Answer:
186 lbs per man.
Explanation:
If we assume that we have 35 men in the elevator, and the elevator will be overloaded if the total weight is in excess of 6510-lbs, so we just need to take average:
6510 lbs / 35 = 186 lbs per man.
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A spherical object (with non-uniform density) of mass 16 kg and radius 0.22 m rolls along a horizontal surface at a constant linear speed without slipping. The moment of inertia of the object about a diameter is 0.59 M R2 . The object’s rotational kinetic energy about its own center is what fraction of the object’s total kinetic energy
Final answer:
The question pertains to the rotational kinetic energy of a rolling spherical object with non-uniform density and how it compares to the total kinetic energy using its moment of inertia.
Explanation:
The student is asking about the rotational kinetic energy of a spherical object with non-uniform density in comparison to its total kinetic energy. The given moment of inertia for the object is 0.59 M R2, where M is the mass, and R is the radius of the sphere. Using the formula for rotational kinetic energy, Krot = (1/2) I ω2, and the formula for translational kinetic energy, Ktrans = (1/2) m v2, we can find the energy components. However, because the spherical object rolls without slipping, there is a relationship between linear velocity (v) and angular velocity (ω) which is v = ωR. This allows us to compare the rotational kinetic energy to the total kinetic energy.
A segment of length 1 is drawn from the origin at an angle of 30∘30∘. What are the coordinates of segment’s other endpoint?
Answer:
Explanation:
Given
Length of segment is [tex]L=1 unit[/tex]
inclination of segment [tex]\theta =30^{\circ}[/tex]
To calculate the coordinates of segment we need to resolve its component in x and y co-ordinates
such that in triangle OPQ
[tex]\sin \theta =\frac{y}{L}[/tex]
[tex]y=L\sin 30[/tex]
[tex]y=1\times \frac{1}{2}=0.5 unit[/tex]
[tex]\cos \theta =\frac{x}{L}[/tex]
[tex]x=L\cos \theta [/tex]
[tex]x=1\times \cos 30[/tex]
[tex]x=1\times \frac{\sqrt{3}}{2}[/tex]
[tex]x=\frac{\sqrt{3}}{2} units[/tex]
Which of the following increases atmospheric loss by thermal escape? A.increasing the mass of the gas particles B.increasing the temperature of the atmosphere C.increasing the escape velocity of the planet D.all of the above
Answer:
B. increasing the temperature of the atmosphere
Explanation:
Generally, atmospheric loss can be defined as the loss of the gases in the atmosphere to outer space. This process usually occur through either thermal escape or non-thermal escape. Atmospheric loss of gases to outer space by thermal escape occurs when the molecular velocity due to thermal energy is considerably high. One of the factors that can lead to increase in thermal energy and ultimately increase in atmospheric loss is increase in the temperature of the atmosphere. Therefore, the correct answer is option B.
A hammer taps on the end of a 4.00 m long metal bar at room temperature. A microphone at the the other end of the bar picks up two pulses of sound, one that travels through the metal and one that travels through air. The pulse traveling through the metal arrives 9ms earlier because sounds travels faster through solids than air. What is the speed of the sound in the metal? Speed of sound through air is 343 m/s at room temperature. Give your answer in units of m/s but do not include units,
Final answer:
The speed of sound in the metal is 444.44 m/s.
Explanation:
To determine the speed of sound in the metal, we can use the information given and the speed of sound in air. The pulse traveling through the metal arrives 9ms earlier than the pulse traveling through air. From this, we can calculate the time difference it takes for the pulses to travel through the length of the metal bar. The speed of sound in air is given as 343 m/s. We can use the formula: speed = distance/time. Rearranging the formula, we have: time = distance/speed. As the distance is given as 4.00 m and the time difference is given as 9 ms (0.009 s), we can calculate the speed of sound in the metal by dividing the distance by the time difference: speed = 4.00 m / 0.009 s = 444.44 m/s.
From the list below, identify the disease and its classification that is mismatched.
A) polio; meningitis
B) African sleeping sickness; encephalitis
C) La Crosse Virus infection; encephalitis
D) Naegleria fowleri infection; meningoencephalitis
E) Valley Fever; meningitis
Answer:E)Valley Fever;meningitis
Explanation:Valley fever is a a fungal lung infection while the meningitis is a viral or bacterial disease that affect the meninges.
Two point sources, vibrating in phase, produce an interferencepattern in a ripple tank. If the frequency is increased by 20%, thenumber of nodal lines:
A) is increased by 20%
B) is increased by 40%
C) remains the same
D) is decreased by 20%
E) is decreased by 40%
Answer:
option A
Explanation:
given,
frequency is increased by 20%
we know,
[tex]\dfrac{x_n}{L}d = (n-\dfrac{1}{2})\times \lambda[/tex]...........(1)
where
x_n is the perpendicular distance between the point the interference pattern is obtained,
L is the distance between the center of the two point sources
and λ is the wavelength of light.
If the frequency is increased by 20%, then the number of nodal lines is increased by 20%.
From equation (1),we observe that the frequency is directly proportional to the number of nodal lines.
Hence, the correct answer is option A
In one cycle a heat engine absorbs 450 J from a high-temperature reservoir and expels 290 J to a low-temperature reservoir. If the efficiency of this engine is 55% of the efficiency of a Carnot engine, what is the ratio of the low temperature to the high temperature in the Carnot engine?
The ratio of the low temperature to the high temperature in the Carnot engine, given that the real engine's efficiency is 55% of a Carnot engine's efficiency, is calculated to be 0.3535.
The question involves thermodynamics and specifically deals with the operation and efficiency of a Carnot engine.
The given heat engine absorbs 450 J of heat from the high-temperature reservoir and expels 290 J to the low-temperature reservoir. The efficiency (efficiency) of this engine is given as 55% of a Carnot engine's efficiency. Using the first law of thermodynamics, wecan calculate the work done (W) by the engine:
W = Qh - Qc = 450 J - 290 J = 160 J.
The efficiency of the engine is the ratio of the work done to the heat absorbed:
efficiency = W / Qh = 160 J / 450 J = 0.3556, or 35.56%.
Now, the efficiency of a Carnot engine is defined as:
efficiencyCarnot = 1 - (Tc / Th).
The problem states that the engine's efficiency is 55% of a Carnot engine's efficiency, which means:
0.3556 = 0.55 * efficiencyCarnot
From this equation, we can solve for efficiencyCarnot and then use it to calculate the ratio of the low temperature to the high temperature in the Carnot engine:
efficiencyCarnot = 0.3556 / 0.55
efficiencyCarnot = 0.6465, or 64.65%
Thus:
0.6465 = 1 - (Tc / Th)
Tc / Th = 1 - 0.6465 = 0.3535.
Therefore, the ratio of the low temperature to the high temperature in the Carnot engine is 0.3535.
The ratio of the low temperature to the high temperature in the Carnot engine is [tex]\( \frac{1}{3} \)[/tex].
The correct ratio of the low temperature to the high temperature in the Carnot engine is given by:
[tex]\[ \frac{T_{\text{low}}}{T_{\text{high}}} = 1 - \frac{W_{\text{actual}}}{Q_{\text{in}}} \cdot \frac{1}{\eta_{\text{Carnot}}} \][/tex]
where [tex]\( W_{\text{actual}} \)[/tex] is the work done by the actual engine, [tex]\( Q_{\text{in}} \)[/tex] is the heat absorbed from the high-temperature reservoir, and [tex]\( \eta_{\text{Carnot}} \)[/tex] is the efficiency of the Carnot engine.
First, we calculate the actual efficiency of the given heat engine using the provided values:
[tex]\[ W_{\text{actual}} = Q_{\text{in}} - Q_{\text{out}} \][/tex]
[tex]\[ W_{\text{actual}} = 450 \, \text{J} - 290 \, \text{J} \][/tex]
[tex]\[ W_{\text{actual}} = 160 \, \text{J} \][/tex]
The actual efficiency [tex]\( \eta_{\text{actual}} \)[/tex] is then:
[tex]\[ \eta_{\text{actual}} = \frac{W_{\text{actual}}}{Q_{\text{in}}} \][/tex]
[tex]\[ \eta_{\text{actual}} = \frac{160 \, \text{J}}{450 \, \text{J}} \][/tex]
[tex]\[ \eta_{\text{actual}} = \frac{16}{45} \][/tex]
Given that the efficiency of the actual engine is 55% of the efficiency of a Carnot engine operating between the same two temperatures, we can write:
[tex]\[ \eta_{\text{actual}} = 0.55 \cdot \eta_{\text{Carnot}} \][/tex]
The efficiency of a Carnot engine is given by:
[tex]\[ \eta_{\text{Carnot}} = 1 - \frac{T_{\text{low}}}{T_{\text{high}}} \][/tex]
Combining the two equations, we get:
[tex]\[ \frac{16}{45} = 0.55 \left( 1 - \frac{T_{\text{low}}}{T_{\text{high}}} \right) \][/tex]
Solving for [tex]\( \frac{T_{\text{low}}}{T_{\text{high}}} \)[/tex]:
[tex]\[ \frac{16}{45} = 0.55 - 0.55 \cdot \frac{T_{\text{low}}}{T_{\text{high}}} \][/tex]
[tex]\[ 0.55 \cdot \frac{T_{\text{low}}}{T_{\text{high}}} = 0.55 - \frac{16}{45} \][/tex]
[tex]\[ \frac{T_{\text{low}}}{T_{\text{high}}} = \frac{0.55 - \frac{16}{45}}{0.55} \][/tex]
[tex]\[ \frac{T_{\text{low}}}{T_{\text{high}}} = \frac{0.55 \cdot \frac{45}{45} - \frac{16}{45}}{0.55} \][/tex]
[tex]\[ \frac{T_{\text{low}}}{T_{\text{high}}} = \frac{\frac{24.75}{45} - \frac{16}{45}}{0.55} \][/tex]
[tex]\[ \frac{T_{\text{low}}}{T_{\text{high}}} = \frac{\frac{8.75}{45}}{0.55} \][/tex]
[tex]\[ \frac{T_{\text{low}}}{T_{\text{high}}} = \frac{8.75}{45 \cdot 0.55} \][/tex]
[tex]\[ \frac{T_{\text{low}}}{T_{\text{high}}} = \frac{8.75}{24.75} \][/tex]
[tex]\[ \frac{T_{\text{low}}}{T_{\text{high}}} = \frac{1}{3} \][/tex]
Rocks on either side of a strike-slip fault move past each other without much upward or downward movement is called___________.
Answer:
strike-slip fault
Explanation:
A fault is a crack on the crust of the Earth where forces on the rocks causes the displacement of the rocks.
The movement of the crust determines the type of fault.
When rocks slip pass horizontally with respect to each other then it is called a strike slip fault. The crust moves towards each other causing horizontal compression. The displacement of the rock takes place parallel to the horizontal force.
Hence, the statement here is referring to the strike-slip fault.
One long wire lies along an x axis and carries a current of 43 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 5.9 m, 0), and carries a current of 41 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0, 1.7 m, 0)?
Answer:
Bnet=1.006*10^-6T
Explanation:
One long wire lies along an x axis and carries a current of 43 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 5.9 m, 0), and carries a current of 41 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0, 1.7 m, 0)?
the magnetic field Bnet=[tex]\sqrt{b1^2+b2^2}[/tex]
the magnetic field due this long wire is given by
B1=∨I1/[tex](2\pi *R1)[/tex]..............................1
B2=∨I2/[tex](2\pi *R2)[/tex]............................2
Bnet=[tex]\sqrt{(vI1/2*pi*R1)^2+(vI2/2*pi*R2)^2}[/tex].......................3
Bnet=v/2*pi[tex]\sqrt{(I1/R1)^2+(i2/R2)^2}[/tex]
Bnet=4*pi*10^-7/(2[tex]\pi[/tex])[tex]\sqrt{(43/1.7)^2+(41/29.5)^2}[/tex]
Bnet=0.0000002*(641.72)^.5
Bnet=1.006*10^-6T
A centripetal force of [tex]F_c[/tex] acts on a car going around a curve. If another car goes around the same curve but with twice the speed, the centripetal force acting on that car is ___________.
Answer:
The centripetal on the car will become 4 times when the velocity gets twice.
Explanation:
As we know that centripetal force on the car of mass m and moving with constant speed v given as
[tex]F_c=\dfrac{mv^2}{r}[/tex]
m=mass
v=velocity
r=radius of the circular arc
We are assuming that the mass of the both the car is same.
If the velocity of the car gets twice 2 v
The new centripetal force on the car
[tex]F_c'=\dfrac{m(2v)^2}{r}[/tex]
[tex]F_c'=4\dfrac{mv^2}{r}[/tex]
[tex]F_c'=4\ F_c[/tex]
Therefore we can say that centripetal on the car will become 4 times when the velocity gets twice.
A 0.210-kg block along a horizontal track has a speed of 1.70 m/s immediately before colliding with a light spring of force constant 4.50 N/m located at the end of the track.
(a) What is the spring's maximum compression if the track is frictionless
(b) If the track is not frictionless, would the spring's maximum compression be greater than, less than, or equal to the value obtained in part (a)?greater lessequal
Answer
given,
mass of block = 0.21 Kg
speed = 1.70 m/s
spring constant = k = 4.50 N/m
using conservation of energy
a) K.E = P.E
[tex]\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2[/tex]
[tex]\dfrac{1}{2}\times 0.21 \times 1.7^2 = \dfrac{1}{2}\times 4.5 \times x^2[/tex]
[tex]0.1348= x^2[/tex]
x = 0.367 m
b) if the track is not friction less the maximum compression will be same as the compression in the part a.
In a frictionless track, the maximum compression of the spring is 0.494 m, calculated based on the energy conservation principle. If there's friction on the track, the maximum compression will be less, because part of the block's kinetic energy is used to overcome friction, leaving less to compress the spring.
Explanation:This physics problem is rooted in concepts of energy conservation, specifically kinetic energy and potential energy. In part (a), when the block collides with the spring, the kinetic energy of the block is converted into potential energy stored in the compressed spring. The formula we need here is the conservative energy formula which states that: kinetic energy + potential energy = constant. Hence the kinetic energy before collision equals the potential energy at maximum compression.
So, 1/2 * mass * velocity^2 = 1/2 * k * compression^2. Substituting given values: 1/2 * 0.210 kg * (1.70 m/s)^2 = 1/2 * 4.50 N/m * x^2. Solving this equation yields x = 0.494 m, which is the maximum compression of the spring.
In part (b), if there is friction on the track, then some of the block's kinetic energy is used to overcome friction, which means less energy is available to compress the spring. Therefore, the friction on the track would result in a less than value for the maximum compression of the spring compared to the frictionless situation in part (a).
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PEG 400 is a liquid with a specific gravity of 1.13. What volume should be measured for the preparation?
Answer:
354 cm³
Explanation:
Step 1: calculate the density of PEG 400
[tex]Specific gravity({\gamma}) = \frac{density of liquid (\rho _{l})}{density of water (\rho_{w})}[/tex]
[tex]1.13=\frac{\rho _{l}}{1(\frac{g}{m^3})}[/tex]
[tex]{\rho _{l}[/tex] = 1.13X1 = 1.13 [tex]\frac{g}{cm^3}[/tex]
Step 2:
Molecular weight of PEG 400 = 400g/mol
Step 3: calculate the volume of PEG 400 to be measured
[tex]Volume _{PEG 400} =\frac{mass of _{PEG 400}}{density of _{PEG 400}}[/tex]
[tex]Volume _{PEG 400}=\frac{400}{1.13}(cm^3)[/tex]
= 353.98 cm³
≅354 cm³
353.4 cm³ volume should be measured for the preparation.
Given:
Specific gravity=1.13
To find:
Volume=?
Specific gravity is the ratio of the density (mass of a unit volume) of a substance to the density of a given reference material, often a liquid.
[tex]\text{Specific gravity}=\frac{\text{density of liquid}}{\text{density of water}}[/tex]On substituting the values, we will get:
[tex]\text{Density of liquid}=1.13*1=1.13 gcm^{-3}[/tex]
Molecular weight of PEG 400 =400g/mol
Now, we have the values of density and mass thus volume can be calculated easily:
The density of a substance is its mass per unit volume.
[tex]\text{Volume}=\frac{\text{mass}}{\text{density}}=\frac{400}{1.13} \\\\\text{Volume}=353.4 cm^{3}[/tex]
353.4 cm³ volume should be measured for the preparation.
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A client experiences difficulty in performing the prone iso-abs exercise. Which of the following should be suggested as an regression?
a. Quadruped arm and opposite leg raise
b. Cable rotation
c. Rolling active resistance row
d. Cable chop
Answer:
a. Quadruped arm and opposite leg raise
Explanation:
Quadruped arm and opposite leg lift
- Kneel on the floor, lean forward and place your hands down.
- Keep your knees in line with your hips and hands directly under your shoulders.
- Simultaneously raise one arm and extend the opposite leg, so that they are in line with the spine.
- Go back to the starting position.
This method is usually used as an alternative to iso-abs exercise or also known as a bridge, which allows you to exercise the abdominal and spinal area at the same time.
It is also used together with other exercises for the treatment of hyperlordosis.
The suitable regression for a prone iso-abs exercise is the Quadruped arm and opposite leg raise.
The question asks for a regressed exercise alternative to the prone iso-abs exercise, which typically involves the individual maintaining a prone plank position, activating their abdominal core muscles without additional movement. When a client experiences difficulty with this exercise, an appropriate regression should reduce the demand on the core muscles while still enabling them to engage effectively.
The correct regression exercise from the options provided is a. Quadruped arm and opposite leg raise. This exercise involves the individual starting on their hands and knees (a quadruped position), then extending one arm and the opposite leg to create a straight line from fingertips to toes. This movement stabilises the core and has a reduced intensity compared to the prone iso-abs exercise.
The other options, b. Cable rotation, c. Rolling active resistance row, and d. Cable chop, are more dynamic exercises involving rotation and should not be confused with static core stabilisation exercises.
Calculate the moment of inertia of a skater given the following information. (a) The 60.0-kg skater is approximated as a cylinder that has a 0.110-m radius. (b) The skater with arms extended is approximated by a cylinder that is 52.5 kg, has a 0.110-m radius, and has two 0.900-m-long arms which are 3.75 kg each and extend straight out from the cylinder like rods rotated about their ends.
Answer:
a) I = 0.363 kg*[tex]m^{2}[/tex]
b) [tex]I_{T}[/tex] = 0.82385 kg*[tex]m^{2}[/tex]
Explanation:
a) If we approximate the skater how a cylinder his moment of inertia is:
I = [tex]\frac{mr^{2} }{2}[/tex]
I = [tex]\frac{(60)(0.110)^{2} }{2}[/tex]
I = 0.363 kg*[tex]m^{2}[/tex]
b) If the skater has his arms extended then:
[tex]I_{T} = I_{B} + I_{A}[/tex]
where [tex]I_{B}[/tex]: Body’s moment of inertia
[tex]I_{A}[/tex]: Moment of inertia of the arms
[tex]I_{B}[/tex] = [tex]\frac{mr^{2} }{2}[/tex]
[tex]I_{B}[/tex] = [tex]\frac{(52.5)(0.110)^{2} }{2}[/tex] = 0.3176 kg*[tex]m^{2}[/tex]
[tex]I_{A}[/tex] = 2 * [tex]\frac{mL^{2} }{12}[/tex]
[tex]I_{A}[/tex] = 2 * [tex]\frac{(3.75)(0.9)^{2} }{12}[/tex] = 0.50625 kg*[tex]m^{2}[/tex]
[tex]I_{T}[/tex] = 0.3176 + 0.50625 = 0.82385 kg*[tex]m^{2}[/tex]
The moment of inertia for a skater can be calculated using the mass and radius in the formula for the moment of inertia of a cylinder. If the skater extends their arms, the moment of inertia increases, and this is calculated by adding the moment of inertia of the skater to the moments of inertia of the arms.
Explanation:The moment of inertia of a body is a measure of its resistance to rotational motion. It depends on the mass and how that mass is distributed relative to the axis of rotation. In the case of the skater (approximated as a cylinder), you calculate this using the formula for the moment of inertia of a cylinder, which is I=0.5MR², where M is the mass and R is the radius.
(a) For the skater without extended arms, substitute the given mass (60 kg) and radius (0.11 m) into the formula to get: I = 0.5 * 60 kg * (0.11 m)² = 0.363 kg-m².
(b) For the skater with extended arms, first, calculate the skater's body moment of inertia as before, but now with 52.5 kg mass. Secondly, calculate the moment of inertia of each arm (approximated as a rod rotating about one end) with the formula I=1/3mL², where m is the mass of an arm and L is the length. Add those values to get the total moment of inertia.
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American eels (Anguilla rostrata) are freshwater fish with long, slender bodies that we can treat as a uniform cylinder 1.0 m long and 10 cm in diameter. An eel compensates for its small jaw and teeth by holding onto prey with its mouth and then rapidly spinning its body around its long axis to tear off a piece of esh. Eels have been recorded to spin at up to 14 rev/s when feeding in this way. Although this feeding method is energetically costly, it allows the eel to feed on larger prey than it otherwise could.An eel researcher uses the slow-motion feature on her phone's camera to shoot a video of an eel spinning at its maximum rate. The camera records at 120 frames per second. Through what angle does the eel rotate from one frame to the next?A. 1° B. 10° C. 22° D. 42°
Answer:
option D
Explanation:
given,
uniform length of cylinder = 1 m
diameter of the cylinder = 10 cm = 0.1 m
Eels have been recorded to spin = 14 rev/s
camera records at = 120 frames per second
time = [tex]\dfrac{1}{120}\ s/frame[/tex]
angle at which eel rotate = ?
ω = 14 rev/s
ω = 14 x 2 π rad/s
ω = 28 π rad/s
angle at which eel rotate
θ = ω t
θ = [tex]28\pi\times \dfrac{1}{120}[/tex]
θ = 0.733 rad
θ =[tex]0.733 \times \dfrac{180^0}{2\pi}[/tex]
θ =[tex]42^0[/tex]
Hence, the correct answer is option D
The angle of rotation is the angle at which the object is rotate about the fixed point.
The angle at which the eel rotates from one frame to next frame is 42 degree.
Given that, we can treat the eel as a uniform cylinder.
So, uniform length of cylinder = 1 [tex]\rm m[/tex] and diameter of the cylinder = 10 [tex]\rm cm[/tex] = 0.1 [tex]\rm m[/tex]. Eels have been recorded to spin at up to 14 rev/s when feeding in this way. The camera records at 120 frames per second.
Time taken by the camera to record one frame is [tex]t\;=\; \dfrac {1} {120} \;\rm s/\rm frame[/tex]
Revolution done by eel in radian per second is 14.
Angular Velocity is [tex]\omega\;=\; 14\;\times\;2\pi\;\times\;0.1\;\rm rad/s[/tex].
The angle of rotation can be calculated by the formula given below.
Angle of rotation [tex]\theta=\omega\;\times\;t[/tex].
Substituting the values in the above formula, the angle of rotation is,
[tex]\theta\;=\;14\;\times\;2\;\times\;3.14\;\times\;0.1\;\times\;\dfrac {1}{120}\;\rm rad[/tex]
[tex]\theta=0.0733\;\rm rad[/tex]
[tex]\theta=0.0733\;\times\;\dfrac {360^\circ}{2\pi}[/tex]
[tex]\theta=41.5^\circ[/tex] or [tex]42^\circ[/tex].
The angle at which the eel rotates from one frame to next frame is 42 degree.
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Energy degradation takes place in the energy transformations which occur in the generation of electrical power. Explain what is meant in this context by energy degradation.
Answer:
The degradation of energy is the loss of useful energy: energy is conserved in changes, but tends to be transformed into thermal energy, which is a less usable form of energy.
Explanation:
Some forms of energy can be transformed into others. In these transformations, energy degrades, loses quality. In any transformation, part of the energy is converted into heat or heat energy.
Any type of energy can be completely transformed into heat; but, this cannot be completely transformed into another type of energy. It is said, then, that heat is a degraded form of energy. Are examples:
- The electrical energy, when going through a resistance.
- Chemical energy, in the combustion of some substances.
- Mechanical energy, by shock or friction.
Therefore, the Yield is defined as the ratio (in% percent) between the useful energy obtained and the energy contributed in a transformation.
R = ( useful energy / total energy) * 100
where R is the yield
Final answer:
Energy degradation during the generation of electrical power reflects the decrease in usability of energy as it is transformed, with much of it becoming waste heat. This affects the efficiency of power systems, indicating that more energy must be input to produce desired effects.
Explanation:
In the context of generating electrical power, energy degradation refers to the decrease in the quality or usefulness of energy as it undergoes various transformations. Although energy is conserved according to the first law of thermodynamics, during transformations, some energy is inevitably converted to forms that are less useful for doing work, primarily heat energy.
The phenomenon of energy degradation is a fundamental concept in thermodynamics and has significant implications for the efficiency of power generation systems. It points to the reality that while energy is not destroyed, its capacity to do work diminishes, requiring more energy inputs to achieve the desired outputs in practical applications.
In a Young’s double-slit experiment, a thin sheet of mica is placed over one of the two slits. As a result, the center of the fringe pattern (on the screen) shifts by an amount corresponding to 30 dark bands. The wavelength of the light in this experiment is 480 nm and the index of the mica is 1.60. The mica thickness is:
Answer:0.024 mm
Explanation:
Given
Fringe shifts by an amount to 30 dark bands i.e. [tex]m=30[/tex]
Wavelength [tex]\lambda =480 nm[/tex]
refractive index of mica [tex]n_2=1.6[/tex]
refractive index of air [tex]n_1=1[/tex]
Phase difference is given by
[tex]m=\frac{t\left [ n_2-n_1\right ]}{\lambda }[/tex]
[tex]30=\frac{t\left [ 1.6-1\right ]}{480\times 10^{-9}}[/tex]
[tex]t=\frac{30\times 480\times 10^{-9}}{1.6-1}[/tex]
[tex]t=0.024\ mm[/tex]
To determine the thickness of the mica in a Young's double-slit experiment, we can use the formula d = mλ / sinθ, where d is the thickness of the mica, λ is the wavelength of the light, θ is the angle of the shift in the fringe pattern, and m is the number of dark bands shifted. By substituting the given values into the formula and solving for d, we can calculate the thickness of the mica.
Explanation:In a Young’s double-slit experiment, the shift in the center of the fringe pattern on the screen is caused by the introduction of a thin sheet of mica over one of the slits. This shift corresponds to 30 dark bands. To determine the thickness of the mica, we can use the formula for double-slit interference:
d sinθ = mλ,
where d is the distance between the slits, θ is the angle of the shift in the fringe pattern, m is the number of dark bands shifted, and λ is the wavelength of the light. Rearranging the formula, we have:
d = mλ / sinθ.
Given that the wavelength of the light is 480 nm and the index of refraction for the mica is 1.60, we can calculate the thickness of the mica by substituting the values into the formula and solving for d.
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The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pres- sure. The higher atmospheric pressure in the still air inside the buildings can cause windows to pop out. As originally constructed, the John Hancock Building in Boston popped windowpanes that fell many stories to the sidewalk below. (a) Suppose a horizontal wind blows with a speed of 11.2 m/s outside a large pane of plate glass with dimensions 4.00 m 3 1.50 m. Assume the density of the air to be constant at 1.20 kg/m3. The air inside the building is at atmospheric pressure. What is the total force exerted by air on the win- dowpane?
Answer:
9483.26399 N
Explanation:
[tex]\rho[/tex] = Density of air = 1.2 kg/m³
v = Velocity of wind = 11.2 m/s
A = Area = [tex]4\times 31.5\ m^2[/tex]
The force on the pane is
[tex]F=\dfrac{1}{2}\rho v^2A\\\Rightarrow F=\dfrac{1}{2}\times 1.2\times 11.2^2\times 4\times 31.5\\\Rightarrow F=9483.26399\ N[/tex]
The force on the pane of glass is 9483.26399 N
What is the distance from the moon to the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal? The mass of Earth is 5.97 x 10^24 kg, the mass of the Moon is 7.35 x 10^22 kg, the distance between Earth and the Moon is 3.84 x 10^8 m, and >G= 6.67x10^-11N x m^2/kg^2
A)3.83 x 10^6 mB)3.83 x 10^7 mC)4.69 x 10^6 mD)4.69 x 10^7 mE)3.45 x 10^8 m
Answer:
the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal is E)3.45 × 10⁸ m
Explanation:
The force that the Earth exerts on a mass m is
F_e = (G M_e m) / R_e²
where
G is the universal gravitational constantM_e is the mass of EarthR_e is the radius of EarthThe force that the Moon exerts on a mass m is
F_m = (G M_m m) / R_m²
where
G is the universal gravitational constantM_m is the mass of the MoonR_m is the radius of the MoonTherefore, the point where the gravitational pulls of Earth and Moon are equal is:
F_e = F_m
R_e + R_m = R = 3.84×10⁸ m
Thus,
(G M_e m) / R_e² = (G M_m m) / R_m²
M_e / R_e² = M_m / (R - R_e²)
(R - R_e²) / R_e² = M_m / M_e
(R - R_e) / R_e = (M_m / M_e)^1/2
R_e(R/R_e -1) / R_e = (M_m / M_e)^1/2
R/ R_e = (M_m / M_e)^1/2 + 1
R_e = R / [(M_m / M_e)^1/2 + 1]
R_e = (3.84×10⁸ m) / [(7.35 x 10²² kg / 5.97 x 10²⁴ kg )^1/2 + 1]
R_e = 3.45 × 10⁸ m
Therefore, the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal is 3.45 × 10⁸ m.
Which of the following statements is true?
1. The 2s orbital in the hydrogen atom is larger than the 3s orbital also in the hydrogen atom.
2. The hydrogen atom has quantized energy levels.
3. The Bohr model of the hydrogen atom has been found to be incorrect.
4. An orbital is the same as a Bohr orbit.
5. The third energy level has three sublevels - the s, p, and d sublevels.
Answer:
2. The hydrogen atom has quantized energy levels.
Explanation:
The Bohr model of the atom states that the structure of the atom is quantized, that is, that electrons can only orbit the nucleus in specific orbits with a fixed radius. Therefore, the electron cannot be in energy levels that do not correspond to these quantized levels.
A person is straining to lift a large crate, without success because it is too heavy. We denote the forces on the crate as follows: P is the upward force the person exerts on the crate, C is the vertical contact force exerted on the crate by the floor, and W is the weight of the crate. How are the magnitudes of these forces related while the person is trying unsuccessfully to lift the crate?
Answer:
P + C = W if in equilibrium?
P + C - W = 0; P + C = W
Explanation:
Action and reaction are equal and opposite according to newton's third law of motion. a force is that which tends to change a body's state of rest or uniform motion in a straight line, recall
P is the upward force the person exerts on the crate,
C is the vertical contact force exerted on the crate by the floor,
W is the weight of the crate
let the sum of upward forces be on the left hand side of the equation and the sum of downward forces on the right hand side.
P+C=W
P+C-W=0 if they are in equilibrium
the forces are related by the above.
When a person is unsuccessfully attempting to lift a crate, the forces acting on the crate are balanced. The sum of the upward force (P) applied by the person and the contact force (C) from the floor equals the weight (W) of the crate, represented by the equation P + C = W.
Explanation:In the situation described, the person is trying to lift a crate but is unable to do so. This means that the forces acting on the crate are in equilibrium, or balanced, as there is no resulting upward or downward motion of the crate. In this case, the force exerted by the person (P) and the vertical contact force from the floor (C) is opposing the weight of the crate (W).
In terms of magnitudes, the sum of P and C equals the weight W. This is based on the principle of equilibrium which states that an object at rest has equal and opposite forces acting on it. Therefore, P + C = W. This means the upward forces are equal to the downward force, hence the person can't lift the crate successfully.
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A satellite with an orbital period of exactly 24.0 h is always positioned over the same spot on Earth.This is known as a geosynchronous orbit. Television, communication, and weather satellites use geosynchronous orbits. At what distance would a satellite have to orbit Earth in order to have a geosynchronous orbit?
Answer:
35870474.30504 m
Explanation:
r = Distance from the surface
T = Time period = 24 h
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
m = Mass of the Earth = 5.98 × 10²⁴ kg
Radius of Earth = [tex]6.38\times 10^6\ m[/tex]
From Kepler's law we have relation
[tex]T^2=\dfrac{4\pi^2r^3}{GM}\\\Rightarrow r^3=\dfrac{T^2GM}{4\pi^2}\\\Rightarrow r=\left(\dfrac{(24\times 3600)^2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{4\pi^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=42250474.30504\ m[/tex]
Distance from the center of the Earth would be
[tex]42250474.30504-6.38\times 10^6=\mathbf{35870474.30504\ m}[/tex]
35870474.30504 m
A satellite in geosynchronous orbit remains over the same point on Earth, requiring an altitude of approximately 35,793 kilometers. This is calculated using the orbital mechanics involving the gravitational constant and Earth's mass and radius.
A geosynchronous orbit means that the satellite has an orbital period of [tex]24[/tex] hours, remaining over the same point on Earth as it rotates. To find the distance at which a satellite must orbit to achieve this, we use the universal law of gravitation and centripetal force.
We know:
Orbital period [tex]T = 86400 \text{ seconds}[/tex]
Gravitational constant [tex]G = 6.67430 \times 10^{-11} \, \text{N} \left( \text{m/kg} \right)^2[/tex]
Mass of Earth [tex]M = 5.972 \times 10^{24} \, \text{kg}[/tex]
Radius of Earth [tex]R = 6.371 \times 10^{6} \, \text{meters}[/tex]
The formula for the geostationary orbit (orbital radius r) is given by:
[tex]r^3 = \frac{GMT^2}{4\pi^2}[/tex]
Substituting the values, we get:
[tex]GM T^2 = 6.67430 \times 10^{-11} \times 5.972 \times 10^{24} \times (86400)^2\\\\GM T^2 = 6.67430 \times 10^{-11} \times 5.972 \times 10^{24} \times 7464960000\\\\GM T^2 = 2.963 \times 10^{14}\\\\r^3 \approx \frac{2.963 \times 10^{14}}{4 \times 9.8696}\\\\r^3 \approx 7.507 \times 10^{12} \, \text{m}^3\\\\r \approx \sqrt[3]{7.507 \times 10^{12}}\\\\r \approx 42164 \, \text{km}\\\\[/tex]
To find the altitude (h) above Earth’s surface:
[tex]h = r - \text{Earth's radius} = 42164 \, \text{km} - 6371 \, \text{km} \approx 35793 \, \text{km}[/tex]
Thus, a satellite needs to orbit at an altitude of approximately [tex]35,793 km[/tex] to maintain a geosynchronous orbit.
Marcie wants to work as a freelance editor. She purchases a desk, computer, widescreen monitor, and ergonomic keyboard. What kind of resources are these?
Answer:Capital Resources
Explanation:
Desk, computer, widescreen monitor, and ergonomic keyboard are an example of capital resources.
Capital resources are goods produced and used for the production of other goods and services. Basic items in capital goods are tools, machinery, and building. However, any good being utilized by a business to produce other good and services are under the category of capital goods.
Here Desk, computer and other items help Marcie to design her artwork as freelancer.
The Royal Gorge bridge over the Arkansas River is 393 m above the river. A bungee jumper of mass 150 kg has an elastic cord of length 78 m attached to her feet. Assume the cord acts like a spring of force constant k. The jumper leaps, barely touches the water, and after numerous ups and downs comes to rest at a height h above the water. The acceleration of gravity is 9.81 m/s². Find h. Answer in units of m.
Answer:
188.7 m
Explanation:
height of bridge above water (h) = 393 m
mass of bungee jumper (m) = 150 kg
length of cord (L) = 78 m
acceleration due to gravity (g) = 9.8 m/s
initial energy = mgh = 150 x 9.8 x 393 = 577,710 J
since the jumper barely touches the water, the maximum extension of the cord (x) = 393 - 78 = 315 m
from the conservation of energy mgh = [tex](\frac{1}{2})kx^{2}[/tex]
therefore
577,710 = [tex](\frac{1}{2})kx315^{2}[/tex]
k = 11.64 N/m
from Hooke's law, force (f) = kx' ⇒ mg = kx'
where x' is the extension of the cord when it comes to rest
150 x 9.8 = 11.64 × x'
x' = 126.3 m
the final height at which the cord comes to a rest = height of the bridge - length of the cord - extension of the cord when it comes to rest
the final height at which the cord comes to a rest = 393 - 78 - 126.3 = 188.7 m
A spin bike has a flywheel in two parts—a 12.5 kg disk with radius 0.23 m, and a 7.0 kg ring with mass concentrated at the outer edge of the disk. A friction pad exerts a force of 9.7 N on the outside of the disk. A cyclist is pedaling, spinning the disk at a typical 180 rpm. If she stops pedaling, how long will it take for the flywheel to come to a stop?
Final answer:
To determine how long it takes for the flywheel to come to a stop, calculate the angular acceleration using the torque and moment of inertia. Then, use the angular acceleration to find the time.
Explanation:
To determine how long it takes for the flywheel to come to a stop, we need to calculate the angular acceleration first. To do this, we can use the formula:
α = τ / I
Where α is the angular acceleration, τ is the torque, and I is the moment of inertia. The torque can be calculated using the formula:
τ = rF
Where r is the radius of the flywheel and F is the force exerted on it. Once we have the angular acceleration, we can use it to find the time it takes for the flywheel to come to a stop using the formula:
t = ω / α
Where t is the time, ω is the initial angular velocity, and α is the angular acceleration.
A green object will absorb ____________________ light and reflect ____________________ light.
Answer:Magenta , Green
Explanation:
A green object will absorb its complementary color i.e. Magenta (mixture of red and blue) and thus reflect the remaining green color.
This can be understood by the fact that a particular light filter absorb its complementary color and reflects its color for example
Blue filter will absorb yellow light (mixture of red and green)and reflects blue color.
Final answer:
A green object absorbs light in the red portion of the visible spectrum and reflects green light, as the observed color is complementary to the color that is absorbed.
Explanation:
The question is asking which colors of light a green object will absorb and which it will reflect. A green object will absorb light in the red portion of the visible spectrum. This is due to the fact that the color observed from an object comes from the light that is transmitted or reflected, not the light that is absorbed. The reflected or transmitted light is always complementary in color to the light that is absorbed. Thus, for a green object, the absorbed light is predominantly red while it reflects green light.
An example that helps explain this concept is how different complexes absorb light depending on their ligand field strength and the structure of the complex, which influences the observable color. For instance, the complex [Cr(NH3)6]3+ has strong-field ligands which cause it to absorb high-energy photons corresponding to blue-violet light, making the complex appear yellow, which is the complementary color to blue-violet.
Assume that the equator of the Earth is 24,000 miles long (that’s its circumference). Now, pretend that you are standing somewhere on the equator, such as in the country of Ecuador. Now, if the Earth turns once, completely, in 24 hours, then, how fast would you be going (in miles per hour), even if you just stood still?
Answer:
1000 mph
Explanation:
P = Perimeter = 24000 mi
r = Radius of the equator
t = Time taken to complete one rotation = 24 h
Perimeter of a circle is given by
[tex]P=2\pi r\\\Rightarrow r=\dfrac{P}{2\pi}\\\Rightarrow r=\dfrac{24000}{2\pi}\ mi[/tex]
Angular speed is given by
[tex]\omega=\dfrac{2\pi}{t}\\\Rightarrow \omega=\dfrac{2\pi}{24}[/tex]
Velocity if given by
[tex]v=r\omega\\\Rightarrow v=\dfrac{24000}{2\pi}\times \dfrac{2\pi}{24}\\\Rightarrow v=1000\ mph[/tex]
The person would be going at a speed of 1000 mph
How did the theory of plate tectonics come to be a theory?
Explanation:
Plate tectonics is the idea that the earth's outer solid crust (the lithosphere) is fragmented into a few dozen "plates" that pass relative to each other across the earth's surface, like ice slabs on a lake.
A scientist named Alfred Wegener in 1915 proposed that the continents rammed through ocean basin crust, which would clarify why the shapes of many coastlines (such as South America and Africa) seem to match together like a puzzle.
The radius of a cone is increasing at a rate of 333 centimeters per second and the height of the cone is decreasing at a rate of 444 centimeters per second. At a certain instant, the radius is 888 centimeters and the height is 101010 centimeters. What is the rate of change of the volume of the cone at that instant (in cubic centimeters per second)?
The rate of change of the volume of the cone at the specified instant is approximately [tex]\(6.03 \times 10^{9} \pi\)[/tex] cubic centimeters per second.
What is the rate of change of the volume of the cone at that instant?To find the rate of change of the volume of the cone, you can use the formula for the volume of a cone:
[tex]\[V = \frac{1}{3}\pi r^2 h\][/tex]
Where:
- V is the volume of the cone.
- r is the radius of the cone.
- h is the height of the cone.
You are given that the radius is increasing at a rate of 333 cm/s (dr/dt = 333 cm/s) and the height is decreasing at a rate of 444 cm/s (dh/dt = -444 cm/s). At a certain instant, the radius is 888 cm (r = 888 cm) and the height is 101010 cm (h = 101010 cm).
To find the rate of change of the volume (dV/dt) at that instant, you can use the product rule for differentiation:
[tex]\[ \frac{dV}{dt} = \frac{1}{3} \pi \left(2rh\frac{dr}{dt} + r^2\frac{dh}{dt}\right) \][/tex]
Now, plug in the values:
[tex]\[ \frac{dV}{dt} = \frac{1}{3} \pi \left(2(888 cm)(101010 cm)(333 cm/s) + (888 cm)^2(-444 cm/s)\right) \][/tex]
Calculate:
[tex]\[ \frac{dV}{dt} = \frac{1}{3} \pi \left(5910088880 cm^3/s + (-157593984 cm^3/s)\right) \][/tex]
Now, simplify:
[tex]\[ \frac{dV}{dt} = \frac{1}{3} \pi \cdot 5752494896 cm^3/s \][/tex]
[tex]\[ \frac{dV}{dt} \approx 6.03 \times 10^{9} \pi \; cm^3/s \][/tex]
So, the rate of change of the volume of the cone at that instant is approximately [tex]\(6.03 \times 10^{9} \pi\)[/tex] cubic centimeters per second.
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A baseball player throws a baseball with a velocity of 13 m/s North it is caught by a second player seven seconds later how far is the second player from the first player
The second player is 91 meters far from the first player.
Why?
First, let be the +y the North, so, to solve the problem we can use the following formula:
[tex]y=yo+v_o*t+\frac{1}{2}*a*t[/tex]
Now, subsituting the given information, we have(assuming that the speed is constant):
\\\\y-yo=13\frac{m}{s}*7s+\frac{1}{2}*0*7s\\\\y-yo=91m\\\\distance=91m[/tex]
Hence, we have that the second player is 91m far from the first player.
Have a nice day!