Thus, the placement of the two converging lenses has an impact on the focal length and, as a result, the location of the image.
When two converging lenses are placed near each other, two distinct points can be obtained where a clear image can be formed, rather than just one. This happens due to the fact that light rays passing through a converging lens bend inwards towards the principal axis and then meet at a specific point, which is known as the focal point. As a result of this property, the point where the light rays converge and the image is formed can vary based on the position of the object being viewed. Since the two lenses were placed near each other in the experiment, it's possible that light rays passing through the first lens were refracted inwards and then converged at a certain point, while light rays passing through the second lens were refracted inwards and converged at a slightly different point. Because of the varying positions of the two lenses, the image is formed at two distinct points instead.
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Consider a light rod of negligible mass and length L = 2.3 m pivoted on a frictionless horizontal bearing at a point O . Attached to the end of the rod is a mass M1 = 6 kg. Also, a second mass M2 = 6 kg of equal size is attached to the rod (3/5 L from the lower end), as shown in the figure below. The acceleration of gravity is 9.8 m/s2. What is the period of this pendulum in the small angle approximation? Answer in units of s.
The period of the pendulum, considering the small angle approximation, is approximately 2.45 seconds (s). This is calculated using the formula T = 2π√(L/g), where L is the effective length of the pendulum and g is the acceleration due to gravity.
Determine how to find the period?To calculate the period, we can use the formula for the period of a simple pendulum, which is given by T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
In this case, the length of the pendulum is 2.3 m. However, we need to consider the effective length of the pendulum due to the position of mass M2. The distance of M2 from the pivot point is (3/5)L = (3/5)(2.3) = 1.38 m.
Therefore, the effective length of the pendulum is L - (1.38) = 0.92 m.
Substituting the values into the formula, we have T = 2π√(0.92/9.8) ≈ 2.45 s.
Thus, the period of this pendulum in the small angle approximation is approximately 2.45 seconds.
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Can you somebody answer this question for me please?
Answer:
The answer is B - the bending of rock layers happens due to stress, and this process is called folding. Faults are when it looks broken/displaced
does the ladybug’s distance from the center of the platform affect the angular velocity? how can you tell?
The distance of the ladybug from the center of the platform does affect the angular velocity, and this can be determined by observing the rotational motion of the ladybug.
Angular velocity is the rate at which an object rotates around a specific axis. In the case of the ladybug on a platform, the distance from the center of the platform will indeed impact the angular velocity.
When the ladybug is closer to the center, it has a smaller radius and therefore a smaller distance to travel in a given time, resulting in a higher angular velocity. Conversely, when the ladybug is farther from the center, it has a larger radius and a greater distance to travel, leading to a lower angular velocity.
To determine the effect of the ladybug's distance on the angular velocity, one can observe the rotational motion of the ladybug. By placing the ladybug at different distances from the center of the platform and measuring the time it takes to complete a full revolution, it becomes evident that the angular velocity varies based on the ladybug's distance.
A shorter time to complete a revolution indicates a higher angular velocity, while a longer time indicates a lower angular velocity. This demonstrates the relationship between the ladybug's distance from the center and its angular velocity.
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Metal sphere A has a charge of -2 units and an identical sphere B has a charge of -4 units. If the spheres are brought into contact with each other and then separted the charge on sphere b will be
Answer:
q1 = q₂= -3
therefore each sphere has the same charge of -3 untis
Explanation:
The metallic spheres have mobile charge, so when the two spheres come into contact the total charge
Q_total = q₁ + q₂
Q_total = -2 -4
Q_total = -6 units
it is distributed in between the two spheres evenly since the charges of the same sign repel each other.
When the spheres separate each one has
q₁ = -6/2
q1 = q₂= -3
therefore each sphere has the same charge of -3 untis
How does creativity affect scientific work?
Answer & Explanation:
In science, rationality and creativity work together. Creativity allows us to view and solve problems with innovation and openness. Scientific theories often came from sparks of creative thinking and bold yet logical processes.
estimate the temperature change (in centigrade) to go from room temperature to water hot enough for a hot shower.50l2.521j0j7
To estimate the temperature change from room temperature to water hot enough for a hot shower, we need more information such as the initial room temperature and the desired temperature of the hot water.
Assuming a typical room temperature of around 20°C and a desired hot water temperature for a shower of around 40-45°C, we can estimate the temperature change as follows: Temperature change = Desired hot water temperature - Initial room temperature. Let's assume the desired hot water temperature is 45°C: Temperature change = 45°C - 20°C = 25°C. Therefore, the estimated temperature change to go from room temperature to hot water for a shower would be approximately 25°C.
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the internal loadings at a section of the beam in (figure 1) are shown.
The reactions at the supports of the beam are [tex]R_1[/tex] = 1150 N and [tex]R_2[/tex] = 1150 N.
Let's denote the reactions at the supports as [tex]R_1[/tex]and [tex]R_2[/tex].
Since the beam is supported at both ends, we can assume that it is a simply supported beam.
The total downward load consists of the distributed load and the concentrated load.
The downward load due to the distributed load can be calculated by integrating the load intensity over the length:
Downward load due to distributed load = (500 N/m) * (3 m) = 1500 N
The total downward load due to the concentrated load is 800 N.
Now, let's consider the equilibrium equation:
[tex]R_1 + R_2[/tex] = 1500 N + 800 N
[tex]R_1 + R_2[/tex] = 2300 N
Since the beam is simply supported, we can assume that the reactions at the supports are equal. Therefore:
[tex]R_1 = R_2[/tex]= 2300 N / 2
[tex]R_1 = R_2[/tex]= 1150 N
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--The complete Question is, The internal loadings at a section of the beam are given as follows: a uniformly distributed load of 500 N/m over a length of 3 meters and a concentrated load of 800 N applied at the midpoint of the same section. Determine the reactions at the supports of the beam.--
what minimum horsepower must a motor have to be able to drag a 370-kg box along a level floor at a speed of 1.20 m/s if the coefficient of friction is 0.45?
The minimum horsepower required to drag the 370-kg box at a speed of 1.20 m/s is the calculated value from the equation above.
To determine the minimum horsepower required, we need to calculate the force needed to overcome friction and move the box at the given speed.
The force required to overcome friction can be calculated using the equation:
F_friction = coefficient of friction * normal force
The normal force can be calculated as the weight of the box:
normal force = mass * gravitational acceleration
Substituting the given values:
normal force = 370 kg * 9.8 m/s^2
Next, we can calculate the force required to maintain a constant speed:
F = mass * acceleration
Since the box is moving at a constant speed, the acceleration is zero. Therefore, the force required to maintain the speed is zero.
The minimum force required is the force to overcome friction, so:
F_required = F_friction
Substituting the values:
F_required = 0.45 * (370 kg * 9.8 m/s^2)
Now, we need to convert this force to horsepower. One horsepower is equal to 745.7 watts. Therefore, we can calculate the minimum horsepower required:
Horsepower = F_required * (1 watt / 745.7) * (1 horsepower / 1 watt)
Finally, substituting the values and calculating:
Horsepower = (0.45 * (370 kg * 9.8 m/s^2)) / 745.7
Hence, the minimum horsepower required to drag the 370-kg box at a speed of 1.20 m/s is the calculated value from the equation above.
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If Earth's mass decreased to one half its original mass, with no change in radius, then your weight would *
1 point
A decrease to one half your original weight
B increase two times.
C stay the same
D decrease to one quarter your original weight
Centripetal acceleration is caused by *
1 point
A the radius of an object’s circular motion.
B constant change in direction.
C a change in object’s tangential speed.
D a change in object’s linear velocity.
If Earth's mass decreased to one half its original mass, with no change in radius, then your weight would decrease to one half your original weight. Hence, option (A) is correct.
Centripetal acceleration is caused by constant change in direction.
What is centripetal acceleration?An attribute of an object moving in a circular route is centripetal acceleration. Any object moving in a circle with an acceleration vector pointing in the direction of the circle's center is said to be experiencing centripetal acceleration.
You must have come across a lot of centripetal acceleration in your daily life. You experience centripetal acceleration as you drive in circles, and a satellite experiences centripetal acceleration when it orbits the planet. Being centered is referred to as being centripetal.
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choose true or false for each statement regarding a converging lens.
False. A converging lens produces an enlarged real image when the object is placed beyond its focal point.
Determine whether converging lens produces an enlarged virtual image?A converging lens is thicker at the center and thinner at the edges. When an object is placed beyond the focal point of a converging lens, a real and inverted image is formed on the opposite side of the lens.
This image is larger than the object, hence producing an enlarged real image. The position of the image depends on the object's distance from the lens and follows the rules of image formation by lenses.
On the other hand, a virtual image is formed when an object is placed between the lens and its focal point. The virtual image is upright and larger than the object. However, for a converging lens, this virtual image is not enlarged but rather diminished compared to the object.
Therefore, a converging lens does not produce an enlarged virtual image when the object is placed just beyond its focal point.
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Complete question here:
Choose true or false for each statement regarding a converging lens.
true false A converging lens produces an enlarged virtual image when the object is placed just beyond its focal point.
show that the difference in decibel levels b1 and b2 of a sound source is related to the ratio of its distances r1 and r2 from the receivers by the formula
The formula relating the difference in decibel levels (b1 and b2) of a sound source to the ratio of its distances (r1 and r2) from the receivers is
b1 - b2 = 20 * log10(r2 / r1)
The difference in decibel levels (b1 and b2) of a sound source can be related to the ratio of its distances (r1 and r2) from the receivers using the inverse square law. The inverse square law states that the intensity of sound decreases proportionally to the square of the distance from the source.
The formula for the difference in decibel levels can be expressed as
b1 - b2 = 10 * log10(I1 / I2)
Where:
b1 and b2 are the decibel levels at distances r1 and r2 respectively.
I1 and I2 are the intensities of sound at distances r1 and r2 respectively.
According to the inverse square law, the relationship between the intensities and distances is:
I1 / I2 = [tex][(r2 / r1)^2][/tex]
Substituting this into the formula for the difference in decibel levels:
b1 - b2 = 10 * log10[tex][(r2 / r1)^2][/tex]
Using the logarithmic property log(a^b) = b * log(a), we can simplify further:
b1 - b2 = 20 * log10(r2 / r1)
Therefore, the formula relating the difference in decibel levels (b1 and b2) of a sound source to the ratio of its distances (r1 and r2) from the receivers is:
b1 - b2 = 20 * log10(r2 / r1)
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Doubling the number of units of a bottleneck resource will double the process capacity True False
The given statement is false, because doubling the number of units of a bottleneck resource does not necessarily double the process capacity.
The capacity of a process is determined by its bottleneck, which is the resource or step with the lowest capacity. Increasing the capacity of the bottleneck resource may improve the overall process capacity, but it depends on the specific circumstances and the nature of the process. Other factors such as dependencies, synchronization, and overall process design can also impact the process capacity. Therefore, simply doubling the units of a bottleneck resource does not guarantee a doubling of the process capacity.
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Mass Number
The mass number of an atom is the sum of the number of protons and the number of neutrons in the nucleus of an atom.
Mass number = number of protons + number of neutrons For example, you can calculate the mass number of the copper atom listed in Table 4. 29 protons
plus 34 neutrons equals a mass number of 63
Also, if you know the mass number and the atomic number of an atom, you can calculate the number of neutrons in the nucleus. The number of neutrons is
equal to the mass number minus the atomic number. In fact, if you know two of the three numbers-mass number, atomic number, number of neutrons-
you can always calculate the third
The mass number of an atom is 35 and it has 16 protons. How many neutrons does this atom contain?
The atom contains
neutrons
Answer:
3
Explanation:
mass number minus the atomic number
35-32
3
Which food provide the best nutrients
Answer: Foods that naturally are nutrient-rich include fruits and vegetables. Lean meats, fish, whole grains, dairy, legumes, nuts, and seeds also are high in nutrients.
A series RLC circuit has components with the following values: L = 18.0 mH, C = 80.0 nF, R = 15.0 Ω, and ΔVmax = 100 V, with Δv = ΔVmax sin ωt. (a) Find the resonant frequency of the circuit. kHz (b) Find the amplitude of the current at the resonant frequency. A (c) Find the Q of the circuit. (d) Find the amplitude of the voltage across the inductor at resonance. kV
For the series of RLC circuits has components,
a. The resonant frequency is 179.1 kHz.
b. The current amplitude at resonance is 6.67 A.
c. The Q of the circuit is 1.35.
d. The voltage amplitude across the inductor is 135 V (or 0.135 kV).
(a) To find the resonant frequency of the circuit, we can use the formula:
ω = 1 / √(LC)
Given:
L = 18.0 mH = 18.0 × [tex]10^{(-3)[/tex] H
C = 80.0 nF = 80.0 × [tex]10^{(-9)[/tex] F
Substituting the values into the formula:
ω = 1 / √((18.0 × [tex]10^{(-3)[/tex]) × (80.0 × [tex]10^{(-9)[/tex]))
Calculating the value of ω:
ω ≈ 1123.6 rad/s
To convert the angular frequency to frequency in kHz, we divide ω by 2π:
f = ω / (2π)
Substituting the value of ω:
f ≈ 1123.6 / (2π) ≈ 179.1 kHz
Therefore, the resonant frequency of the circuit is approximately 179.1 kHz.
(b) At the resonant frequency, the impedance of the circuit is at a minimum, and the current amplitude is at maximum. The current amplitude can be calculated using the formula:
I = ΔVmax / R
Given:
ΔVmax = 100 V
R = 15.0 Ω
Substituting the values into the formula:
I = 100 / 15.0 ≈ 6.67 A
Therefore, the amplitude of the current at the resonant frequency is approximately 6.67 A.
(c) The Q of the circuit can be calculated using the formula:
Q = ωL / R
Given:
L = 18.0 mH = 18.0 × [tex]10^{(-3)[/tex] H
R = 15.0 Ω
ω = 1123.6 rad/s
Substituting the values into the formula:
Q = (1123.6 × (18.0 × [tex]10^{(-3)[/tex])) / 15.0 ≈ 1.35
Therefore, the Q of the circuit is approximately 1.35.
(d) The amplitude of the voltage across the inductor at resonance can be calculated using the formula:
VL = Q × VR
Given:
Q = 1.35
VR = ΔVmax = 100 V
Substituting the values into the formula:
VL = 1.35 × 100 ≈ 135 V
Therefore, the amplitude of the voltage across the inductor at resonance is approximately 135 V (or 0.135 kV).
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Geronimo wants to move an object 12 meters. Calculate the net work done by the object with an applied force of 150 N and a friction force of 37 N.
Answer:
1476 J
Explanation:
From the question,
Net Work done = Net force× distance moved by net force.
W' = (F-F')×d................... Equation 1
Where W' = Net work done, F = force applied, F' = Frictional force, d = distance moved.
Given: F = 150 N, F' = 37 N, d = 12 m
Substitute these values into equation 1
W' = (150-37)×12
W' = 123×12
W' = 1476 J.
hence the Net Work done by the object is 1476 J
Can someone please help me
Answer:
I don't know the answer but I needed the answer to that on a quiz and I downloaded sorcatic and it brings u to an app or website with the answer I hope this helps if you can't find the app than just tell me
This force on compass dials is an example of a force that _______.
EMERGENCY
Parallax
Find the distance to the following stars:
.768”
.09”
.63”
.25”
.125”
Rank the following types of electromagnetic radiation from lowest to highest energy per photon. To rank items as equivalent, overlap them. lowest highest
1. radio waves 2. microwaves 3. infrared radiation 4. ultraviolet radiation
The correct order of electromagnetic radiation from lowest to highest energy per photon is- Radio waves < Microwaves < Infrared radiation > Visible light < Ultraviolet radiation < and x-rays. So the order is 1,2,4,3.
Radio waves contain low-energy photons; microwave photons have slightly higher energy than radio waves; infrared photons have more energy than visible, ultraviolet, and x-rays.
Gamma irradiation is very penetrating, and it interacts with matter by ionization in three ways; photoelectric effects, Compton scattering, or pair generation. These radiations are referred to as non-ionizing radiations as they can ionize the molecules due to high penetration power.
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Which is a characteristic of the image formed by an
object between 2F and F?
O The image is virtual.
O The image is bigger than the object.
O The image is inverted,
O
When the object is placed between 2F and F in front of a concave lens characteristic of the image formed by an object is virtual, therefore the correct option first option that the image is virtual.
What is refraction?
It is the phenomenon of bending of light when it travels from one medium to another medium. The bending towards or away from the normal depends upon the medium of travel as well as the refractive index of the material.
Snell's law,
n₁sin(θ₁) = n₂sin(θ₂)
Where n is the refractive index and θ represents angles
A concave lens is used to diverge the incident rays of light falling on it. because of this, the image formed by the concave lens is virtual.
These concave lenses are used in several days to day life applications such as cameras, telescopes, and eye glasses.
When the object is placed between 2F and F in front of a concave lens the characteristic of the image formed by an object is virtual. therefore the correct option first option is that the image is virtual.
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Answer:
the image is virtual
Explanation:
I got it right
Consider the system shown in the figure below. Block A weighs 43.2 N and block B weighs 29.0 N. Once block B is set into downward motion, it descends at a constant speed.
Consider the system shown in the figure below. Blo
(a) Calculate the coefficient of kinetic friction between block A and the tabletop.
(b) A cat, also of weight 43.2 N, falls asleep on top of block A. If block B is now set into downward motion, what is its acceleration?
magnitude m/s2
direction ---Select---
The coefficient of kinetic friction between block A and the tabletop is 0.336.
The weight of block A = 43.2 N
The weight of block B = 29.0 N
(a) The downward motion of block B is constant
(b) The acceleration of block B is -0.00069 m/s²
(a)
The net force acting on the block B will be,
F_net = T - f_fric = m_b × a
Where
T is the tension in the string,
f_fric is the frictional force acting on the block A,
m_b is the mass of block B and
a is the acceleration of block B.
Also,
T = m_b × g = 29.0 N
where g is the acceleration due to gravity.
And as the block is moving with constant velocity, the acceleration of block B is zero.
So, F_net = 0
T - f_fric = 0
f_fric = T
The frictional force f_fric can be expressed as
f_fric = μ_k × N
where N is the normal force.
The normal force on block A is the weight of block A + the weight of the cat,
so,
N = m_Ag + m_catg
The mass of the cat is also 43.2 N.
Thus, N = 43.2 N + 43.2 N = 86.4 N
Therefore,
μ_k × N = T
μ_k = T/N
μ_k = 29.0/86.4
μ_k = 0.336
The coefficient of kinetic friction between block A and the tabletop is 0.336.
(b)
The net force acting on the block B is F_net = T - f_fric
F_net = m_b × a
Where T is the tension in the string,
f_fric is the frictional force acting on the block A,
m_b is the mass of block B and
a is the acceleration of block B.
T = 29.0 N
f_fric = μ_k × N
f_fric = 0.336 × 86.4
f_fric = 29.02 N
F_net = T - f_fric
F_net = 29.0 - 29.02
F_net = -0.02 N
Thus, F_net = m_b × a
-0.02 N = 29.0 N × a
a = -0.02/29.0
a = -0.00069 m/s²
The acceleration of block B is negative and it is slowing down.
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The orbital radius of the Earth (the average Earth-Sun distance) is 1.496 × 1011 m. Mercury’s orbital radius is 5.79 × 1010 m and Pluto’s is 5.91 × 1012 m. Calculate the time required for light to travel from the Sun to each of the three celestial bodies
Answer:
Earth: [tex]t = 498.667\,s[/tex], Mercury: [tex]t = 193\,s[/tex], Pluto: [tex]t = 19700\,s[/tex]
Explanation:
The light travels at a constant speed of approximately [tex]3\times 10^{8}[/tex] meters per second. The time ([tex]t[/tex]), in seconds, required for light to travel a given distance is:
[tex]t = \frac{x}{v_{l}}[/tex] (1)
Where:
[tex]x[/tex] - Travelled distance, in meters.
[tex]v_{l}[/tex] - Speed of light, in meters per second.
Now, we calculate the time for light to travel to each planet:
Earth ([tex]v_{l} = 3\times 10^{8}\,\frac{m}{s}[/tex], [tex]x = 1.496\times 10^{11}\,m[/tex])
[tex]t = \frac{x}{v_{l}}[/tex]
[tex]t = 498.667\,s[/tex]
Mercury ([tex]v_{l} = 3\times 10^{8}\,\frac{m}{s}[/tex], [tex]x = 5.79\times 10^{10}\,m[/tex])
[tex]t = \frac{x}{v_{l}}[/tex]
[tex]t = 193\,s[/tex]
Pluto ([tex]v_{l} = 3\times 10^{8}\,\frac{m}{s}[/tex], [tex]x = 5.91\times 10^{12}\,m[/tex])
[tex]t = \frac{x}{v_{l}}[/tex]
[tex]t = 19700\,s[/tex]
What angle is necessary to keep a 10 kg box motionless if the coefficient of static friction between the box and the ramp is 0.55?
a.33.4°
b.28.8°
c.56.6°
d.45.0°
The angle necessary to keep a 10 kg box motionless, given a coefficient of static friction of 0.55 between the box and the ramp, is 33.4°, which corresponds to Option A.
To determine the angle, we can use the relationship between the coefficient of static friction, the angle of the incline, and the gravitational force acting on the box. The maximum static friction force can be calculated using the formula:
Friction force = coefficient of static friction * Normal force
The Normal force can be found by decomposing the gravitational force acting on the box into components parallel and perpendicular to the incline. The perpendicular component (Normal force) is equal to the weight of the box (mass * gravitational acceleration).
Since the box is motionless, the friction force must be equal to the component of the gravitational force acting parallel to the incline:
Friction force = Component of weight parallel to incline
By substituting the given values and solving for the angle, we find:
coefficient of static friction = tan(angle)
angle = arctan(coefficient of static friction)
angle = arctan(0.55) ≈ 33.4°
Therefore, the correct answer is Option A, 33.4°.
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how does the umts channel structure of the air interface differ from gsm?
The UMTS (Universal Mobile Telecommunications System) and GSM (Global System for Mobile Communications) are two different cellular technologies used for mobile communication. The channel structure of the air interface in UMTS differs from GSM in several ways.
GSM:
In GSM, the air interface channel structure is based on a combination of time division multiple access (TDMA) and frequency division multiple access (FDMA). The spectrum is divided into multiple frequency channels, and each channel is further divided into time slots. Each time slot supports one user at a time, allowing multiple users to share the same frequency but with different time slots. This TDMA/FDMA combination is known as the TDMA frame structure.
UMTS:
UMTS, on the other hand, utilizes a different channel structure called wideband code division multiple access (WCDMA). WCDMA is a spread spectrum technique that employs a wider bandwidth compared to GSM. The entire available spectrum is shared among all users simultaneously, using different codes to differentiate between different users. This enables multiple users to access the same frequency at the time, resulting in a more efficient utilization of the spectrum.
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Currents of devices that are in a series circuit ar the same, but the __________can be different, which causes __________to be different as well.
Answer: its flowing, reaction
Explanation: this is because currents in a device have a flowing object inside
Animals in cold climates often depend on two layers of insulation: a layer of body fat [of thermal conductivity 0. 200W/(m⋅K) ] surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus) as a sphere 1. 60m in diameter having a layer of fat 3. 90cm thick. (Actually, the thickness varies with the season, but we are interested in hibernation, when the fat layer is thickest. ) In studies of bear hibernation, it was found that the outer surface layer of the fur is at 2. 80∘C during hibernation, while the inner surface of the fat layer is at 30. 9∘C a) What should the temperature at the fat-inner fur boundary be so that the bear loses heat at a rate of 51. 4W ? b) How thick should the air layer (contained within the fur) be so that the bear loses heat at a rate of 51. 4W ?
a) Calculation of temperature at the fat-inner fur boundaryThe rate of heat flow is given by:
[tex]q =\frac{kA\Delta T}{d}[/tex]
where, k = thermal conductivity; A = surface area; ΔT = temperature difference and d = thicknessSince the rate of heat flow is given to be 51.4 W, we can obtain the temperature difference from the given data.
[tex]ΔT = \frac{30.9 - 2.8}{\ln \frac{3.9}{1.6/2}} ≈ 3.6°C[/tex]
Now, substituting the given values of A, d and k, we get
[tex]51.4 = \frac{0.200 \pi (1.6)^{2} \times 3.6}{0.039} × (T1 - 30.9)[/tex]
where T1 is the required temperature at the fat-inner fur boundarySimplifying, we getT1 ≈ -9.7°Cb) Calculation of thickness of air layerAssuming the layer of air to be stationary and isothermal, the rate of heat flow can be calculated using the following equation:q = hAΔTwhere, h = heat transfer coefficientThe heat transfer coefficient, h can be calculated using the relation:
[tex]q = [\frac{kA\Delta T}{d} = hAΔT ⇒ h =\frac{k}{d}\\[/tex]
Using this, we can obtain the heat transfer coefficient, which is approximately 0.7 W/(m².K)Using the relation above, we can write:
[tex]51.4 = 0.7 × (4π(1.6/2)²) × ΔT × d[/tex]
where ΔT is the temperature difference and d is the thickness of the air layerSolving for d, we getd ≈ 1.2 cmTherefore, the thickness of the air layer should be around 1.2 cm so that the bear loses heat at a rate of 51.4 W.
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PLZZ HELP
If you have two objects moving at the same velocity, would the object with bigger mass have higher or lower kinetic energy?
Answer:
The kinetic energy of a moving object is directly proportional to its mass and directly proportional to the square of its velocity. This means that an object with twice the mass and equal speed will have twice the kinetic energy while an object with equal mass and twice the speed will have quadruple the kinetic energy.
The smallest molecules are made up of -
a. 1 atom
b. 2 atoms
c. 3 atoms
The largest molecules are made up of -
a. billions
b. millions
c. hundreds
d. thousands
- of atoms.
You have a 40-Hz sound wave and a 5,000-Hz sound wave. Both are traveling
through steel. Which sound wave will travel faster?
The waves will travel at the same speed as one another.
The 40-Hz wave will travel the fastest.
The 5,000-Hz wave will travel the fastest.
The louder of the two sound waves with travel the fastest.
Answer:
5,000-Hz
Explanation: