Answer:
b.) The car travels around a circular track at 30 m/s
Explanation:
To find - In which situation is the speed of the car constant while its velocity is changing?
a.) The car travels down a straight track at 30 m/sec
b.) The car travels around a circular track at 30 m/s
c.) The car begins from rest and accelerates to 20 m/s
d.) The car begins traveling at 30 m/s and decreases to 15 m/s
Proof -
We know that,
Speed is distance covered in a particular time
Velocity is distance covered by body in particular time in defined direction.
Now,
In a.)
Speed and direction is not changing.
So, Option a) is wrong.
In c.) and d.)
There exists a positive and negative acceleration.
Even If direction is constant, but the magnitude of speed is changing here, So, Option c) and d) are wrong.
Now,
In b.)
We can see that,
Due to circular motion, the speed of the car is constant at 30 m/s
But Due to circular motion, direction is changing continuously.
So, here speed is constant while its velocity is changing.
∴
The correct option is - b.) The car travels around a circular track at 30 m/s
A six-lane divided multilane highway (three lanes in each direction) has a measured free-flow speed of 50 mi/h. It is on mountainous terrain with a traffic stream consisting of 7% large trucks and buses and 3% recreational vehicles. The driver population adjustment in 0.92. One direction of the highway currently operates at maximum LOS C conditions and it is known that the highway has PHF = 0.90.
Required:
How many vehicles can be added to this highway before capacity is reached, assuming the proportion of vehicle types remains the same but the peak-hour factor increases to 0.95?
Answer:
2901 vehicles
Explanation:
We are given;
Percentage of large trucks & buses; p_t = 7% = 0.07
Percentage of recreational vehicles; p_r = 3% = 0.03
PHF = 0.90
Driver population adjustment; f_p = 0.92
First of all, let's Calculate the heavy vehicle factor from the formula;
f_hv = 1/[1 + p_t(e_t - 1) + p_r(e_r - 1)]
Where;
e_t = passenger car equivalents for trucks and buses
e_r = passenger car equivalents for recreational vehicles
From the table attached, for a mountainous terrain, e_t = 6 and e_r = 4. Thus;
f_hv = 1/[1 + 0.07(6 - 1) + 0.03(4 - 1)]
f_hv = 1.44
Let's now calculate the initial hourly volume from the formula;
v_p = V1/(PHF × N × f_hv × f_p)
Where;
v_p = 15-minute passenger-car equivalent flow rate
V1 = hourly volume
N = number of lanes in each direction
From online tables of LOS criteria for multilane freeway segments, v_p = 1300 pc/hr/ln
Thus;
1300 = V1/(0.9 × 3 × 1.44 × 0.92)
V1 = 1300 × (0.9 × 3 × 1.44 × 0.92)
V1 = 4650 veh/hr
Now, let's Calculate the final hourly volume;
From online sources, the maximum capacity of a 6 lane highway with free-flow speed of 50 mi/h is 2000 pc/hr/ln.
We are told the online peak-hour factor increases to 0.95 and so PHF = 0.95.
Thus;
2000 = V2/(0.95 × 3 × 1.44 × 0.92)
V2 = 2000(0.95 × 3 × 1.44 × 0.92)
V2 = 7551 veh/hr
Number of vehicles added to the highway = V2 - V1 = 7551 - 4650 = 2901 vehicles
. An ideal vapor compression refrigeration cycle operates with a condenser pressure of 900 kPa. The temperature at the inlet to the compressor is -5oC. If this device operates using R134a as the working fluid, calculate the actual COP of this device as well as the maximum theoretical COP.
Answer:
Actual COP = 5.368
Maximum theoretical COP = 6.368
Explanation:
Given - An ideal vapor compression refrigeration cycle operates with a condenser pressure of 900 kPa. The temperature at the inlet to the compressor is -5oC.
To find - If this device operates using R134a as the working fluid. Calculate the actual COP of this device as well as the maximum theoretical COP.
Proof -
Given that,
An ideal vapor compression refrigeration cycle operates with a condenser pressure of 900 kPa.
From Refrigerant 134-a Table
At T1 = -5°C
h1 = 247.505 KJ/kg
S1 = 0.93434 KJ/kg
At P2 = 900 KPa
S1 = S2
h2 = 274.679 Kj/Kg
h3 = h4 = 101.61 KJ/g
So,
Compressor work (Wc) = h2 - h1
= 274.679 - 247.505
= 27.174
⇒Compressor work (Wc) = 27.174 KJ/kg
Now,
Heat out (Qout) = h2 - h3
= 274.679 - 101.61
= 173.069
⇒Heat out (Qout) = 173.069 KJ/kg
Now,
Heat input (Qin) = h1 - h4
= 274.505 - 101.61
= 145.895
⇒Heat input (Qin) = 145.895 KJ/kg
So,
Actual COP at the refrigerator is -
(COP)R = (Qin)/(Wc)
= (145.895)/ (27.174)
= 5.368
⇒Actual COP = 5.368
Now,
Maximum theoretical COP is -
(COP) = (Qout)/(Wc)
= (173.069)/ (27.174)
= 6.368
⇒Maximum theoretical COP = 6.368
A mixture of pulverized fuel ash and Portland cement to be used for grouting should have a true average compressive strength of more than 1300 KN/m2. The mixture will not be used unless experimental evidence indicates conclusively that the strength specification has been met. State the relevant hypotheses, and describe the type I and type II errors in the context of this problem.
Answer:
Null Hypothesis - True average compressive strength of mixture of pulverized fuel ash and Portland cement is more than 1300 KN/m2.
Type I error (false positive) - this occur when the True average compressive strength of mixture of pulverized fuel ash and Portland cement is not more than 1300 KN/m2 in actual scenario but the measurements shows an incorrect reading of True average compressive strength greater than 1300 KN/m2.
Type II error (false-negative)- his occur when the True average compressive strength of mixture of pulverized fuel ash and Portland cement is more than 1300 KN/m2 in actual scenario but the measurements show an incorrect reading of True average compressive strength less than 1300 KN/m2.
Explanation:
The hypothesis for this piece of information is as follows -
Null Hypothesis - True average compressive strength of mixture of pulverized fuel ash and Portland cement is more than 1300 KN/m2.
Alternate hypothesis - True average compressive strength of mixture of pulverized fuel ash and Portland cement is less than or equal to 1300 KN/m2
Type I error (false positive) - this occur when the True average compressive strength of mixture of pulverized fuel ash and Portland cement is not more than 1300 KN/m2 in actual scenario but the measurements shows an incorrect reading of True average compressive strength greater than 1300 KN/m2.
Type II error (false-negative)- his occur when the True average compressive strength of mixture of pulverized fuel ash and Portland cement is more than 1300 KN/m2 in actual scenario but the measurements show an incorrect reading of True average compressive strength less than 1300 KN/m2.
Forget it because I almost have it
Answer:
hahahaha
Explanation:
DUE IN 15 MINUTES
How is meteor related to a meteoroid?
What is the force applied to a bone with a cross-sectional area of 0.0004 m2 if it is being stressed with 5,000 Pa?
Answer:
[tex]2\ \text{N}[/tex]
Explanation:
[tex]A[/tex] = Cross sectional area of bone = [tex]0.0004\ \text{m}^2[/tex]
[tex]P[/tex] = Pressure = [tex]5000\ \text{Pa}[/tex]
[tex]F[/tex] = Force applied to the bone
Force is given by
[tex]F=PA[/tex]
[tex]\Rightarrow F=5000\times 0.0004[/tex]
[tex]\Rightarrow F=2\ \text{N}[/tex]
The force applied to the bone is [tex]2\ \text{N}[/tex].
Consider a machine of mass 70 kg mounted to ground through an isolation system of total stiffness 30,000 N/m, with a measured damping ratio of 0.2. The machine produces a harmonic force of 450 N at 13 rad/s during steady-state operating conditions. Determine
Complete Question
Consider a machine of mass 70 kg mounted to ground through an isolation system of total stiffness 30,000 N/m, with a measured damping ratio of 0.2. The machine produces a harmonic force of 450 N at 13 rad/s during steady-state operating conditions. Determine
(a) the amplitude of motion of the machine,
(b) the phase angle of the motion,
(c) the transmissibility ratio,
(d) the maximum dynamic force transmitted to the floor, and
(e) the maximum velocity of the machine.
Answer:
a) [tex]X=0.0272m[/tex]
b) [tex]\phi=22.5 \textdegree[/tex]
c) [tex]T_r=1.57[/tex]
d) [tex]F=706.5N[/tex]
e) [tex]V_{max}=0.35m/s[/tex]
Explanation:
From the question we are told that:
Mass [tex]M=70kg[/tex]
Total Stiffness [tex]\mu=30000[/tex]
Damping Ratio [tex]r=0.2[/tex]
Force [tex]F=450N[/tex]
Angular velocity [tex]\omega =13rad/s[/tex]
Generally the equation for vibration in an isolated system is mathematically given by
[tex]\omega_n=\sqrt{\frac{k}{m}}[/tex]
[tex]\omega_n=\sqrt{\frac{30000}{70}}[/tex]
[tex]\omega_n=20.7rad/s[/tex]
a)
Generally the equation for Machine Amplitude is mathematically given by
[tex]X=\frac{F_O/m}{(\omega_n^2-\omega^2)^2-(2*r*\omega)*\omega_n*\omega^2)^{1/2}}[/tex]
[tex]X=\frac{450}{70}}{(20.7^2-(137^2)^2-(2*0,2*(20.7(13)))^2)^{1/2}[/tex]
[tex]X=0.0272m[/tex]
b)
Generally the equation for Phase Angle is mathematically given by
[tex]\phi=tan^{-1}\frac{2*r*\omega_n*\omega}{\omega_n^2*\omega^2}[/tex]
[tex]\phi=tan^{-1}\frac{2*0.2*20.7*13}{\20.7^2*13^2}[/tex]
[tex]\phi=22.5 \textdegree[/tex]
c)
Generally the equation for transmissibility ratio is mathematically given by
[tex]T_r=\sqrt{\frac{1+(2r\beta)^2}{(1-r^2)^2+(2*\beta*r)^2}}[/tex]
Where
[tex]\beta=Ratio\ of\ angular\ velocity[/tex]
[tex]\beta=\frac{13}{20.7}\\\beta=0.638[/tex]
Therefore
[tex]T_r=\sqrt{\frac{1+(2*(0.2)(0.638))^2}{(1-(0.2)^2)^2+(2*0.2*0.638)^2}}[/tex]
[tex]T_r=1.57[/tex]
d)
Generally the equation for Maximum dynamic force transmitted to the floor is mathematically given by
[tex]F=(T_r)*F_o[/tex]
[tex]F=(1.57)*450[/tex]
[tex]F=706.5N[/tex]
e)
Generally the equation for Maximum Velocity of Machine is mathematically given by
[tex]V_{max}=\omega*x[/tex]
[tex]V_{max}=13*0.0272[/tex]
[tex]V_{max}=0.35m/s[/tex]
