Answer:
a) [tex]\angle r_{ag}=38.79^{\circ}[/tex]
b) [tex]\angle r_{gw}=44.95^{\circ}[/tex]
c) not possible
Explanation:
Given:
angle of incidence on the air-glass interface, [tex]\angle i_{ag}=90-20=70^{\circ}[/tex]
refractive index of glass with respect to air, [tex]n_g=1.5[/tex]
refractive index of water with respect to air, [tex]n_a=1.33[/tex]
a)
For angle of refraction in glass we use Snell's law:
[tex]n_g=\frac{sin\ i_{ag}}{sin\ r{ag}}[/tex]
[tex]1.5=\frac{sin\ 70}{sin\ r_{ag}}[/tex]
[tex]\angle r_{ag}=38.79^{\circ}[/tex]
b)
Now we have angle of incident for glass-water interface, [tex]\angle i_{gw}=\angle r_{ag}=38.79^{\circ}[/tex]
And the refractive index of water with respect to glass:
[tex]n_{gw}=\frac{n_w}{n_g}[/tex]
[tex]n_{gw}=\frac{1.33}{1.5}[/tex]
[tex]n_{gw}=0.8867 [/tex]
Therefore, angle of refraction in the water:
[tex]n_{gw}=\frac{sin\ i_{gw}}{sin\ r_{gw}}[/tex]
[tex]0.8867 =\frac{sin\ 38.79}{sin\ r_{gw}}[/tex]
[tex]\angle r_{gw}=44.95^{\circ}[/tex]
c)
For total internal reflection through water the light must enter the glass-water interface at an angle greater than the critical angle.
So,
[tex]n_{gw}=\frac{sin\ i_{(gw)_c}}{sin\ 90}[/tex]
[tex]0.8867 =\frac{sin\ i_{(gw)_c}}{1}[/tex]
[tex]i_{(gw)_c}=62.46^{\circ}[/tex]
Now this angle will become angle of refraction for the air-glass interface.
Hence,
[tex]n_g=\frac{sin\ i_{ag}}{sin\ i_{(gw)_c}}[/tex]
[tex]1.5=\frac{sin\ i_{ag}}{0.8867 }[/tex]
[tex]sin\ i_{ag}=1.33005[/tex]
Since we do not have any angle for sine for which the value exceeds 1, therefore it is not possible that the light will reflect from the water.
A 0.6-m3 rigid tank contains 0.6 kg of N2 and 0.4 kg of O2 at 300 K. Determine the partial pressure of each gas and the total pressure of the mixture. The gas constant for N2 is 0.2968 kPa·m3/kg·K and the gas constant for O2 is 0.2598 kPa·m3/kg·K.
Answer:
Pnitrogen=3.18 kPa, Poxygen=1.62 kPa , Ptotal= 4.80 kPa
Explanation:
partial pressure equation becomes Ptotal = Pnitrogen + Poxygen
Partial pressure of Nitrogen
Pnitrogen= nRT/V
n=no of moles =mass/molar mass
mass of nitrogen=0.6kg
Molar mass of nitrogen gas=28gmol^-1
n=0.6/28=0.0214moles
R=0.2968 kPa·m3/kg·K
T=300k
V=0.6m^3
Pnitrogen=(0.0214 * 0.2968 * 300)/0.6
Pnitrogen=3.18 kPa
Likewise
Poxygen=nRT/V
n=0.4/32=0.0125moles
R=0.2598 kPa·m3/kg·K
T=300k
V=0.6m^3
Poxygen=(0.0125 * 0.2598 * 300)/0.6
Poxygen=1.62 kPa
Ptotal= 3.18+1.62= 4.80 kPa
Using the ideal gas law, the partial pressure for Nitrogen ([tex]N_{2}[/tex]) and Oxygen ([tex]O_{2}[/tex]) is 148.4 kPa and 86.6 kPa respectively. The total pressure of the mixture is the sum of these two partial pressures, equaling 235 kPa.
Explanation:The pressure exerted by individual gases in a mixture is known as partial pressure. The calculation of the partial pressure of each gas, and the total pressure of the mixture involves using the ideal gas law. In this case, the ideal gas equation is P = mRT/V, where P represents the pressure, m is the mass of gas, R is the gas constant, T is the temperature and V is the volume. Thus, for Nitrogen ([tex]N_{2}[/tex]), the partial pressure is (0.6 kg × 0.2968 kPa·m3/kg·K × 300K) / 0.6 m3 = 148.4 kPa, and for Oxygen ([tex]O_{2}[/tex]), the partial pressure is (0.4 kg× 0.2598 kPa·m3/kg·K × 300K) / 0.6 m3 = 86.6 kPa. The total pressure, then, is the sum of these partial pressures, which equals 148.4 kPa + 86.6 kPa = 235 kPa.
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what consistent physiological pattern are more common in women's teeth? A. Women's canines are generally sharper. B. Women have straighter teeth. C. Women have 1 less set of molar teeth. D. Women normally have longer pre-molars.
Answer:
B
Explanation:
B. Women have straighter teeth.
A circuit with current increasing at a rate of 4 A/s contains an inductor, L. If the induced emf is -2 V, what is the inductance of the inductor?
Answer:
L=500 mH
Explanation:
Here di/dt = 4A/s, ε= -2V
Inductance of inductor, induced emf and rate of change of current have the following relation.
ε= [tex] - L\frac{di}{dt}[/tex]
⇒L= - ε/[tex]\frac{di}{dt}[/tex]
⇒L= -(-2)/ 4
⇒ L= 0.5 H or
⇒ L= 500 mH
A titanium (c = 520 J/kg*K) satellite of mass m = 500 kg at a temperature of 10 K is in geostationary orbit above the equator. It is impacted by a meteorite, which knocks it off course but does not affect its speed. Because of the change in trajectory, the satellite will now crash-land on Earth in a small pond containing 5 × 10^5 kg of water (c = 4186 J/kg*K) at 20°C. Ignoring air resistance, and assuming that the water stays in the pond somehow, what is the final temperature of the water? Please indicate if any water boils, and justify any simplifying assumptions you make in your solution.
Bandar Industries Berhad of Malaysia manufactures sporting equipment. One of the company’s products, a football helmet for the North American market, requires a special plastic. During the quarter ending June 30, the company manufactured 3,200 helmets, using 2,368 kilograms of plastic. The plastic cost the company $15,629. According to the standard cost card, each helmet should require 0.68 kilograms of plastic, at a cost of $7.00 per kilogram. Required: 1. What is the standard quantity of kilograms of plastic (SQ) that is allowed to make 3,200 helmets? 2. What is the standard materials cost allowed (SQ × SP) to make 3,200 helmets? 3. What is the materials spending variance? 4. What is the materials price variance and the materials quantity variance? (For requirements 3 and 4, indicate the effect of each variance by selecting "F" for favorable, "U" for unfavorable, and "None" for no effect (i.e., zero variance). Input all amounts as positive values. Do not round intermediate calculations.)
Answer:
1. 2176 kilograms of plastic 2. $15,232 3. $397 (U) 4. $947 (F) $1,344 (U)
Explanation:
Generally, cost variance analysis can be used to estimate the difference between the actual cost and the expected costs. However, if the actual cost is more than the expected cost, then the variance is said to be unfavorable and vice versa.
1. The standard quantity of kilograms of plastic (SQ) that is allowed to make 3,200 helmets?
We know that each helmet requires 0.68 kilograms of plastic. Thus, to make 3200 helmets, we will need 0.68*3200 = 2176 kilograms of plastic
2. The standard materials cost allowed (SQ × SP) to make 3,200 helmets.
We also know that each helmet costs $7.00 per kilogram. Therefore, to make 3200 helmets, the standard materials cost = $7.00*2176 = $15,232
3. The material spending variance = difference between the actual cost and the standard cost = $15,629 - $15,232 = $397 U
4. The materials price variance and the materials quantity variance?
The materials price variance is the actual cost- (standard cost per kilogram x actual number of plastic used) . Therefore:
Materials price variance = $15,629 - ($7 x 2,368 kg)
Materials price variance = $15,629 - $16,576 = ($947) F
Since the budgeted cost is relatively higher than the actual cost, the materials price variance is favorable (F) by $947.
The materials quantity variance = (Actual number of plastic used x Standard cost per kilogram) - Standard cost
Materials quantity variance = ($7 x 2,368 kg) - $15,232
Materials quantity variance = $16,576 - $15,232 = $1,344 U
Since the budgeted cost is relatively higher than the standard cost, the materials quantity variance is unfavorable (U) by $1,344.
To find the standard quantity of plastic and standard materials cost allowed, multiply the standard quantity per helmet by the number of helmets produced and multiply the standard quantity by the standard price per kilogram, respectively. The materials spending variance is the difference between the actual cost and the standard cost that should have been incurred. The materials price variance is the difference between the actual quantity of plastic used multiplied by the standard price per kilogram and the actual cost, while the materials quantity variance is the difference between the standard quantity of plastic allowed multiplied by the standard price per kilogram and the actual cost.
Explanation:To calculate the standard quantity of plastic (SQ) allowed to make 3,200 helmets, multiply the standard quantity per helmet by the number of helmets produced. In this case, SQ = 0.68 kg/helmet * 3,200 helmets = 2,176 kg. To calculate the standard materials cost allowed (SQ × SP), multiply the standard quantity by the standard price per kilogram. In this case, the materials cost allowed is $15,232 (2,176 kg * $7.00/kg).
To calculate the materials spending variance, subtract the actual cost from the standard cost that should have been incurred. In this case, the materials spending variance is $397 (Standard Cost - Actual Cost = $15,232 - $15,629). The materials price variance is calculated by subtracting the actual quantity of plastic used multiplied by the standard price per kilogram from the actual cost.
The materials quantity variance is calculated by subtracting the standard quantity of plastic allowed multiplied by the standard price per kilogram from the actual cost. In this case, the materials price variance is $7,774 (Actual Cost - Actual Quantity * Standard Price = $15,629 - 2,368 kg * $7.00/kg) and the materials quantity variance is -$2,145 (Actual Quantity * Standard Price - Standard Cost = 2,368 kg * $7.00/kg - $15,232).
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You are lowering two boxes, one on top of the other, down a ramp by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 15.0 cm/s. The coefficient of kinetic friction between the ramp and the lower box is 0.444, and the coefficient of static friction between the two boxes is 0.800.
(a) What force do you need to exert to accomplish this? (b) What are the magnitude and direc- tion of the friction force on the upper box?
Answer:
(a) 57.17 N
(b) 146.21 N up the ramp
Explanation:
Assume the figure attached
(a)
The angle of ramp, [tex]\theta=tan ^{-1} \frac {2.5}{4.75}=27.75854^{\circ}\approx 27.76^{\circ}[/tex]
[tex]N=(32+48)*9.81*cos 27.76^{\circ}=694.4838 N[/tex]
m=32+48=80 kg
[tex]T+ \mu N= mg sin \theta[/tex]
[tex]T=mg sin \theta - \mu N= mg sin \theta- \mu mg cos \theta= mg (sin \theta - \mu cos \theta) [/tex] where [tex]\mu[/tex] is coefficient of kinetic friction
[tex]T=80*9.81(sin 27.76^{\circ} -0.444 cos 27.76^{\circ})=57.16698 N \approx 57.17 N[/tex]
(b)
Upper box doesn’t accelerate
[tex]F_r= mgsin\theta= 32*9.81sin 27.76^{\circ}=146.2071\approx 146.21 N[/tex]
The direction will be up the ramp
To lower the boxes down the ramp, a force is needed that is equal to the force of kinetic friction on the lower box. The magnitude and direction of the friction force on the upper box is the same as the force of kinetic friction on the lower box.
Explanation:To determine the force required to lower the two boxes down the ramp, we can use the equation: force = friction force + weight force. Since the boxes are moving at a constant speed, the force of kinetic friction on the lower box is equal to the force applied to it. Therefore, the force needed to lower the boxes is equal to the force of kinetic friction on the lower box which can be calculated using the equation: force = coefficient of kinetic friction × normal force.
For the magnitude and direction of the friction force on the upper box, we can use the equation: friction force = coefficient of static friction × normal force. Since the boxes are moving together, we can assume that the friction force on the upper box is equal to the force of kinetic friction on the lower box. Therefore, the magnitude and direction of the friction force on the upper box is the same as the force of kinetic friction on the lower box.
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The ballistic pendulum consists of a block of dense material suspended from a cord and allowed to swing freely. A bullet is fired into the block, lodging itself into the block in a __________________. The (block+bullet) having now acquired a velocity, swings upward to some maximum height. (See lab manual)
The subject of this question is Physics and it pertains to the concept of a ballistic pendulum.
Explanation:The subject of the question is Physics. Specifically, it is related to the concept of a ballistic pendulum.
A ballistic pendulum consists of a block of dense material suspended from a cord. When a bullet is fired into the block, it lodges itself into the block in a specific manner, either elastically or by sticking together. The (block+bullet) swings upward to some maximum height due to the acquired velocity.
For example, if the block and bullet stick together after the collision, the system forms a ballistic pendulum. The bullet's initial kinetic energy is converted into potential energy as the (block+bullet) rises.
The ballistic pendulum consists of a block of dense material suspended from a cord and allowed to swing freely. A bullet is fired into the block, lodging itself into the block in a sticky collision. The (block+bullet) having now acquired a velocity, swings upward to some maximum height.
A ballistic pendulum is a device that measures the speed of a projectile, such as a bullet. When a bullet of mass m is fired into a block of mass M suspended from a cord, it lodges itself into the block in a sticky collision. The combined mass then swings to a maximum height h, where the kinetic energy is converted into potential energy.
First, understand that the bullet embeds itself in the block. This scenario is a completely inelastic collision, also known as a sticky collision.Upon impact, the kinetic energy of the bullet and block system is high, which gets converted into potential energy at the maximum height of the swing.To find the initial speed of the bullet use the conservation of momentum and energy principles:1. Conservation of momentum: m[tex]v_{i}[/tex] = (m + M)[tex]v_{f}[/tex]
2. Conservation of energy: 1/2(m + M)[tex]v_{f}^{2}[/tex] = (m + M)gh.
From these equations, [tex]v_{i}[/tex] can be derived:
[tex]v_{i}[/tex] = (1 + M/m) √(2gh)
Thus, the ballistic pendulum helps measure the speed of the bullet before the collision.
A particle of mass M moves along a straight line with initial speed vi. A force of magnitude F pushes the particle a distance D along the direction of its motion.
By what multiplicative factor RK does the initial kinetic energy increase, and by what multiplicative factor RWdoes the work done by the force increase (with respect to the case when the particle had a mass M)?
If one of the quantities doubles, for instance, it would increase by a factor of 2. If a quantity stays the same, then the multiplicative factor would be 1.
Answer:
Explanation:
Initial kinetic energy of M = 1/2 M vi²
let final velocity be vf
v² = u² + 2a s
vf² = vi² + 2 (F / M) x D
Kinetic energy
= 1/2 Mvf²
= 1/2 M ( vi² + 2 (F / M) x D
1/2 M vi² + FD
Ratio with initial value
1/2 M vi² + FD) / 1/2 M vi²
RK = 1 + FD / 2 M vi²
A row of five 1.2 N wooden blocks is being pushed across a tabletop at a constant speed by a toy tractor that exerts a force of 1.7 N on the row. What is the coefficient of kinetic friction between the wooden blocks and the tabletop?
To solve this problem we will apply the concepts related to the Kinetic Friction Force for which we define as
[tex]F = \mu_k N[/tex]
Where,
N = Normal Force (Mass for gravity)
[tex]\mu_k =[/tex] Kinetic frictional coefficient
The total force applied is 1.7N and the Force from the (normal) weight is equivalent to five times 1.2N, therefore:
[tex]F = 5(N)\mu_k[/tex]
[tex]1.7 = 5(1.2)(\mu_k)[/tex]
[tex]\mu_k = \frac{1.7}{5*1.2}[/tex]
[tex]\mu_k =0.283[/tex]
Therefore the coefficient of kinetic friction between the wooden blocks and the tabletop is 0.283
A continuous and aligned fiber reinforced composite having a cross-sectional area of 1130 mm^2 (1.75 in.^2) is subjected to an external tensile load. If the stresses sustained by the fiber and matrix phases are 156 MPa (22,600 psi) and 2.75 MPa (400 psi), respectively; the force sustained by the fiber phase is 74,000 N (16,600 lbf); and the total longitudinal strain is 1.25 10^ -3, determine: (a)the force sustained by the matrix phase, (b)the modulus of elasticity of the compos- ite material in the longitudinal direction, and (c) the moduli of elasticity for fiber and matrix phases.
Answer:
a) [tex]F_m=1803.013\ N[/tex]
b) [tex]E=53665.84\ MPa[/tex]
c) [tex]E_f=124800\ MPa[/tex]
[tex]E_m=2200\ MPa[/tex]
Explanation:
Given:
cross-sectional area of reinforced composite, [tex]A=1130\ mm^2[/tex]stress sustained by the fiber phase, [tex]\sigma_f=156\ MPa[/tex]force sustained by the fiber phase, [tex]F_f=74000\ N[/tex]Total strain on the composite, [tex]\epsilon=1.25\times 10^{-3}[/tex]stress sustained in the matrix phase, [tex]\sigma_m=2.75\ MPa[/tex]Now, the area of fiber phase:
[tex]A_f=\frac{F_f}{\sigma_f}[/tex]
[tex]A_f=\frac{74000}{156}[/tex]
[tex]A_f=474.359\ mm^2[/tex]
∴Area of matrix phase:
[tex]A_m=A-A_f[/tex]
[tex]A_m=1130-474.359[/tex]
[tex]A_m=655.641\ mm^2[/tex]
(a)
Now the force sustained by the matrix phase:
[tex]F_m=\sigma_m\times A_m[/tex]
[tex]F_m=2.75\times 655.641[/tex]
[tex]F_m=1803.013\ N[/tex]
(b)
Total stress on the composite:
[tex]\sigma=\frac{(F_f+F_m)}{A}[/tex]
[tex]\sigma=\frac{(74000+1803.013)}{1130}[/tex]
[tex]\sigma=67.082\ MPa[/tex]
Now,Modulus of elasticity of the composite:
[tex]E=\frac{\sigma}{\epsilon}[/tex]
[tex]E=\frac{67.082}{1.25\times 10^{-3}}[/tex]
[tex]E=53665.84\ MPa[/tex]
(c)
Since, strain will be same in this case throughout the material.
Now the modulus of elasticity of fiber phase:
[tex]E_f=\frac{\sigma_f}{\epsilon}[/tex]
[tex]E_f=\frac{156}{1.25\times 10^{-3}}[/tex]
[tex]E_f=124800\ MPa[/tex]
Now the modulus of elasticity of matrix phase:
[tex]E_m=\frac{\sigma_m}{\epsilon}[/tex]
[tex]E_m=\frac{2.75}{1.25\times 10^{-3}}[/tex]
[tex]E_m=2200\ MPa[/tex]
The force sustained by the matrix phase is 3125 N, the modulus of elasticity of the composite material in the longitudinal direction is 124800 MPa, and the moduli of elasticity for the fiber and matrix phases cannot be calculated without information about the volume fractions.
Explanation:To determine the force sustained by the matrix phase, we can use the equation:
Force sustained by matrix phase = Stress in matrix phase * Cross-sectional area
Using given values, the force sustained by the matrix phase is:
Force sustained by matrix phase = 2.75 MPa * 1130 mm2 = 3125 N
To determine the modulus of elasticity of the composite material in the longitudinal direction, we can use the equation:
Modulus of elasticity = Stress / Strain
Using given values, the modulus of elasticity of the composite material in the longitudinal direction is:
Modulus of elasticity = 156 MPa / (1.25 x 10-3) = 124800 MPa
To determine the moduli of elasticity for the fiber and matrix phases, we can multiply the modulus of elasticity of the composite material by the volume fractions of the fiber and matrix phases. In this case, we don't have information about the volume fractions, so we cannot calculate the moduli of elasticity for the fiber and matrix phases.
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A(n) 59 kg astronaut becomes separated from
the shuttle, while on a spacewalk. She finds
herself 63.2 m away from the shuttle and moving with zero speed relative to the shuttle. She
has a(n) 0.695 kg camera in her hand and decides to get back to the shuttle by throwing
the camera at a speed of 12 m/s in the direction away from the shuttle.
How long will it take for her to reach the
shuttle? Answer in minutes.
1. 10.06 min
2. 9.687 min
3. 8.569 min
4. 8.197 min
5. 7.452 min
6. 7.824 min
7. 10.43 min
8. 8.942 min
Answer:
5. 7.452 min
Explanation:
Momentum is conserved, so:
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
0 = (0.695 kg) (-12 m/s) + (59 kg) v
v = 0.1414 m/s
She is 63.2 m away, so the time it takes to reach the shuttle is:
t = 63.2 m / 0.1414 m/s
t = 447.1 s
t = 7.452 min
To find how long it will take for the astronaut to reach the shuttle, the conservation of momentum is used to calculate her velocity after she throws the camera. She will travel at a velocity of 0.141 m/s. It will take her approximately 7.47 minutes to cover the 63.2-meter distance to the shuttle.
Explanation:The given scenario can be solved using the principle of conservation of momentum. The momentum of the system (the astronaut and the camera) before and after the camera is thrown must be equal since there are no external forces acting on the system. The momentum of the astronaut can be determined using the equation p = m × v, where p is momentum, m is mass, and v is velocity.
To find the velocity of the astronaut after throwing the camera, we use the following relationship:
Solving for the velocity of the astronaut, we get:
0 = (59 kg × velocity of astronaut) + (0.695 kg × (-12 m/s))
Velocity of astronaut = (0.695 kg × 12 m/s) / 59 kg = 0.141 m/s
To calculate how long it will take for the astronaut to reach the shuttle, we divide the distance by her velocity:
Time = Distance / Velocity = 63.2 m / 0.141 m/s = 448.227 s
Convert seconds to minutes:
Time in minutes = 448.227 s / 60 = 7.47 minutes
The closest answer from the options provided is 7.452 min, but rounding should be done judiciously, and the calculated value is between 7.452 min and 7.824 min, so it's important to check the context in which the answer will be used, as the rounding might affect the selection of the correct option.
An average nerve axon is about 4x10-6m in radius, and the axoplasm that composes the interior of the axon has a resistivity of about 2.3 Ω ⋅m.
What is the resistance of just 2-cm length of this axon?
Provide your answer in mega-ohms (1 mega-ohm = 106 ohms or "millions of ohms").
(Note, this value is so large -- it corresponds to the resistance of tens of thousands of miles of the thinnest copper wire normally manufactured(!) -- that it explains why a nerve pulse traveling down an axon CANNOT simply be a current traveling along the axon. The voltage required to achieve a perceptible current in the axon would have to be gigantic! We will investigate how voltage pulses -- not current -- travel down axons in a future lab.)
For the calculation of resistance there are generally two paths. The first is through Ohm's law and the second is through the relationship
[tex]R = \frac{pL}{A}[/tex]
Where
p = Specific resistance of material
L = Length
A = Area
The area of nerve axon is given as
[tex]A = \pi r^2[/tex]
[tex]A = \pi (5*10^{-6})^2[/tex]
[tex]A = 7.854*10^{-11}m^2[/tex]
The rest of values are given as
[tex]p= 2 \Omega\cdot m[/tex]
[tex]L = 2cm = 0.02m[/tex]
Therefore the resistance is
[tex]R = \frac{pL}{A}[/tex]
[tex]R = \frac{2*0.02}{7.854*10^{-11}}[/tex]
[tex]R = 509.3*10^6\Omega[/tex]
[tex]R = 509.3M\Omega[/tex]
We had a homework problem in which the Arrhenius equation was applied to the blinking of fireflies. Several other natural phenomena also obey that equation, including the temperature dependent chirping of crickets. A particular species, the snowy tree cricket, has been widely studied. These crickets chirp at a rate of 178 times per minute at 25.0°C, and the activation energy for the chirping process is 53.9 kJ/mol. What is the temperature if the crickets chirp at a rate of 126 times per minute?
Answer:
Temperature = 20.35 °C
Explanation:
Arrhenius equation is as follows:
k = A*exp(-Ea/(R*T)) , where
k = rate of chirps
Ea = Activation Energy
R = Universal Gas Constant
T = Temperature (in Kelvin)
A = Constant
Given Data
Ea = 53.9*10^3 J/mol
R = 8.3145 J/(mol.K)
T = 273.15 + 25 K
k = 178 chirps per minutes
Calculation
Using the Arrhenius equation, we can find A,
A= 4.935x10^11
Now we can apply the same equation with the data below to find T at k=126,
k = A*exp(-Ea/(R*T))
Ea = 53.9*10^3
R = 8.3145
k = 126
T = 20.35 °C
A 100 kg box is hanging from two strings. String #1 pulls up and left, making an angle of 80o with the horizontal on the left, and string #2 pulls up and to the right, making an angle of 65o with the horizontal on the right. FInd the tension in string #2 (on the right side)
Answer:296.76 N
Explanation:
Given
mass of box [tex]m=100 kg[/tex]
Let [tex]T_1[/tex] be the Tension in left side and [tex]T_2[/tex] be the Tension in the right side
From diagram
[tex]T_1\cos 80=T_2\cos 65[/tex]
[tex]T_1=T_2\cdot \frac{\cos 65}{\cos 80}[/tex]
and
[tex]T_1\sin 80+T_2\sin 65=100\cdot g[/tex]
[tex]T_2\left [ \tan 80\cdot \cos 65+\sin 65\right ]=100\cdot g[/tex]
[tex]T_2=\frac{100\cdot g}{\left [ \tan 80\cdot \cos 65+\sin 65\right ]}[/tex]
[tex]T_2=\frac{980}{3.3023}=296.76 N[/tex]
A concrete pump of 120-cy/hr max. output is used to place 250 cy of concrete. It is supplied by 8 cy transit-mix trucks arriving to the site every 15 mins. Barring interruptions how many hours will the operation require, considering 80% efficiency factor.
Answer:
Completing the operation will require 9.765625 hours.
Explanation:
First, let's check which of the stages will be the process limitation:
Regarding the pump, it can perform up to 120 cy/h. The trucks will deliver 8 cy every 15 min, i.e., 32 cy/h, then this is the smaller capacity within the whole process.
Now, let's establish a simple relation:
Placement speed x Efficiency = Placed quantity / time,
solving for time. we have
[tex]t = \frac{q}{\epsilon s} = \frac{250 cy}{0.8\times 32\frac{cy}{h}} = \mathbf{9.765625 h}.[/tex]
The operation will require approximately 2.6 hours, taking into account the efficiency factor.
Explanation:To determine how many hours the operation will require, we need to calculate the total time it takes for all the trucks to arrive at the site and unload the concrete. Since each truck delivers 8 cy of concrete every 15 mins, we can calculate the time it takes for all the trucks to deliver 250 cy of concrete: 250 cy / (8 cy/truck * 15 mins/truck) = 250 cy / (120 mins) = 2.08 hours.
However, we need to account for the 80% efficiency factor, which means the pump is only operating at 80% of its maximum output. Therefore, the total time required would be 2.08 hours / 0.8 = 2.6 hours.
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A cylinder contains 0.300 mol of carbon dioxide (CO2) gas at a temperature of 30.0 ∘C. The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 atm on the gas. The cylinder is placed on a hot plate and a 920 J of heat flows into the gas, thereby raising its temperature to 129 ∘C. Assume that the CO2 may be treated as an ideal gas. How much work is done by the gas in this process?
Answer:246.92 kJ
Explanation:
Given
Initial Temperature [tex]T_i=30^{\circ}C[/tex]
Final Temperature [tex]T_f=129^{\circ}C[/tex]
no of moles [tex]n=0.3[/tex]
Pressure is constant [tex]P=1 atm [/tex]
As Pressure is constant therefore work done is given by
[tex]W=P\Delta V[/tex]
[tex]P\Delta V[/tex] can also be written as [tex]P\Delta V=nR\Delta T[/tex]
[tex]W=nR\Delta T[/tex]
[tex]W=0.3\times 8.314\times (129-30)[/tex]
[tex]W=246.92 kJ[/tex]
Which one of the following statements is not a characteristic of a plane mirror?
A) The image is the same size as the object.
B) The image is always upright.
C) The image is real.
D) The image is reversed left to right compared to the object.
E) The image is the same distance behind the mirror as the object is in front of the mirror.
Answer:
C) false. The image is formed by the prolongation of the rays, so it is VIRTUAL
Explanation:
Let's review each of the statements
A) True. The image is the same size as the object in a flat mirror, m = 1
B) True. The rays are not inverted, so the right images
C) false. The image is formed by the prolongation of the rays, so it is VIRTUAL
D) True. Flat mirrors reverse left and right
E) True. When using trigonometry the angles are equal, therefore two triangles formed have the same leg, and the distance to the object and the image are equal
Answer:
The image is real is not a characteristics of a plane mirror.
Explanation:
A plane mirror is a reflector with a flat reflective covering. For light beams hitting a plane mirror, the angle of reflection matches the angle of incidence. The angle of the incidence is the angle within the incident beam and the surface normal flat mirror.
Reason for Image is not real in the plane mirror:
A real image is that picture that is created when the light beams originating from an object meet each other after reflection or refraction.A real image can be captured on the screen. The real image is constantly inverted. A common example of a real image is the image formed on the cinema screen.Thus we can say, real image is not a characteristic of a plane mirror.
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The system consists of a 20-lb disk A, 4-lb slender rod BC, and a 1-lb smooth collar C. If the disk rolls without slipping, determine the velocity of the collar at the instant the rod becomes horizontal. The system is released from rest when != 45°.
The velocity of the collar at the instant the rod becomes horizontal can be determined using the conservation of angular momentum principle. By calculating the moment of inertia of the disk and equating the initial and final angular momenta, we can solve for the final angular velocity. Since the disk rolls without slipping, the final linear velocity of the disk can be determined using the equation v' = Rω'.
Explanation:To determine the velocity of the collar at the instant the rod becomes horizontal, we can use conservation of angular momentum. Since the disk rolls without slipping, its angular momentum is conserved. The angular momentum is given by the product of the moment of inertia and the angular velocity. When the rod becomes horizontal, the angular velocity of the disk is equal to the velocity of the collar.
We can calculate the moment of inertia of the disk using the formula I = (1/2)MR^2, where M is the mass of the disk and R is its radius. Substituting the values, we get I = (1/2)(20 lb)(2 ft)^2. We also know that the angular momentum is conserved, so the initial angular momentum is equal to the final angular momentum. The initial momentum is Iω, where ω is the initial angular velocity. The final momentum is Iω', where ω' is the final angular velocity.
Since the disk rolls without slipping, the linear velocity of the disk is equal to the radius times the angular velocity, v = Rω. So, the final linear velocity of the disk is equal to the final angular velocity times the radius, v' = Rω'. Substituting the values, we can solve for ω'.
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NASA operates a 2.2-second drop tower at the Glenn Research Centre in Cleveland, Ohio. At this facility, experimental packages are dropped from the top of the tower, on the 8th floor of the building. During their 2.2 seconds of free fall, experiments experience a microgravity environment similar to that of a spacecraft in orbit,
(a) What is the drop distance of a 2.2-s tower?
(b) How fast are the experiments travelling when they hit the air bags at the bottom of the tower?
(c) If the experimental package comes to rest over a distance of 0.75 m upon hitting the air bags, what is the average stopping acceleration?
Answer:
23.7402 m
21.582 m/s
310.521816 m/s²
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
g = Acceleration due to gravity = 9.81 m/s² = a
Equation of motion
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 2.2^2\\\Rightarrow s=23.7402\ m[/tex]
The drop distance is 23.7402 m
[tex]v=u+at\\\Rightarrow v=0+9.81\times 2.2\\\Rightarrow v=21.582\ m/s[/tex]
When they hit the air bags at the bottom of the tower the speed of the experiments is 21.582 m/s
The final speed of the fall will be the initial velocity of stopping
[tex]v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-21.582^2}{2\times 0.75}\\\Rightarrow a=-310.521816\ m/s^2[/tex]
The average stopping acceleration is 310.521816 m/s²
Final answer:
Calculations reveal a drop distance of 23.65 meters, a final velocity of 21.56 m/s, and an average stopping acceleration of 311.43 m/s² for the experimental package in the 2.2-second drop tower scenario.
Explanation:
The question encompasses the calculation of drop distance, final velocity, and average stopping acceleration of an experiment package in a microgravity test environment. To solve these, we apply the equations of motion under uniform acceleration.
To find the drop distance (d) for a 2.2-second fall, we use the equation d = 0.5 * g * t, where g is the acceleration due to gravity (9.8 m/s²) and t is time in seconds. This gives us d = 0.5 * 9.8 * 2.22 = 23.65 meters.To calculate the final velocity (v) when the experiments hit the air bags, we employ v = g * t, resulting in v = 9.8 * 2.2 = 21.56 m/s.The average stopping acceleration (a) can be derived from the equation v² = 2 * a * d, rearranged to a = v² / (2*d), where d is the stopping distance of 0.75 m, leading to a = (21.562) / (2*0.75) = 311.43 m/s².This analysis presupposes a vacuum environment in the drop tower to negate air resistance, thereby approximating conditions of free fall and achieving microgravity for the duration of the drop.
A special system is set up in a lab that lets its user select any wavelength between 400nm and 700 nm with constant intensity. This light is directed at a thin glass film (n =1.53) with a thickness of 350 nm and that is surrounded by air. As one scans through these possible wavelengths, which wavelength of light reflected from the glass film will appear to be the brightest, if any?
a) 428 nm
b) 535 nm
c) 657 nm
d) 700 nm
e) Since the intensity of the light is constant,
all wavelengths of light reflected from the glass will appear to be the same.
Answer:
Explanation:
The case relates to interference in thin films , in which we study interference of light waves reflected by upper and lower surface of a medium or glass.
For constructive interference , the condition is
2μt = ( 2n+1)λ/2
μ is refractive index of glass , t is thickness , λ is wavelength of light.
putting the given values
2 x 1.53 x 350 x 10⁻⁹ = ( 2n+1) λ/2
λ = 2142nm / ( 2n+1)
For n = 2
λ = 428 nm
This wave length will have constructive interference making this light brightest of all .
For n = 1
λ = 714 nm
So second largest brightness will belong to 700 nm wavelength.
8. A 40.0kg block of metal is suspended from a scale and is immersed in water. The dimensions of the block are 12.0cm x 10.0cm x 10.0cm. The 12.0cm dimension is vertical, and the top of the block is 5.00cm below the surface of the water. (a) What are the forces exerted by the water on the top and bottom of the block? (Take P0 = 1.0130 X 105 N/m2 .) (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block.
Final answer:
The forces exerted by the water on the submerged block are due to pressure differences and can be calculated using the dimensions of the block, the density of water, and the depth of immersion. The scale reading reflects the difference between the block's weight and the buoyant force. Lastly, the buoyyant force is confirmed to be the difference between the top and bottom forces on the block, in alignment with Archimedes' principle.
Explanation:
The student's question involves calculating the forces exerted by water on a submerged block and the reading of a scale when the block is immersed in water. To calculate these forces, one must consider the forces exerted by the water on the top and bottom of the block, which are due to the differences in pressure at these points, as well as the buoyant force acting on the block. Using the principle of Archimedes, which states that the upthrust or buoyant force on a submerged object is equal to the weight of the fluid it displaces, we can find the buoyant force which in turn will help to determine the scale reading.
The weight of the water displaced can be calculated using the volume of the block and the density of the water. To answer part (a), the pressure difference at the top and bottom of the block due to the water column can be calculated using the given atmospheric pressure and depth of the block in the water. For part (b), the scale reading would be the weight of the block minus the buoyant force exerted by the water. Part (c) involves showing that the buoyant force is indeed the difference in forces at the top and bottom of the block, confirming Archimedes' principle.
A tube has a length of 0.025 m and a cross-sectional area of 6.5 x 10-4 m2. The tube is filled with a solution of sucrose in water. The diffusion constant of sucrose in water is 5.0 x 10-10 m2/s. A difference in concentration of 5.2 x 10-3 kg/m3 is maintained between the ends of the tube. How much time is required for 5.7 x 10-13 kg of sucrose to be transported through the tube?
Answer:
The time required for sucrose transportation through the tube is 8.4319 sec.
Explanation:
Given:
L = 0.025 m
A = 6.5×10^-4 m^2
D = 5×10^-10 m^2/s
ΔC = 5.2 x 10^-3 kg/m^3
m = 5.7×10^-13 kg
Solution:
t = m×L / D×A×ΔC
t = (5.7×10^-13) × (0.025) / (5×10^-10)×(6.5×10^-4)×(5.2 x 10^-3)
t = 8.4319 sec.
A hypodermic needle can be thought of as a large area pipe leading to a small pipe that opens into the vein. The diameter of the plunger is 1.2 cm and the needle diameter is 0.15 cm. If the gauge pressure in the vein is 25 mmHg, what minimum is needed on the plunger in order for a fluid flow into the vein to occur
According to the statement and the data presented, it is presumed that the variable to look for is force. To solve this problem it is necessary to apply the concepts related to the Force according to the pressure and the Area.
Mathematically the Force can be expressed as
[tex]F = PA[/tex]
Where,
P = Pressure (At this case, the plunger pressure)
A = Cross-sectional Area ( At this case the plunger area), defined for a circle.
Our values are given as,
[tex]r = 0.006m[/tex]
[tex]P = 25mmHg[/tex]
Replacing we have that
[tex]F = (25mmHg \big [\frac{133N/m^2}{1mm\cdot Hg} \big ])(\pi 0.006^2)[/tex]
[tex]F = 0.376N[/tex]
Therefore the minimum force needed on the plunger inorder for a fluid flow into the vein to occur is 0.376
The driver of a pickup truck accelerates from rest to a speed of 37 mi/hr over a horizontal distance of 215 ft with constant acceleration. The truck is hauling an empty 460-lb trailer with a uniform 72-lb gate hinged at O and held in the slightly tilted position by two pegs, one on each side of the trailer frame at A. Determine the maximum shearing force developed in each of the two pegs during the acceleration.
Final answer:
To find the maximum shearing force in the pegs during the truck's acceleration, Newton's second law is applied, and the weight and acceleration of the gate are considered. The final shear force for each peg is 7.7021 lbs.
Explanation:
To find the maximum shearing force in each peg during the acceleration of the pickup truck with the trailer, we need to apply Newton's second law.
First, let's convert the speed from miles per hour to feet per second (1 mi/hr = 1.46667 ft/s):
37 mi/hr × 1.46667 ft/s/mi/hr = 54.2667 ft/s
Using the kinematic equation v^2 = u^2 + 2as (where v is final velocity, u is initial velocity, s is distance, and a is acceleration), we can solve for acceleration (a) since the truck starts from rest (u = 0):
54.2667^2 = 0 + 2 × a × 215
a = × 54.2667^2 / (2 × 215) = 6.88776 ft/s^2
The force affecting the gate pegs comes from the horizontal component of the gravitational force and the force due to acceleration. Assume that the mass of the gate is 72 lbs; the gravitational force is:
F_gravity = m × g = 72 lbs (Since g in lbs already accounts for Earth's gravitational acceleration)
To find the force due to acceleration, we use F = m × a:
F_acceleration = 72 lbs × 6.88776 ft/s^2
However, since lbs already include g, we convert lbs to mass in slugs:
m = 72 lbs / 32.2 ft/s^2 = 2.23602 slugs
F_acceleration = 2.23602 slugs × 6.88776 ft/s^2 = 15.4042 lbs
The total force on each peg is thus the shearing force due to the acceleration, but as there are two pegs, we divide the total force by 2, assuming the load is distributed evenly:
Shear force per peg = F_acceleration / 2 = 15.4042 lbs / 2 = 7.7021 lbs (rounded to four significant figures)
A bowling ball of mass 5.8 kg moves in a
straight line at 1.59 m/s.
How fast must a Ping-Pong ball of mass
1.528 g move in a straight line so that the two
balls have the same momentum?
Answer in units of m/s.
Answer:
Velocity of the ping pong ball must be = V2= 6,035.34m/s
Explanation:
M1= momentum of the bowling ball
m1 = mass of the bowling ball= 5.8kg
v1= velocity of the bowling ball= 1.59m/s
M2= momentum of the ping pong ball
m2= mass of the ping pong ball= 1.528 g/1000= 0.001528kg
v2= velocity of the ping pong ball
Momentum of the bowling ball= M1= m1v1= 5.8* 1.59= 9.222 kg-m/s
Momentum of the ping pong ball = M2= M1= m2v2
= 0.001528 *v2= 9.222
v2= 9.222/0.001528= 6,035.34 m/s
The roads are icy, and you observe a head-on collision on Summit, at the corner with Rhodes: a 1ton car swerves out of his lane and slides through a stop sign at 41 mph straight into a 3 ton SUV traveling at 16mph in the other direction. The car and the SUV crumple from the collision, and stick together.
1. What is the final velocity, in MPH (you don't need to enter MPH in your answer) , of the SUV/car entanglement (the positive direction is the direction the car was initially going)
Answer:
final velocity = - 1.75 mph
Explanation:
given data
mass m1 = 1 ton
mass m2 = 3 ton
velocity v = 41 mph
velocity u = 16 mph
to find out
what is final velocity V
solution
we will apply here Conservation of momentum that is express as
mv + Mu = (m + M) × V ...........................1
put here value we get
1 × 41 - 3 × 16 = (1 + 3 ) × V
solve it we get
41 - 48 = 4 V
V = [tex]\frac{-7}{4}[/tex]
final velocity = - 1.75 mph
Final answer:
The final velocity of the SUV/car entanglement is 9.9 m/s.
Explanation:
In order to determine the final velocity of the SUV/car entanglement, we need to first calculate the momentum of each vehicle before the collision.
Momentum is determined by the product of an object's mass and velocity. So, the momentum of the 1 ton car before the collision is 1 ton (or 1000 kg) multiplied by its velocity of 41 mph (which is equivalent to 18.3 m/s). Therefore, the momentum of the car is 1000 kg imes 18.3 m/s = 18300 kg*m/s.
Similarly, the momentum of the 3 ton SUV before the collision is 3 ton (or 3000 kg) multiplied by its velocity of 16 mph (which is equivalent to 7.1 m/s). Therefore, the momentum of the SUV is 3000 kg imes 7.1 m/s = 21300 kg*m/s.
Since momentum is conserved in collisions, the total momentum before the collision is equal to the total momentum after the collision. This means that the final velocity of the SUV/car entanglement can be calculated by dividing the total momentum by the total mass. The total momentum is 18300 kg*m/s + 21300 kg*m/s = 39600 kg*m/s. The total mass is 1000 kg + 3000 kg = 4000 kg. So, the final velocity of the SUV/car entanglement is 39600 kg*m/s / 4000 kg = 9.9 m/s.
Choose the scenario under which each of the given Doppler shift effects will be seen.a. The source and observer are approaching one anotherb. The source and observer are moving away from one anotherc. The source and observer are stationary relative to one another
Answer:
scenario A and B
Explanation:
The Doppler effect is the change in frequency of a wave by the relative movement of the source and the observer. It is described by the expression
f ’= (v + v₀) / (v - [tex]v_{s}[/tex]) f₀
Where f₀ is the emitted frequency, v the speed of the wave, v₀ and [tex]v_{s}[/tex] the speed of the observer and the source, respectively, the signs are for when they are approaching and in the case of being away the signs are changed.
Consequently, from the above for the Doppler effect to exist there must be a relative movement of the source and the observer.
Let's examine the scenarios
A) True. You agree with the equation shown
B) True. Only the signs should be changed and it is described by the equation shown
C) False. If there is no relative movement there is no Doppler effect
The second law of thermodynamics states that spontaneous processestend to be accompanied by entropy increase. Consider, however, thefollowing spontaneous processes:
the growth of plants from simple seeds to well-organizedsystems
the growth of a fertilized egg from a single cell to a complexadult organism
the formation of snowflakes from molecules of liquid water withrandom motion to a highly ordered crystal
the growth of organized knowledge over time
In all these cases, systems evolve to a state of less disorder andlower entropy, apparently violating the second law ofthermodynamics. Could we, then, consider them as processesoccurring in systems that are not isolated?
True or False?
Answer:
True
Explanation:
According to the definition of a closed system, It is true because it's not precisely a closed system. A closed system is a physical system that doesn't let certain types of transfers (such as transfer of mass in or out of the system), though the transfer of energy is allowed.
A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it has a speed of 12 m/s. Which of the following correctly identifies whether the object-Earth system is open or closed and describes the net external force?
A. The system is closed, and the net external force is zero.
B. The system is open, and the net external force is zero.
C. The system is closed, and the net external force is nonzero.
D. The system is open, and the net external force is nonzero.
Answer:D
Explanation:
Given
mass of object [tex]m=5 kg[/tex]
Distance traveled [tex]h=10 m[/tex]
velocity acquired [tex]v=12 m/s[/tex]
conserving Energy at the moment when object start falling and when it gains 12 m/s velocity
Initial Energy[tex]=mgh=5\times 9.8\times 10=490 J[/tex]
Final Energy[tex]=\frac{1}{2}mv^2+W_{f}[/tex]
[tex]=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}[/tex]
where [tex]W_{f}[/tex] is friction work if any
[tex]490=360+W_{f}[/tex]
[tex]W_{f}=130 J[/tex]
Since Friction is Present therefore it is a case of Open system and net external Force is zero
An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .
A parallel plate capacitor is being charged by a constant current i. During the charging, the electric field within the plates is increasing with time.
Which one of the following statements concerning the magnetic field between the plates is true?
A) The magnetic field within a parallel plate capacitor is always equal to zero teslas.
B) The induced magnetic field is directed antiparallel to the increasing electric field.
C) The induced magnetic field strength has its largest value at the center of the plates and decreases linearly toward the edges of the plates.
D) At a given moment, the induced magnetic field strength has the same magnitude everywhere within the plates of the capacitor, except near the edges.
E) The induced magnetic field strength is zero teslas near the center of the plates and increases as r increases toward the edges of the plates.
Answer:
E
Explanation:
The parallel plate capacitor is being charged by a steady current i. If the constant current is flowing along a straight wire, what happens inside the capacitor, between the plates is that the induced magnetic field strength is zero Tesla near the center of the plates and increases as r increases toward the edges of the plates.
From Ampère–Maxwell law, the magnetic field between the capacitor plates assuming that the capacitor is being charged at a constant rate by a steady current is
B = μ₀r /2A [tex]\frac{dQ}{dt} e_{o}[/tex]
where;
μ₀ is permeability
Q(t) is the instantaneous charge on the positive plate,
A is the cross-sectional area of a plate and
e₀ is a unit vector