[/tex]
Magnitude of [tex]\overrightarrow{E}[/tex] = [tex]\sqrt{6^{2}+6^{2}}[/tex]
= 8.485 m
Let θ be the direction of vector D
[tex]tan\theta =\frac{6}{-6}[/tex]
θ = 135°
a) For vector [tex]\( \mathbf{D} \)[/tex]: Magnitude: [tex]\( 4 \, \text{meters} \)[/tex] ,Direction: [tex]\( 90^\circ \)[/tex]
b) For vector [tex]\( \mathbf{E} \)[/tex]: Magnitude: [tex]\( 7.21 \, \text{meters} \)[/tex] ,Direction: [tex]\( 123.69^\circ \)[/tex]
To find the resultant vectors [tex]\( \mathbf{D} = \mathbf{A} + \mathbf{B} + \mathbf{C} \) and \( \mathbf{E} = -\mathbf{A} - \mathbf{B} + \mathbf{C} \)[/tex] using the component method, we need to break each vector into its [tex]\( i \)[/tex] and [tex]\( j \)[/tex] components, sum these components, and then find the magnitude and direction of the resulting vectors.
Given vectors:
[tex]\[ \mathbf{A} = 3\mathbf{i} + 3\mathbf{j} \][/tex]
[tex]\[ \mathbf{B} = -\mathbf{i} - 4\mathbf{j} \][/tex]
[tex]\[ \mathbf{C} = -2\mathbf{i} + 5\mathbf{j} \][/tex]
a) Vector [tex]\(\mathbf{D} = \mathbf{A} + \mathbf{B} + \mathbf{C}\)[/tex]
1. Sum the components:
[tex]\[ \mathbf{D} = (3\mathbf{i} + 3\mathbf{j}) + (-\mathbf{i} - 4\mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) \][/tex]
Combine like terms:
[tex]\[ D_i = 3 - 1 - 2 = 0 \][/tex]
[tex]\[ D_j = 3 - 4 + 5 = 4 \][/tex]
So, the components of [tex]\( \mathbf{D} \)[/tex] are:
[tex]\[ \mathbf{D} = 0\mathbf{i} + 4\mathbf{j} \][/tex]
2. Magnitude of [tex]\( \mathbf{D} \)[/tex]:
[tex]\[ |\mathbf{D}| = \sqrt{D_i^2 + D_j^2} = \sqrt{0^2 + 4^2} = 4 \, \text{meters} \][/tex]
3. Direction of [tex]\( \mathbf{D} \)[/tex]:
The angle [tex]\( \theta_D \)[/tex] from the positive x-axis is:
[tex]\[ \theta_D = \tan^{-1}\left(\frac{D_j}{D_i}\right) = \tan^{-1}\left(\frac{4}{0}\right) = 90^\circ \][/tex]
b) Vector [tex]\(\mathbf{E} = -\mathbf{A} - \mathbf{B} + \mathbf{C}\)[/tex]
1. Sum the components:
[tex]\[ \mathbf{E} = - (3\mathbf{i} + 3\mathbf{j}) - (-\mathbf{i} - 4\mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) \][/tex]
Simplify the negative signs:
[tex]\[ \mathbf{E} = (-3\mathbf{i} - 3\mathbf{j}) + (\mathbf{i} + 4\mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) \][/tex]
Combine like terms:
[tex]\[ E_i = -3 + 1 - 2 = -4 \][/tex]
[tex]\[ E_j = -3 + 4 + 5 = 6 \][/tex]
So, the components of [tex]\( \mathbf{E} \)[/tex] are:
[tex]\[ \mathbf{E} = -4\mathbf{i} + 6\mathbf{j} \][/tex]
2. Magnitude of [tex]\( \mathbf{E} \)[/tex]:
[tex]\[ |\mathbf{E}| = \sqrt{E_i^2 + E_j^2} = \sqrt{(-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} \approx 7.21 \, \text{meters} \][/tex]
3. Direction of [tex]\( \mathbf{E} \)[/tex]:
The angle [tex]\( \theta_E \)[/tex] from the positive x-axis is:
[tex]\[ \theta_E = \tan^{-1}\left(\frac{E_j}{E_i}\right) = \tan^{-1}\left(\frac{6}{-4}\right) = \tan^{-1}\left(-1.5\right) \][/tex]
Since [tex]\( E_i \)[/tex] is negative and [tex]\( E_j \)[/tex] is positive, [tex]\( \theta_E \)[/tex] is in the second quadrant:
[tex]\[ \theta_E = 180^\circ - \tan^{-1}(1.5) \approx 180^\circ - 56.31^\circ = 123.69^\circ \][/tex]