Answer:
Difference in Length of steel is 0.000246m
Difference in Length of invar is 0.00001845m
Difference in Length of steel surveyor's tape is 0.00738m
Difference in Length of invar surveyor's tape is 0.0005535m
Explanation:
Linear expansivity of steel is volume expansivity ÷ 3
Linear expansivity of steel = 0.000036/°C ÷ 3 = 0.000012/°C
Difference in Length of steel= Linear expansivity × initial length × temperature change
= 0.000012 × 1 × (20.5-0)
= 0.000012×1×20.5 = 0.000246m
Linear expansivity of invar = volume expansivity of invar ÷ 3
Linear expansivity of invar= 0.0000027/°C ÷ 3= 0.0000009/°C
Difference in Length of invar = 0.0000009×1×(20.5-0) = 0.00001845m
Difference in Length of steel surveyor's tape= 0.000012×30×(20.5-0) = 0.00738m
Difference in Length of invar surveyor's tape = 0.0000009×30×(20.5-0) = 0.0005535m
The difference in length between a steel and invar meter stick at 20.5°C is 6.83 mm, and for two 30-m-long surveyor's tapes, the difference at the same temperature is 20.5 cm.
Explanation:When a steel meter stick and an invar meter stick both expand due to an increase in temperature, we can determine their difference in length using the coefficients of linear expansion for each material.
The difference in length at 20.5°C can be determined using the formula:
ΔL = αLΔT
For steel, the coefficient of linear expansion (α) is 3.6 × 10^-5 /°C, and for invar it is 2.7 × 10^-6 /°C. Plugging in the values:
ΔL_steel = 3.6 × 10^-5 /°C × 1.00 m × 20.5 °C = 7.38 × 10^-3 mΔL_invar = 2.7 × 10^-6 /°C × 1.00 m × 20.5 °C = 5.54 × 10^-4 mThe difference in length is therefore 7.38 × 10^-3 m - 5.54 × 10^-4 m = 6.83 × 10^-3 m or 6.83 mm.
For two 30.00-m-long surveyor’s tapes:
ΔL_steel = 3.6 × 10^-5 /°C × 30.00 m × 20.5 °C = 2.21 × 10^-1 mΔL_invar = 2.7 × 10^-6 /°C × 30.00 m × 20.5 °C = 1.66 × 10^-2 mThe difference in length for the surveyor's tapes is 2.21 × 10^-1 m - 1.66 × 10^-2 m = 2.05 × 10^-1 m or 20.5 cm.
Interactive Solution 9.63 illustrates one way of solving a problem similar to this one. A thin rod has a length of 0.620 m and rotates in a circle on a frictionless tabletop. The axis is perpendicular to the length of the rod at one of its ends. The rod has an angular velocity of 0.185 rad/s and a moment of inertia of 1.43 x 10-3 kg·m2. A bug standing on the axis decides to crawl out to the other end of the rod. When the bug (whose mass is 5 x 10-3 kg) gets where it's going, what is the change in the angular velocity of the rod?
Answer:
w = 7.89 10⁻² rad/s
Explanation:
We will solve this exercise with the conservation of the annular moment, let's write it in two moments
Initial. With the insect in the center
L₀ = I w₀
End with the bug on the edge
[tex]L_{f}[/tex]= I w + [tex]I_{bug}[/tex] w
The moments of inertia are
For a rod
I = 1/3 M L²
For the insect, taken as a particle
I = m L²
The system is formed by the rod and the insect, this is isolated, therefore the external torque is zero and the angular momentum is conserved
L₀ = [tex]L_{f}[/tex]
I w₀ = I w + [tex]I_{bug}[/tex] w
w = I / (I + [tex]I_{bug}[/tex]) w₀
w = I / (I + m L²) w₀
Let's calculate
w = 1.43 10⁻³ / (1.43 10⁻³ + 5 10⁻³ 0.620²)² 0.185
w = 1.43 10⁻³ / 3.352 10³ 0.185
w = 7.89 10⁻² rad/s
A 0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of cold water at 0∘C (cwater=4186J/kg⋅∘C) and sealed. At equilibrium, the temperature of the water and block are measured to be 20∘C.If the original experiment is repeated with a 1.0 kg aluminum block, what is the final temperature of the water and block?If the original experiment is repeated with a 1.0 kg copper (ccopper=387 J/kg⋅∘C) block, what is the final temperature of the water and block?If the original experiment is repeated but 100 g of the 0∘C water is replaced with 100 g of 0∘C ice, what is the final temperature of the water and block? If the original experiment is repeated but 100 g of the 0∘C water is replaced with only 25 g of 0∘C ice, what is the final temperature of the water and block?
These scenarios involve the transfer of heat in a closed system by utilizing calorimetry. The final temperature is affected by the mass of the metal block, the specific heat capacity of the metal, and the introduction of phase changes with the addition of ice into the system.
Explanation:These problems revolve around the principle of conservation of energy, and more specifically the concept of heat transfer, which is often studied using calorimetry. We can solve these problems by using the formula for heat transfer, Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.
For the first scenario, using a 1.0 kg aluminum block instead of a 0.5 kg block, the heat capacity of the system has simply doubled. The final temperature for the aluminum block and the water can be determined by setting the heat gained by the water equal to the heat lost by the aluminum. Since the mass of the block has doubled, the final temperature will be lower than the original experiment.
In the second scenario with 1.0 kg copper block, copper has a lower specific heat capacity than aluminum. This means that when heated to the same temperature, the copper block will contain less energy than the aluminum block. Consequently, when put in contact with the water, the final temperature of the water and block will be lower than in the original experiment.
In the final two scenarios, adding ice to the system adds an additional phase change. This means, it’s necessary to add in the energy used to convert the ice at 0 degrees Celsius into water at 0 degrees Celsius before heating the water to the final temperature. The amount of ice being added in these instances changes the energy dynamics of the system and thus resulting in different final temperatures.
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A heat engine takes thermal energy QH from a hot reservoir and uses part of this energy to perform work W. Assuming that QH cannot be changed, how can the efficiency of the engine be improved?
a) Increase the work W, the thermal energy QC rejected to the cold reservoir increasing as a result.
b) Decrease the work W, the rejected QC decreasing as a result.
c) Increase the work W, the rejected QC remaining unchanged.
d) Increase the work W, the rejected QC decreasing as a result.
e) Decrease the work W, the rejected QC remaining unchanged
Answer:
d) Increase the work W, the rejected QC decreasing as a result.
Explanation:
By the second law of thermodynamics the efficiency of a heat engine working between two reservoirs is:
[tex]\eta=\frac{W}{Q_{H}} [/tex] (1)
With W the work and [tex] Q_{H} [/tex] the heat of the hot reservoir, note in (1) that efficiency is directly proportional to the work and inversely proportional to the heat of the hot reservoir, so if we remain [tex]Q_{H} [/tex] constant we should increase the work to increase the efficiency.
Also, efficiency is:
[tex] \eta=1-\frac{Q_{C}}{Q_{H}}[/tex] (2)
With [tex]Q_{C} [/tex] the heat released to the cold reservoir, it is important to note that because second law of thermodynamics the efficiency of a heat engine should be between 0 and 1 ([tex]0\leq\eta\leq1 [/tex]), so the ratio [tex]\frac{Q_{C}}{Q_{H}} [/tex] always is positive and its maximum value is 1, that implies if [tex]Q_{H} [/tex] remains constant and efficiency increases, [tex]Q_{C} [/tex] will decrease and the ratio [tex]\frac{Q_{C}}{Q_{H}} [/tex] too.
So, the correct answer is d)
Final answer:
The efficiency of a heat engine is increased by increasing the work output while decreasing the rejected heat to the cold reservoir, in accordance with the first and second laws of thermodynamics.
Explanation:
To improve the efficiency of a heat engine, one must increase the work output, W, while simultaneously decreasing the heat rejected to the cold reservoir, Qc. The correct choice is:
d) Increase the work W, the rejected Qc decreasing as a result.
The efficiency, η, of a heat engine is defined as the ratio of the work done, W, to the heat absorbed from the hot reservoir, QH. By the first law of thermodynamics, QH = W + Qc, meaning that the efficiency can be increased either by increasing W or decreasing Qc. According to the second law of thermodynamics, there is a minimum amount of QH that cannot be used for work and must be rejected as Qc. Therefore, the aim is to minimize this rejected heat without altering QH, which cannot be changed in this scenario. The ideal is to approach the efficiency of a Carnot engine, which has the maximum possible efficiency between two given temperatures by operating in a reversible manner and reducing entropy generation.
My Notes A container is divided into two equal compartments by a partition. One compartment is initially filled with helium at a temperature of 240 K; the other is filled with nitrogen at a temperature of 315 K. Both gases are at the same pressure. If we remove the partition and allow the gases to mix, what will be their final temperature?
Answer:
[tex]final-temperature = T_{f} = 252.51K[/tex]
Explanation:
we can solve this problem by using the first law of thermodynamics.
[tex]\Delta U= Q-W[/tex]
Q= heat added
U= internal energy
W= work done by system
[tex]E_{final}= E_{initial}[/tex]
[tex]C_{v} (N_{2}) C_{v}(He) T_{f} =C_{v}( N_{2}) T_{N_{2} } C_{v} (He) T_{He}[/tex] (1)
[tex]C_{v}(N_{2})=1.04\frac{KJ}{Kg K}[/tex]
[tex]C_{v}(He)=5.193\frac{KJ}{Kg K}[/tex]
now
From equation 1
[tex]T_{f}=\frac{C_{v}( N_{2}) T_{N_{2} } C_{v} (He) T_{He}}{C_{v} (N_{2}) C_{v}(He)}[/tex]
[tex]T_{f} = \frac{315\times1.04+5.193\times240}{1.04+5.193}[/tex]
[tex]T_{f} = 252.51K[/tex]
You are baking and run out of an important ingredient. Rather than fire up the car, you decide to use a bicycle. The store is 1/2 mile away. Assume your car gets 30 miles to the gallon. If every driver on the planet made a similar decision once per week for a year, approximately how much gas would be saved for that year? (As this is an estimate, you may make the arithmetic easier by assuming a year has 50 weeks rather than 52 weeks. What number do you use for 'every driver on the planet'?) a. 100 million gallons b. 2 billion gallons c. 20,000 gallons d. 200,000 gallons e. 2 million gallons
Answer: this one is tough. There is a estimated 1.2 billion drivers to help best I could. I did the math the best I could and didn’t get an answer close to the choices. Just guessing I’d guess C. 20,000 because of logic.
Explanation: I can’t help much but would love to hear how to work this out.
A steam of hot dry nitrogen flows through a process unit that contains liquid acetone. A substantial part of the acetone vaporizes and is carried off by the nitrogen. The combined gases leave the unit at 205 ◦C and 1.1 bar, are compressed and then enter a condenser in which a portion of the acetone is liquefied. The remaining gas leaves the condenser at 10 ◦C and 40 bar. The partial pressure of acetone in the feed to the condenser is 0.100 bar, and that in the effluent gas from the condenser is 0.379 bar. Assume ideal gas behavior.
(a) Calculate, for a basis of m3 of gas fed to the condenser, the mass of ace- tone condensed (kg) and the volume of gas leaving the condenser (m3).
(b) Suppose the volumetric flow rate of the gas leaving the condenser is 20.0 m3/h. Calculate the rate (kg/h) at which acetone is vaporized in the solvent recovery unit.
The problem is solved by applying the Ideal Gas Law and Dalton's Law of Partial Pressures. First, we calculate the mass of acetone condensed using these laws and then determine the rate at which acetone is vaporized using the given volumetric flow rate.
Explanation:The given problem involves a number of gas law principles, but its main focus is on the application of the Ideal Gas Law and Dalton's Law of partial pressures. Initially, we calculate the moles of acetone in the feed using the Ideal Gas Law, and then we find out the moles of acetone in the effluent using Dalton's law. Subtracting gives us the moles of acetone condensed.
(a) Using Ideal Gas Law, we have PV=nRT. Hence, n (acetone, feed) = P (acetone, feed) * V(feed) /RT(feed). To find the moles of acetone in the effluent, we use Dalton's law and the Ideal Gas Law to get n (acetone, effluent) = P(acetone, effluent) * V (effluent) / RT (effluent). Subtracting moles in effluent from moles in feed gives moles condensed. Multiplying by the molar mass of acetone, we get mass of acetone condensed.
(b) The question tells us the volumetric flow rate of the gas leaving the condenser. Therefore, number of moles of acetone vaporized per hour can be calculated using the Ideal Gas Law and then can be converted into mass by using the molar mass of acetone. Hence, rate at which acetone is vaporized = moles (acetone, vaporized per hour) * molar mass (acetone).
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At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate heat input of the second. The operating temperatures of the first are 750◦C and 440◦C respectively, and of the second 415◦C and 270◦C.
Answer:
The rate at which coal burned is 111.12 kg/s.
Explanation:
Given that,
First initial temperature =750°C
First final temperature =440°C
Second initial temperature =415°C
Second final temperature =270°C
Suppose If the heat of combustion of coal is [tex]2.8×10^{7}\ J/kg[/tex], at what rate must coal be burned if the plant is to put out 950 MW of power? Assume the efficiency of the engines is 65% of the Carnot efficiency.
The work done by first engine is
[tex]W=eQ[/tex]
The work done by second engine is
[tex]W'=e'Q'[/tex]
[tex]W'=e'Q(1-e)[/tex]
Total out put of the plant is given by
[tex]W+W'=950\ MW[/tex]
Put the value into the formula
[tex]eQ+e'Q(1-e)=950\times10^{6}[/tex]....(I)
We need to calculate the efficiency of first engine
Using formula of efficiency
[tex]e=65%(1-\dfrac{T_{L}}{T_{H}})[/tex]
[tex]e=0.65(1-\dfrac{440+273}{750+273})[/tex]
[tex]e=0.196[/tex]
We need to calculate the efficiency of second engine
Using formula of efficiency
[tex]e'=65%(1-\dfrac{T_{L}}{T_{H}})[/tex]
[tex]e'=0.65(1-\dfrac{270+273}{415+273})[/tex]
[tex]e'=0.136[/tex]
Put the value of efficiency for first and second engine in the equation (I)
[tex]Q(0.196+0.136(1-0.196))=950\times10^{6}[/tex]
[tex]Q=\dfrac{950\times10^{6}}{(0.196+0.136(1-0.196))}[/tex]
[tex]Q=3111.24\times10^{6}\ W[/tex]
We need to calculate the rate at which coal burned
Using formula of rate
[tex]R=\dfrac{Q}{H_{coal}}[/tex]
[tex]R=\dfrac{3111.24\times10^{6}}{2.8×10^{7}}[/tex]
[tex]R=111.12\ kg/s[/tex]
Hence, The rate at which coal burned is 111.12 kg/s.
The oscillating current in an electrical circuit is as follows, where I is measured in amperes and t is measured in seconds. I = 7 sin(60πt) + cos(120πt). Find the average current for each time interval. (Round your answers to three decimal places.) (a) 0 ≤ t ≤ 1 60 amps
The average current in the given oscillating current function, I = 7 sin(60πt) + cos(120πt), can be found by integrating each component of the function over the given time interval and taking the average. Due to the complexity of the function, it is recommended to use mathematical software or a calculator with integral computation capability.
Explanation:The given function describes the oscillating current in an electrical circuit: I = 7 sin(60πt) + cos(120πt). The average current, Iave, is conceptually the net charge, ΔQ, that passes through a given cross-sectional area per unit time, Δt. Due to the complexity of this function, we must split it into two integrals to find the average currents.
For the time interval 0 ≤ t ≤ 1/60 seconds, we have two integrals for each component of the current: ∫01/607 sin(60πt) dt and ∫01/60cos(120πt) dt. Calculating these will give us the average current for this time interval. Due to the complexity of the function and the requirement of calculus to solve it, it's recommended to use mathematical software or a calculator with integral computation capability to get the numerical values, always rounding your answers to three decimal places.
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What is the object’s velocity, in meters per second, at time t = 2.9? Calculate the object’s acceleration, in meters per second squared, at time t = 2.9. What is the magnitude of the object’s maximum acceleration, in meters per second squared? What is the magnitude of the object’s maximum velocity, in meters per second?
Answer:
the question is incomplete, below is the complete question
"An object is undergoing simple harmonic motion along the x-axis. Its position is described as a function of time by x(t) = 5.5 cos(4.4t - 1.8), where x is in meters, the time, t, is in seconds, and the argument of the cosine is in radians.
a.What is the object's velocity, in meters per second, at time t = 2.9?
b.Calculate the object's acceleration, in meters per second squared, at time t = 2.9.
c. What is the magnitude of the object's maximum acceleration, in meters per second squared?
d.What is the magnitude of the object's maximum velocity, in meters per second?"
a.[tex]v(t)==24.1m/s[/tex]
b.[tex]a(t)=3.79m/s^{2}[/tex]
c.[tex]a_{max}=106.48m/s^{2}[/tex]
d.[tex]v_{max}=24.2m/s[/tex]
Explanation:
the gneral expression for the displacement of object in simple harmonic motion is represented by
[tex]x(t)=Acos(wt- \alpha)\\[/tex]
while the velocity is express as
[tex]v(t)=-Awcos(4.4t-1.8)\\[/tex]
and the acceleration is
[tex]a(t)=-aw^{2}cos(wt- \alpha )\\[/tex]
Note: the angle is in radians
The expression for the displacement from the question is [tex]x(t)=5.5cos(4.4t-1.8)\\[/tex]
comparing, A=5.5, w=4.4,α=1.8
a.To determine the object velocity at t=2.9secs,
we substitute for t in the velocity equation
[tex]v(t)=-5.5*4.4sin(4.4*2.9-1.8)\\v(t)=-24.2sin(10.96)\\[/tex]
[tex]v(t)=-24.2*(-0.9993)\\v(t)==24.1m/s[/tex]
b.To determine the object acceleration at t=2.9secs,
we substitute for t in the acceleration equation
[tex]a(t)=-5.5*4.4^{2} cos(4.4*2.9-1.8)\\a(t)=-106.48cos(10.96)\\[/tex]
[tex]a(t)=-106.48*0.0356\\a(t)=3.79m/s^{2}[/tex]
c. The acceleration is maximum when the displacement equals the amplitude. hence magnitude of the object acceleration is
[tex]a_{max}=-w^{2}A\\ a_{max}=-4.4^{2}*5.5\\ a_{max}=106.48m/s^{2}[/tex]
d.The maximum velocity is expressed as
[tex]v_{max}=wA\\v_{max}=4.4*5.5\\v_{max}=24.2m/s[/tex]
A spring that is stretched 23 cm from its equilibrium point experiences a force of 103 N.a.) How much energy is stored in the spring?
Answer:
U = 11.85 J
Explanation:
given,
spring is stretched = 23 cm
x = 0.23 m
Force experiences = 103 N
we know,
F = k x
where k is the spring constant
[tex]k =\dfrac{F}{x}[/tex]
[tex]k =\dfrac{103}{0.23}[/tex]
k = 447.83 N/m
energy stored in the spring
[tex]U =\dfrac{1}{2}kx^2[/tex]
[tex]U =\dfrac{1}{2}\times 447.83 \times 0.23^2[/tex]
U = 11.85 J
hence, energy stored in the spring is equal to U = 11.85 J
The elastic potential energy stored in the spring when stretched to a length of 23 cm, given a spring constant of 4 N/cm, is 0.18 J.
Explanation:The question pertains to how much energy is stored in a spring when it is stretched 23 cm from its equilibrium point, where it experiences a force of 103 N. The potential energy stored in the spring can be calculated using the formula U = 1/2kx². Given that the spring constant is 4 N/cm and the displacement of the spring from its unstretched length is 3 cm (since the unstretched length is 20 cm), the potential energy can be calculated as follows: U = 1/2 * 4 N/cm * (3 cm)² = 0.18 J. Thus, the elastic potential energy contributed by the spring when it is stretched to a length of 23 cm is 0.18 J.
A "Barrel of Fun" consists of a large vertical cylinder that spins about its axis fast enough so that any person inside will be held against the wall. At an angular speed ω1, an upward frictional force F1 holds a person against the wall without slippingIf you double the angular speed (i.e., ω2 = 2 ω1) what is the new friction force F2?
Answer:
Friction force F₂ after doubling the angular speed is same as the friction force at angular speed ω₁
Explanation:
Consider the fig attached below. Forces acted on person are Centripetal force (-mv²/r) exerted in x direction and reversal normal force N wall exerted on person.
[tex]\sum F_{x} =0\\N+ ma_{x}\\-N=m(-\frac{v^{2}}{r})\\-N=m(-r\omega^{2})N=m(r\omega^{2})[/tex] ---(1)
In y direction there is frictional force Fs exerted in upward direction that keeps the person standing without falling which is balanced by weight of person in downward direction.
[tex]\sum F_{y} = 0\\F_{s}-mg=0\\F_{s}=mg[/tex]----(2)
from eq 2 it can e seen that frictional force is equal to weight of person exerted in upward direction, it does not depends on angular speed ω₁. So when the angular speed is doubled i.e ω₂ = 2ω₁, frictional force Fs remains same.
A car of mass 900 kg is traveling at 20 m/s when the brakes are applied. The car then comes to a complete stop in 5 s. What is the average power that the brakes produce in stopping the car?
A. 36,000 W
B. 7200 W
C. 3600 W
D. 1800 W
Answer:
A. 36,000 W
Explanation:
[tex]m[/tex] = mass of the car = 900 kg
[tex]v_{o}[/tex] = Initial speed of the car = 20 m/s
[tex]v_{f}[/tex] = Final speed of the car = 0 m/s
[tex]W[/tex] = Work done by the brakes on the car
Magnitude of work done on the car by the brakes is same as the change in kinetic energy of the car.hence
[tex]W = (0.5) m (v_{o}^{2} - v_{f}^{2})\\W = (0.5) (900) ((20)^{2} - (0)^{2})\\W = 180000 J[/tex]
[tex]t[/tex] = time taken by the car to come to stop = 5 s
[tex]P[/tex] = Average power produced by the car
Average power produced by the car is given as
[tex]P = \frac{W}{t} =\frac{180000}{5} \\P = 36000 W[/tex]
How much heat energy is required to convert 93.4 g of solid ethanol at − 114.5 ° C to gasesous ethanol at 149.8 ° C ? The molar heat of fusion of ethanol is 4.60 kJ/mol , and its molar heat of vaporization is 38.56 kJ/mol . Ethanol has a normal melting point of − 114.5 ° C and a normal boiling point of 78.4 ° C . The specific heat capacity of liquid ethanol is 2.45 J / g ⋅ ° C , and that of gaseous ethanol is 1.43 J / g ⋅ ° C .
Answer:
Q' = 140.859 kJ
Explanation:
Given that, 93.4 g of solid ethanol at − 114.5 °C is converted to gasesous ethanol at 149.8 ° C.
The molar heat of fusion of ethanol is, ΔH(f) = 4.60 kJ/mol , and its molar heat of vaporization is ΔH(v) = 38.56 kJ/mol .
And also Ethanol has a normal melting point of − 114.5 ° C and a normal boiling point of 78.4 ° C .
The specific heat capacity of liquid ethanol is S(l) = 2.45 J / g ⋅°C , and that of gaseous ethanol is S(g) = 1.43 J / g ⋅°C .
Lets solve this step wise ;
Given 93.4 g of ethanol is taken, but 1 mole of ethanol consists of 46.06 g
⇒ moles of ethanol given = [tex]\frac{93.4}{46.06}[/tex] = 2.02 moles
step 1: solid ethanol to liquid ethanol at melting point of − 114.5 ° C
⇒ 1 mole requires ΔH(f) = 4.60 kJ/mol of heat
⇒ heat required = 4.60 × 2.02 = 9.292 kJ.
step 2: liquid ethanol at -114 °C to liquid ethanol at 78.4 °C
Q = m×S×ΔT ; Q = heat required
m = mass of the substance
S = specific heat of the substance
ΔT = change in temperature
Here S = S(l);
⇒ Q = 93.4×2.45×(78.4-(-114.5))
= 44.141 kJ
step 3: liquid ethanol at 78.4°C to gaseous ethanol at 78.4°C
1 mol of liquid ethanol requires ΔH(v) = 38.56 kJ/mol of heat
⇒ required heat = 38.56×2.02 = 77.89 kJ
step 4: gaseous ethanol at 78.4 °C to gaseous ethanol at 149.8 °C
Q = m×S×ΔT
Here, S = S(g)
Q = 93.4×1.43×(149.8-78.4)
= 9.536 kJ
⇒ Total heat required = 9.292 + 44.141 + 77.89 + 9.536
= 140.859 kJ
⇒ Q' = 140.859 kJ
The total heat energy required to convert 93.4 g of solid ethanol at − 114.5 ° C to gaseous ethanol at 149.8 ° C is 140.859 kJ.
Given data:,
93.4 g of solid ethanol at − 114.5 °C is converted to gaseous ethanol at 149.8 ° C.
The molar heat of fusion of ethanol is, ΔH(f) = 4.60 kJ/mol.
Molar heat of vaporization is ΔH(v) = 38.56 kJ/mol .
And also Ethanol has a normal melting point of − 114.5 ° C and a normal boiling point of 78.4 ° C .
The specific heat capacity of liquid ethanol is S(l) = 2.45 J / g ⋅°C , and that of gaseous ethanol is S(g) = 1.43 J / g ⋅°C .
Since, 93.4 g of ethanol is taken, but 1 mole of ethanol consists of 46.06 g
⇒ moles of ethanol given is
⇒93.04/46.06 = 2.02 moles
Heat required for conversion of solid ethanol to liquid ethanol at melting point of − 114.5 ° C
⇒ 1 mole requires ΔH(f) = 4.60 kJ/mol of heat
⇒ Q = 4.60 × 2.02 = 9.292 kJ.
Heat required to convert liquid ethanol at -114 °C to liquid ethanol at 78.4 °C
Q' = m×s×ΔT
Here,
m = mass of the substance
s = specific heat of the substance
ΔT = change in temperature
Solving as,
Q' = 93.4×2.45×(78.4-(-114.5))
Q' = 44.141 kJ
Heat required to convert liquid ethanol at 78.4°C to gaseous ethanol at 78.4°C
1 mol of liquid ethanol requires ΔH(v) = 38.56 kJ/mol of heat
Q'' = 38.56×2.02 = 77.89 kJ
Heat required to convert gaseous ethanol at 78.4 °C to gaseous ethanol at 149.8 °C
Q''' = m×S×ΔT
Q''' = 93.4×1.43×(149.8-78.4)
= 9.536 kJ
⇒ Total heat required = 9.292 + 44.141 + 77.89 + 9.536
= 140.859 kJ
Thus, we can conclude that the total heat energy required to convert 93.4 g of solid ethanol at − 114.5 ° C to gaseous ethanol at 149.8 ° C is 140.859 kJ.
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In an atom, an electron is confined to a space of roughly 10?10 meters. If we take this to be the uncertainty in the electron's position, what is the minimum uncertainty ?p in its momentum?
Answer:
[tex]5.2728\times 10^{-25}\ kgm/s[/tex]
Explanation:
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
[tex]\Delta x[/tex] = Uncertainity in position = [tex]10^{-10}\ m[/tex]
[tex]\Delta p[/tex] = Uncertainty in momentum
According to the Heisenberg uncertainity principle we have
[tex]\Delta x\Delta p=\dfrac{h}{4\pi}\\\Rightarrow \Delta p=\dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p=\dfrac{6.626\times 10^{-34}}{4\pi\times 10^{-10}}\\\Rightarrow \Delta p=5.2728\times 10^{-25}\ kgm/s[/tex]
The minimum uncertainty in its momentum is [tex]5.2728\times 10^{-25}\ kgm/s[/tex]
Which of the following statements concerning the nuclear force is false? O The nuclear force is attractive and not repulsive. O The nuclear is one of only four known types of forces in nature. O The nuclear force is very short-ranged. O The nuclear force acts on both protons and neutrons. O The nuclear force is very weak and much smaller in relative magnitude than the electrostatic and gravitational forces. Submit Request
Answer:
The nuclear force is attractive and not repulsive.The nuclear force is very weak and much smaller in relative magnitude than the electrostatic and gravitational forces.Explanation:
Nuclear force is the strongest existing force in the nature. It has the shortest range.Its main function is to hold the subatomic particles together in nature.The nuclear force is created by the exchange of pi mesons between the nucleons of an atom, but for this exchange to happen the particles must be close to one another of the order of few femtometer.At about 1 femtometer the nuclear force is very strongly attractive in nature but at distance greater than 2.5 femtometer it fades away. The force becomes repulsive in nature at distance less than 0.7 femtometer.This force holds the likely charged protons together in the nucleus.For a new TV series "Stupidity Factor contestants are dropped into the ocean (p 1030 kg/ms) along with a Styrofoam soo kg/ms block that is mm x 3 mm X 20 cm thick. If too many contestants climb aboard the block and it sinks below the water surface, they are declared "stupid" and abandoned at sea. Find the maximum number of 7o kg contestants that the block can hold. (No fractional contestants, please. They hate it when that happens.
Answer:
Explanation:
Heres the correct and full question:
For a new TV series "Stupidity Factor contestants are dropped into the ocean (p=1030 kg/m³) along with a Styrofoam p=3oo kg/m³ block that is 2m x 3 m X 20 cm thick. If too many contestants climb aboard the block and it sinks below the water surface, they are declared "stupid" and abandoned at sea. Find the maximum number of 7o kg contestants that the block can hold. (No fractional contestants, please. They hate it when that happens.
answer:
volume of styrofoam block=V=2m x 3m x 0.20m =1.2m³
density of styrofoam=ρs=300kg/m³
mass of styrofoam=ms=v(ρs)=1.2 x 300=360kg
weight of styrofoam = ws=(ms)g=360 x 9.8=3528N
consider number of contestants= n
toatal weight of ccontestants=W=n(70 x 9.8)=n(686N)
since, styrofoam fully submerged into water, then bat force,
B=ρ(vs)g=1030 x 1.2 x 9.8 = 12112.8N
At equilibrium,
B - W - Ws = 0
12112.8 - 686n - 3528 = 0
[tex]n=\frac{12112.8-3528}{686}=12.5[/tex]
n=12 person(contestants)
A student in an undergraduate physics lab is studying Archimede's principle of bouyancy. The student is given a brass cylinder and, using a triple beam balance, finds the mass to be 3.21 kg. The density of this particular alloy of brass is 8.62 g/cm 3 . The student ties a massless string to one end of the cylinder and submerges it into a tank of water where there is an apparent reduction in the weight of the cylinder. With this information, calculate the volume, V , of the cylinder and the tension, T , in the string when it is submerged in the tank of water. The density of water is 1.00 g/cm 3 , and the acceleration due to gravity is gVolume of the cylinder: _____ cm3Tension in the string: ______ N
Answer:
V = 0.3724 m³
T = 27.836 N
Explanation:
Given :
m = 3.21 kg , W= 3.21 * 9.81 m / s² = 31.4901 N
ρ = 8.62 g / cm ³ = 8620 kg / m³
V = m / ρ = 3.21 kg / 8620 kg / m³
V = 0.3724 m³
when submerged the weight of brass cylinder is equal to the tension in string:
F = (0.3724m³) * (1000 kg / m³) * (9.81 m/s²²) = 3.653 ≈ 3.65 N
T = 31.4901 N - 3.65 N
T = 27.836 N
Snorkeling by humans and elephants. When a person snorkels, the lungs are connected directly to the atmosphere through the snorkel tube and thus are at atmospheric pressure. In atmospheres, what is the difference Δp between this internal air pressure and the water pressure against the body if the length of the snorkel tube is
(a)24 cm (standard situation) and
(b)4.1 m (probably lethal situation)?
Answer:
2354.4 Pa
40221 Pa
Explanation:
[tex]\rho[/tex] = Density = 1000 kg/m³
g = Acceleration due to gravity = 9.81 m/s²
h = Depth
The pressure difference would be
[tex]\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 0.24\times 9.81\\\Rightarrow \Delta P=2354.4\ Pa[/tex]
The pressure difference in the first case is 2354.4 Pa
[tex]\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 4.1\times 9.81\\\Rightarrow \Delta P=40221\ Pa[/tex]
The pressure difference in the second case is 40221 Pa
A simple piping system consisting of a small pipe (diameter 6 cm) connected to a larger pipe (diameter 30 cm) is used to transfer and magnify force from one end to the other. The system is full of water and completely enclosed by two pistons, one on each end. If a force 11.2 N is exerted (additional to the forces already present) on the small piston, then how much additional force is exerted at the larger piston
Answer:
280N
Explanation:
Pascal's law states that the pressure in a fluid is transmitted across every point in the fluid system.
Hence, the pressure in both tubes must remain same;
So, pressure = F1/A1 = F2/A2
where F1 = initial force on small pipe, A1 = area of small pipe
F2 = force on larger piston, A2 = area of larger piston
Given:
F1 = 11.2N
D1 = diameter of smaller pipe = 6 cm
D2 = diameter of larger piston = 15 cm
A1 = π*(r1)² = π*(6/2)² =π*9 (As radius, r = diameter/2)
A2 = π*(r2)² = π*(30/2)² =π*225
Hence 11.2/π*9 = F2/π*225
solving, we have
F2 = 280N
A 1.6-kg block is attached to the end of a 2.0-m string to form a pendulum. The pendulum is released from rest when the string is horizontal. At the lowest point of its swing when it is moving horizontally, the block is hit by a 10-g bullet moving horizontally in the opposite direction. The bullet remains in the block and causes the block to come to rest at the low point of its swing. What was the magnitude of the bullet's velocity just before hitting the block?
Answer:
1002.2688 m/s
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
h = The length of a string = 2 m
m = Mass of block = 1.6 kg
[tex]m_2[/tex] = Mass of bullet = 0.01 kg
Here, the potential energy of the fall will balance the kinetic energy of the bullet
[tex]mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 2}\\\Rightarrow v=6.26418\ m/s[/tex]
Velocity of block is 6.26418 m/s
As the momentum of system is conserved we have
[tex]mv=m_2u\\\Rightarrow u=\dfrac{mv}{m_2}\\\Rightarrow u=\dfrac{1.6\times 6.26418}{0.01}\\\Rightarrow u=1002.2688\ m/s[/tex]
The magnitude of velocity just before hitting the block is 1002.2688 m/s
(FLUID MECHANICS)
The apparent weight of an object when a system is not in equilibrium is represented by:
A. Fg (object)
B. Fnet (system)
C. Fb
D. rhoV
Answer:
option B
Explanation:
When a body is immersed in liquid there will be two force is acting on the body.
First one force acting downward due to weight of the body.
And the second force acting on the object will be buoyant force.
If the object is not in equilibrium the apparent weight will be equal to net force acting on the object.
[tex]F_{net} = W - F_b[/tex]
W is the weight of the object acting downward
Fb is the buoyancy force acting upward on the object.
Hence, the correct answer is option B
You are unloading a refrigerator from a delivery van. The ramp on the van is 5.0 m long, and its top end is 1.4 m above the ground. As the refrigerator moves down the ramp, you are on the down side of the ramp trying to slow the motion by pushing horizontally against the refrigerator with a force of 270N .
Part A
How much work do you do on the refrigerator during its trip down the ramp?
W = ???
Final answer:
The work done on the refrigerator during its trip down the ramp is 1350 J.
Explanation:
To calculate the work done while unloading the refrigerator, we need to determine the displacement of the refrigerator and the force applied.
The displacement is given by the length of the ramp, which is 5.0 m.
The force applied is given as 270 N.
To calculate the work done, we use the formula:
Work = Force x Displacement x Cosine(angle)
In this case, the angle is 0 degrees since the force is applied horizontally.
Therefore, the work done on the refrigerator is:
Work = 270 N x 5.0 m x Cos(0°) = 1350 J.
A 10-kg block on a rough horizontal surface is attached to a light spring (force constant = 1400 N/m). The block is pulled 10.0 cm to the right from its equilibrium position and released from rest. The frictional force between the block and surface has a magnitude of 30 N. What is the kinetic energy of the block as it passes through its equilibrium position?
To develop this problem it is necessary to apply the concepts related to the conservation of Energy. In this case the definition concerning kinetic energy from the simple harmonic movement.
From the conservation of energy we know that the kinetic energy would be conserved through the work done by the frictional force and the simple harmonic potential energy, in other words:
[tex]KE = PE_s +W_f[/tex]
[tex]KE = \frac{1}{2}kA^2 + W_f[/tex]
Where,
[tex]KE =[/tex] Kinetic Energy
[tex]PE_s =[/tex]Potential Harmonic Simple Energy
[tex]W_f =[/tex] Work made by friction.
Our values are given as,
[tex]m = 10Kg \rightarrow[/tex] mass
[tex]k = 1400N/m \rightarrow[/tex] Spring constant
[tex]A = 0.1m \rightarrow[/tex]Amplitude
[tex]f_f = 30N \rightarrow[/tex] Frictional Force
Replacing we have,
[tex]KE = \frac{1}{2}kA^2 + W_f[/tex]
[tex]KE =\frac{1}{2} 1400 * 0.1^2 + ( - 30 * 0.1)[/tex]
[tex]KE = 4 J[/tex]
Therefore the Kinetic Energy of the block as it passes through its equlibrium position is 4J.
The kinetic energy of the 10-kg block as it passes through its equilibrium position is 4 Joules. This is calculated by converting the potential energy stored in the spring to kinetic energy and then subtracted the energy lost due to friction.
Explanation:The first step in solving this problem is to understand the two forces acting on the block in this question: the spring force and the frictional force. The spring potential energy when the block is pulled 10 cm to the right is given by the formula U = 1/2kx^2, where k is the force constant and x is the displacement. Substituting the given values, we have U = 1/2(1400 N/m)(0.1 m)^2 = 7 Joules. This is the initial potential energy stored in the spring. As the block passes through its equilibrium position, this potential energy is fully converted to kinetic energy.
We also need to take into account the work done against frictional force which is equal to the frictional force times the displacement, i.e., W_friction = Friction * displacement = 30N * 0.1m = 3 Joules. This is the energy lost due to friction.
Finally, we subtract the work done by the frictional force from the potential energy to achieve the kinetic energy. Therefore, the kinetic energy of the block as it passes through its equilibrium position is K = U - W_friction = 7J - 3J = 4 Joules.
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Billiard ball A of mass mA = 0.117 kg moving with speed vA = 2.80 m/s strikes ball B, initially at rest, of mass mB = 0.135 kg . As a result of the collision, ball A is deflected off at an angle of θ′A = 30.0∘ with a speed v′A = 2.10 m/s, and ball B moves with a speed v′B at an angle of θ′B to original direction of motion of ball A.Part A Taking the x axis to be the original direction of motion of ball A , choose the correct equation expressing the conservation of momentum for the components in the x direction.A. 0=mAv′Asinθ′A−mBv′Bsinθ′BB. mAvA=mAv′Acosθ′A+mBv′Bcosθ′BC. mAvA=mAv′Acosθ′A−mBv′Bsinθ′BD. 0=(mAvA+mBv′B)sinθ′BPart B Taking the x axis to be the original direction of motion of ball A , choose the correct equation expressing the conservation of momentum for the components in the y direction.A. mAvA=mAv′Acosθ′A−mBv′Bsinθ′BB. 0=(mAvA+mBv′B)sinθ′BC. 0=mAv′Asinθ′A−mBv′Bsinθ′BD. mAvA=mAv′Acosθ′A+mBv′Bcosθ′BPart C Solve these equations for the angle, θ′B , of ball B after the collision. Do not assume the collision is elastic.Part D Solve these equations for the speed, v′B , of ball B after the collision. Do not assume the collision is elastic.
Answer:
Part A:
B) mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B
Part B:
C) 0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B
since vAy = 0 m/s
Part C:
θ′B = tan⁻¹(1.0699) = 46.94°
Part D:
v′B = 1.246 m/s
Explanation:
Given:
mA = 0.117 kg
vA = vAx = 2.80 m/s
mB = 0.135 kg
vB = 0 m/s
θ′A = 30.0°
v′A = 2.10 m/s
Part A: Taking the x axis to be the original direction of motion of ball A, the correct equation expressing the conservation of momentum for the components in the x direction is
B) mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B
Part B: Taking the x axis to be the original direction of motion of ball A, the correct equation expressing the conservation of momentum for the components in the y direction is
C) 0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B
since vAy = 0 m/s
Part C: Solving these equations for the angle, θ′B , of ball B after the collision and assuming that the collision is not elastic:
mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B
⇒ (0.117)(2.80) = (0.117)(2.10)Cos 30° + (0.135)*v′B*cosθ′B
⇒ v′B*cosθ′B = 0.8505 ⇒ v′B = 0.8505/cosθ′B
then
0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B
⇒ 0 = (0.117)(2.10)Sin 30° - (0.135)*v′B*sinθ′B
⇒ v′B*sinθ′B = 0.91 ⇒ v′B = 0.91/sinθ′B
if we apply
0.8505/cosθ′B = 0.91/sinθ′B
⇒ tanθ′B = 0.91/0.8505 = 1.0699
⇒ θ′B = tan⁻¹(1.0699) = 46.94°
Part D: Solving these equations for the speed, v′B , of ball B after the collision and assuming that the collision is not elastic:
if v′B = 0.91/sinθ′B
⇒ v′B = 0.91/sin 46.94°
⇒ v′B = 1.246 m/s
Conservation of momentum can be used to solve the equations for the angle and the speed of ball B after the collision, using the initial conditions, final conditions, and trigonometric identities. This involves the application of physics concepts, combined with the mathematics of trigonometry.
Explanation:Part A: The correct equation expressing the conservation of momentum for the components in the x direction would be option B, mAvA=mAv′Acosθ′A+mBv′Bcosθ′B. This equation shows that the initial momentum of ball A (mAvA) equals the sum of the momentum of ball A and ball B after the collision in the x direction.
Part B: For the components in the y direction, the right answer is C, 0=mAv′Asinθ′A−mBv′Bsinθ′B. Since ball B was initially at rest and ball A was moving along the x-axis, there was no momentum in the y direction before the collision. Therefore, the total momentum in the y direction after the collision should also be 0.
Part C and D: To solve these equations for the angle and the speed of ball B after the collision, you need to use these equations in combination with the conservation of kinetic energy formula (1/2*m*v^2) and the trigonometric identities. Detailed solution steps require knowledge of the involved mathematics.
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For atomic hydrogen, the Paschen series of lines occurs when nf = 3, whereas the Brackett series occurs when nf = 4 in the equation
1/?= 2p2 mk2 e4/ h3c (Z2) (1/n2f - 1/n2i)
Using this equation, show that the ranges of wavelengths in these two series overlap.
Shortest Wavelength (m) Longest Wavelength (m)
Paschen Series
Final answer:
The Paschen and Brackett series of lines in atomic hydrogen can overlap in terms of their wavelength ranges.
Explanation:
The Paschen series of lines in atomic hydrogen occurs when nf = 3, and the Brackett series occurs when nf = 4.
The range of wavelengths in these two series can overlap because the wavelength equation 1/λ = 2π²mk²e⁴/(h³c)(Z²)(1/nf² - 1/ni²) depends on the values of nf and ni, which can be different for each series
For example, if nf = 3 in the Paschen series and nf = 4 in the Brackett series, the wavelengths in these two series can overlap depending on the values of ni.
A visible disturbance propagates around a crowded soccer stadium when fans, section by section, jump up and then sit back down.
Answer:
Transverse wave formation.
Explanation:
This type of wave formation is called transverse wave formation.
Transverse wave is a wave arising when the medium's pulse is perpendicular to the wave's path of propagation. The crowd was standing vertically, but they were arranged horizontally in the stadium, according to the question. Hence the wave form was transverse.
The free-fall acceleration on Mars is 3.7 m/s2.
(a) What length of pendulum has a period of 1.2 s on Earth? cm
(b) What length of pendulum would have a 1.2-s period on Mars? cm An object is suspended from a spring with force constant 10 N/m.
(c) Find the mass suspended from this spring that would result in a period of 1.2 s on Earth. kg
(d) Find the mass suspended from this spring that would result in a period of 1.2 s on Mars.
Answer:
(a) The length of the pendulum on Earth is 36.8cm
(b) The length of the pendulum on Mars is 13.5cm
(c) Mass suspended from the spring on Earth is 0.37kg
(d) Mass suspended from the spring on Mars is 0.36kg
Explanation:
Period = 1.2s, free fall acceleration on Earth = 9.8m/s^2, free fall acceleration on Mars = 3.7m/s^2
( a) Length of pendulum on Earth = [( period ÷ 2π)^2] × acceleration = (1.2 ÷ 2×3.142)^2 × 9.8 = 0.0365×9.8 = 0.358m = 35.8cm
(b) Length of the pendulum on Mars = (1.2÷2×3.142)^2 × 3.7 = 0.0365×3.7 = 0.135cm = 13.5m
(c) Mass suspended from the spring on Earth = (force constant×length in meter) ÷ acceleration = (10×0.358) ÷ 9.8 = 0.37kg
(d) Mass suspended from the spring on Mars = (10×0.135)÷3.7 = 0.36kg
The length of a pendulum with a period of 1.2 s on Earth is approximately 36.95 cm, while on Mars it is around 16.99 cm. The mass suspended from a spring that would result in a period of 1.2 s on Earth is approximately 0.722 kg, and on Mars it is approximately 0.329 kg.
Explanation:(a) What length of pendulum has a period of 1.2 s on Earth? cm
Using the equation for the period of a pendulum, T = 2π √(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity, we can solve for L. Rearranging the equation, we have L = (T/2π)² * g.
Given that the free-fall acceleration on Earth is approximately 9.8 m/s², substituting the values into the equation, we have:
L = (1.2/2π)² * 9.8 = 0.3695 m = 36.95 cm
(b) What length of pendulum would have a 1.2-s period on Mars? cm
Using the same equation, L = (T/2π)² * g, we can substitute the values for the period and acceleration due to gravity on Mars:
L = (1.2/2π)² * 3.7 = 0.1699 m = 16.99 cm.
(c) Find the mass suspended from this spring that would result in a period of 1.2 s on Earth. kg
For a spring-mass system, the period is given by T = 2π √(m/k), where T is the period, m is the mass, and k is the spring constant. Rearranging the equation, we have m = (T/2π)² * k.
Given that the spring constant is 10 N/m, substituting the values into the equation, we have:
m = (1.2/2π)² * 10 = 0.722 kg.
(d) Find the mass suspended from this spring that would result in a period of 1.2 s on Mars. kg
Using the same equation, m = (T/2π)² * k, we can substitute the values for the period and spring constant:
m = (1.2/2π)² * 10 = 0.329 kg.
A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and a layer of Styrofoam insulation 2.4 cm thick on the inside wall surface. The wood has k=0.080W/(m?K), and the Styrofoam has k= 0.010 W/(m?K). The interior surface temperature is 20.0 ?C , and the exterior surface temperature is -13.0 ?C
A.)What is the temperature at the plane where the wood meets the Styrofoam? _______ Celsius
B.)What is the rate of heat flow per square meter through this wall? ______W/m^2
Answer:
Explanation:
Given
thermal conductivity of wood [tex] K_w=0.08 W/m^2-K[/tex]
thermal conductivity of insulation [tex]K_i=0.01 W/m^2-K[/tex]
thickness of wood [tex]t_2=3 cm[/tex]
thickness of insulation [tex]t_1=2.4 cm[/tex]
[tex]T_i=20^{\circ}C[/tex]
[tex]T_o=-13^{\circ}C[/tex]
we know heat Flow is given by
[tex]Q=kA\frac{dT}{dx}[/tex]
[tex]dT=[/tex] change in temperature
[tex]dx=[/tex] thickness
K=thermal conductivity
A=Area of cross-section
A is same
Suppose T is the temperature of Junction
as heat Flow is same thus
[tex]\frac{k_w(20-T)}{3}=\frac{k_i(T-(-13))}{2.4} [/tex]
[tex]\frac{0.08(20-T)}{3}=\frac{0.01(T+13)}{2.4}[/tex]
[tex]T=19.36 ^{\circ}C[/tex]
(b)Rate of heat flow
[tex]Q=\frac{k_w(T+13)}{3\times 10^{-2}}[/tex]
[tex]Q=\frac{0.08\times 32.36}{0.03}[/tex]
[tex]Q=86.303 W/m^2[/tex]
The temperature at which the wood meets the Styrofoam is determined using the proportional temperature drops across each material, considering their respective thermal resistances. The rate of heat flow per square meter through the wall is calculated using Fourier's law, considering the combined thermal resistance of both layers.
Explanation:Temperature at the Wood-Styrofoam Plane and Heat Flow Rate
To find the temperature at the plane where the wood meets the Styrofoam, we need to use the concept of thermal resistance and the fact that the heat flow through both materials is the same. With the given thermal conductivity (k values) and thicknesses of the wood and Styrofoam, we can calculate their respective thermal resistances (R = thickness/k). Then, by setting up a proportion, we can find the temperature drop across the wood (ΔTw) and the Styrofoam (ΔTs) using the formula ΔT = (k*A* ΔT)/(thickness), where A is the area (which cancels out as it's the same for both layers). Knowing the temperature drop across each and the outer and inner surface temperatures, we can find the temperature at the plane where the layers meet.
To calculate the rate of heat flow per square meter through this wall, we'll use Fourier's law of thermal conduction, which is given by Q = k*A* ΔT/d, where Q is the heat flow rate, A is the area, ΔT is the temperature difference, and d is the thickness. Since the wall consists of two layers with different thermal conductivities and thicknesses, we need to calculate the equivalent thermal resistance for the combined wall and then use the overall temperature difference to find the heat flow rate.
A particle moves according to the law of motion s(t) = t^{3}-8t^{2}+2t, t\ge 0, where t is measured in seconds and s in feet. a.) Find the velocity at time t. Answer: b.) What is the velocity after 3 seconds? Answer: c.) When is the particle at rest? Enter your answer as a comma separated list. Enter None if the particle is never at rest. At t_1= and t_2= with t_1
Answer:
Explanation:
Given
displacement is given by
[tex]s(t)=t^3-8t^2+2t[/tex]
so velocity is given by
[tex]v(t)=\frac{\mathrm{d} s(t)}{\mathrm{d} t}[/tex]
[tex]v(t)=3t^2-16t+2[/tex]
(b)velocity after [tex]t=3 s[/tex]
[tex]v(3)=3(3)^2-8\cdot 3+2[/tex]
[tex]v(3)=19 m/s[/tex]
(c)Particle is at rest
when its velocity will become zero
[tex]v(t)=0[/tex]
i.e. [tex]3t^2-16t+2=0[/tex]
[tex]t=\frac{16\pm \sqrt{16^2-4\cdot 3\cdot 2}}{2\cdot 3}[/tex]
[tex]t=\frac{16\pm 15.23}{6}[/tex]
[tex]t=5.20 s[/tex]
A uniform disk, a uniform hoop, and a uniform solid sphere are released at the same time at the top of an inclined ramp. They all roll without slipping. In what order do they reach the bottom of the ramp?A uniform disk, a uniform hoop, and a uniform solid sphere are released at the same time at the top of an inclined ramp. They all roll without slipping. In what order do they reach the bottom of the ramp?hoop, sphere, diskhoop, disk, spheredisk, hoop, spheresphere, disk, hoopsphere, hoop, disk
Answer:
First let's write down the moment of inertia of the objects.
[tex]I_{sphere} = \frac{2}{5}mR^2\\I_{disk} = \frac{1}{2}mR^2\\I_{hoop} = mR^2[/tex]
If they all roll without slipping, then the following relation is applied to all ot them:
[tex]v = \omega R[/tex]
where v is the translational velocity and ω is the rotational velocity.
We will use the conservation of energy, because we know that their initial potential energies are the same. (Here, I will assume that all the objects have the same mass and radius. Otherwise we couldn't determine the difference. )
[tex]K_1 + U_1 = K_2 + U_2\\0 + mgh = \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2[/tex]
For sphere:
[tex]\frac{1}{2}\frac{2}{5}mR^2(\frac{v}{R})^2 + \frac{1}{2}mv^2 = mgh\\\frac{1}{5}mv^2 + \frac{1}{2}mv^2 = mgh\\\frac{7}{10}mv^2 = mgh\\v_{sphere} = \sqrt{\frac{10gh}{7}}[/tex]
For disk:
[tex]\frac{1}{2}\frac{1}{2}mR^2(\frac{v}{R})^2 + \frac{1}{2}mv^2 = mgh\\\frac{1}{4}mv^2 + \frac{1}{2}mv^2 = mgh\\\frac{3}{4}mv^2 = mgh\\v_{disk} = \sqrt{\frac{4gh}{3}}[/tex]
For hoop:
[tex]\frac{1}{2}mR^2(\frac{v}{R})^2 + \frac{1}{2}mv^2 = mgh\\\frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mgh\\mv^2 = mgh\\v_{hoop}= \sqrt{gh}[/tex]
The sphere has the highest velocity, so it arrives the bottom first. Then the disk, and the hoop arrives the last.
Explanation:
The moment of inertia can be defined as the resistance to the rotation. If an object has a high moment of inertia, it resist to rotate more so its angular velocity would be lower. In the case of rolling without slipping, the angular velocity and the linear (translational) velocity are related by the radius, so the object with the highest moment of inertia would arrive the bottom the last.
The order in which a sphere, disk, and hoop reach the bottom of an incline when released from the same height is determined by their moments of inertia. The solid sphere arrives first, followed by the disk, and finally the hoop.
Explanation:In the scenario where a uniform disk, a uniform hoop, and a uniform solid sphere are released from the top of an inclined ramp and roll without slipping, the order in which they reach the bottom depends on their moments of inertia and the distribution of mass. The solid sphere has the smallest moment of inertia relative to its mass (I = 2/5 MR²), which means it will accelerate faster than the other shapes and hence get to the bottom first. The uniform disk, with a moment of inertia of I = 1/2 MR², will follow. The uniform hoop has the largest moment of inertia (I = MR²) for a given mass and radius, so it will accelerate the slowest of the three and reach the bottom last.
Therefore, the objects reach the bottom of the ramp in the following order: sphere, disk, hoop.