Answer:
"150000 N/m²" is the right approach.
Explanation:
According to the question, the pressure on the two spheres 1 and 2 is same.
Sphere 1 and 2:
Then,
⇒ [tex]P_1=P_2[/tex]
⇒ [tex]\frac{\Delta V_1}{V_1}=\frac{\Delta V_2}{V_2}[/tex]
and the bulk modulus be,
⇒ [tex]B_1=B_2[/tex]
Sphere 3:
⇒ [tex]\frac{\Delta V_3}{V_3} =\frac{\frac{\Delta V_1}{V_1} }{\frac{\Delta V_2}{V_2} } =1[/tex]
then,
⇒ [tex]P_3=B\times \frac{\Delta V_3}{V_3}[/tex]
⇒ [tex]=B\times 1[/tex]
⇒ [tex]=150000\times 1[/tex]
⇒ [tex]=150000 \ N/m^2[/tex]
The part in the photo below is
A. an evaporator.
B. an accumulator.
C. a condenser.
D. a compressor.
Answer: a accumulator
Explanation:
Knowing that the central portion of the link BD has a uniform cross-sectional area of 800 2, determine the magnitude of the load P for which the normal stress in that portion of BD is 50 .
Answer: 50
Explanation:
a) For Well A, provide a cross-section sketch that shows (i) ground elevation, (ii) casing height, (iii) depth to
water table, (iv) sampling depth, (v) elevation of the well top of casing, (vi) water table elevation, (vii) elevation
head of the water sampled for bromide, and (viii) pressure head of the water sampled for bromide. Label each of
these distances with the above phrases, plus a unique variable.
b) Calculate the following for each well: (i) elevation of the well top of casing, (ii) water table elevation,
(iii) sampling port elevation, (iv) elevation head of the water sampled for bromide, and (v) pressure head of the
water sampled for bromide. Use sea level as your vertical datum. Write out all calculations (including equations
with variables) for Well A.
Combinations of velocity and acceleration
Answer:
acceleration=change in velocity/ time
Explanation:
The velocity of an object is its speed in a particular direction. Velocity is a vector quantity because it has both a magnitude and an associated direction. To calculate velocity, displacement is used in calculations, rather than distance.
Define;
i) Voltage
ii) Current
iii) Electrical Power
iv) Electrical Energy
Answer:
I) Voltage - is the pressure from an electrical circuit's power source that pushes charged electrons (current) through a conducting loop, enabling them to do work such as illuminating a light. In brief, voltage = pressure, and it is measured in volts (V).
II) Current - is the movement of electrons through a wire. Electric current is measured in amperes (amps) and refers to the number of charges that move through the wire per second. If we want current to flow directly from one point to another, we should use a wire that has as little resistance as possible.
III) Electrical Power - is the rate, per unit time, at which electrical energy is transferred by an electric circuit. The SI unit of power is the watt, one joule per second. Electric power is usually produced by electric generators, but can also be supplied by sources such as electric batteries.
IV) Electrical Energy - is a form of energy resulting from the flow of electric charge. Energy is the ability to do work or apply force to move an object. In the case of electrical energy, the force is electrical attraction or repulsion between charged particles.
Explanation:
I hope ot helps to you a lot! Correct me if I'm wrong.
Which of these parts converts the spinning motion of the driveshaft 90° to turn the wheels?
A. Transmission
B. Axle
C. Differential
D. Engine
If it is struck by a rigid block having a weight of 550 lblb and traveling at 2 ft/sft/s , determine the maximum stress in the cylinder. Neglect the mass of the cylinder. Express your answer to three significant figures and include appropriate units.
This question is incomplete, The missing image is uploaded along this answer below;
Answer:
the maximum stress in the cylinder is 3.23 ksi
Explanation:
Given the data in the question and the diagram below;
First we determine the initial Kinetic Energy;
T = [tex]\frac{1}{2}[/tex]mv²
we substitute
⇒ T = [tex]\frac{1}{2}[/tex] × (550/32.2) × (2)²
T = 34.16149 lb.ft
T = ( 34.16149 × 12 ) lb.in
T = 409.93788 lb.in
Now, the volume will be;
V = [tex]\frac{\pi }{4}[/tex]d²L
from the diagram; d = 0.5 ft and L = 1.5 ft
so we substitute
V = [tex]\frac{\pi }{4}[/tex] × ( 0.5 × 12 in )² × ( 1.5 × 12 in )
V = 508.938 in³
So by conservation of energy;
Initial energy per unit volume = Strain energy per volume
⇒ T/V = σ²/2E
from the image; E = 6.48(10⁶) kip
so we substitute
⇒ 409.93788 / 508.938 = σ²/2[6.48(10⁶)]
508.938σ² = 5,312,794,924.8
σ² = 10,438,982.5967
σ = √10,438,982.5967
σ = 3230.9414
σ = 3.2309 ksi ≈ 3.23 ksi { three significant figures }
Therefore, the maximum stress in the cylinder is 3.23 ksi
A cylindrical rod of brass originally 10 mm in diameter is to be cold worked by drawing. The circular cross section will be maintained during deformation. A cold-worked tensile strength in excess of 380 MPa and a ductility of at least 15 %EL are desired. Furthermore, the final diameter must be 7.5 mm. Explain how this may be accomplished. Use the graphs given in previous question.
Answer:
Explanation:
From the information given:
original diameter [tex]d_o[/tex] = 10 mm
final diameter [tex]d_f =[/tex] 7.5 mm
Cold work tensile strength of brass = 380 MPa
Recall that;
[tex]\text {The percentage CW }= \dfrac{\pi (\dfrac{d_o}{2})^2 - \pi(\dfrac{d_f}{2})^2 }{\pi(\dfrac{d_o}{2})^2} \times 100[/tex]
[tex]\implies \dfrac{\pi (\dfrac{10}{2})^2 - \pi(\dfrac{7.5}{2})^2 }{\pi(\dfrac{10}{2})^2} \times 100[/tex]
[tex]\implies43.87\% \ CW[/tex]
→ At 43.87% CW, Brass has a tensile strength of around 550 MPa, which is greater than 380 MPa.
→ At 43.87% CW, the ductility is less than 5% EL, As a result, the conditions aren't met.
To achieve 15% EL, 28% CW is allowed at most
i.e
The lower bound cold work = 15%
The upper cold work = 28%
The average = [tex]\dfrac{15+28}{2}[/tex] = 21.5 CW
Now, after the first drawing, let the final diameter be [tex]d_o^'[/tex]; Then:
[tex]4.5\% \ CW = \dfrac{\pi (\dfrac{d_o^'}{2})^2 - (\dfrac{7.5}{2})^2}{\pi (\dfrac{d_o^'}{2})^2}\times 100[/tex]
By solving:
[tex]d_o^'} = 8.46 mm[/tex]
To meet all of the criteria raised by the question, we must first draw a wire with a diameter of 8.46 mm and then 21.5 percent CW on it.
Water from an upper tank is drained into a lower tank through a 5 cm diameter iron pipe with roughness 2 mm. The entrance to the pipe has minor loss coefficient 0.4 and the exit has minor loss coefficient of 1, both referenced to the velocity in the pipe. The water level of the upper tank is 4 m above the level of the lower tank, and the pipe is 5 m long. You will find the drainage volumetric flow rate. a) What is the relative roughness
Answer:
Relative roughness = 0.04
Explanation:
Given that:
Diameter = 5 cm
roughness = 2 mm
At inlet:
Minor coefficient loss [tex]k_{L1} = 0.4[/tex]
At exit:
Minor coefficient loss [tex]k_{L2} = 1[/tex]
Height h = 4m
Length = 5 m
To find the relative roughness:
Relative roughness is a term that is used to describe the set of irregularities that exist inside commercial pipes that transport fluids. The relative roughness can be evaluated by knowing the diameter of the pipe made with the absolute roughness in question. If we denote the absolute roughness as e and the diameter as D, the relative roughness is expressed as:
[tex]e_r = \dfrac{e}{D}[/tex]
[tex]e_r = \dfrac{0.2 }{5}[/tex]
[tex]\mathbf{e_r = 0.04}[/tex]
A satellite at a distance of 36,000 km from an earth station radiates a power of 10 W from an
antenna with a gain of 25 dB. What is the received power if the effective aperture area of the
receiving antenna is 20 m2?
The received power if the effective aperture area of the receiving antenna is 20 m2 is 177.77 m2.
What is Power?In physics, power is referred to as the rate of energy conversion or transfer over time. The unit of power in the SI system, often known as the International System of Units, is the Watt (W). A single joule per second is one watt.
Power was formerly referred to as activity in some research. A scalar quantity is power. As power is always a function of labor done, it follows that if a person's output varies during the day depending on the time of day, so will his power.
A measure of the pace at which energy is transferred, power is a physical quantity. As a result, it can be described as the pace of job completion relative to time.
Therefore, The received power if the effective aperture area of the receiving antenna is 20 m2 is 177.77 m2.
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The evaporator:
A. directs airflow to the condenser.
B. absorbs heat from the passenger compartment.
C. removes moisture from the refrigerant.
D. restricts refrigerant flow.
Answer:
Option B
Explanation:
An evaporator along with cold low pressure refrigerant absorbs heat from the air within the passenger compartment thereby supplying cool air for the occupants.
Hence, option B is correct
In warm climates, a vapor barrier is placed on the exterior side of the insulation, and in cold climates it is installed on the interior side of the
insulation. Which of the following explains this placement of the barrier?
The barrier should always be placed on the side opposite from where the water condenses.
The barrier should always be placed on the side opposite where rain or snow hit.
The barrier should always be placed on the side where rain or snow hit.
The barrier should always be placed on the side where the water condenses
Answer: its c
Explanation:
In a certain company the cost of software depends on the license type which could be Individual or Enterprise. Write a program that reads License Type wanted (just the first character of each type: I, i, E, e). Number of Users to use the software. Type Price/User Minimum number of users Individual 500$ 1 Enterprise 300$ 5 Your program should: Check if the number of users is greater than or equal than Minimum Number of Users allowed Compute the cost: (for example cost = Cost per user x Number of Users)
Solution :
import [tex]$\text{java}.$[/tex]util.*;
public [tex]$class$[/tex] currency{
public static [tex]$\text{void}$[/tex] main(String[tex]$[]$[/tex] args) {
Scanner input [tex]$=$[/tex] new Scanner(System[tex]$\text{.in}$[/tex]);
System[tex]$\text{.out.}$[/tex]print("Enter [tex]$\text{number of}$[/tex] quarters:");
int quarters = input.nextInt();
System.out.print("Enter number of dimes:");
int [tex]$\text{dimes =}$[/tex] input.nextInt();
System[tex]$\text{.out.}$[/tex]print("Enter number of nickels:");
int nickels = input.nextInt();
System[tex]$\text{.out.}$[/tex]print("Enter number of pennies:");
int [tex]$\text{pennies = }$[/tex] input.nextInt();
// computing dollors
double dollars = (double) ((quarters*0.25)+(dimes*0.10)+(nickels*0.05)+(pennies*0.01));
System[tex]$\text{.out.}$[/tex]format("You have : $%.2f",dollars);
}
}
Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a velocity of 5 m/s. At the exit, the refrigerant is a saturated vapor at -16oC. The evaporator flow channel has constant diameter of 1.7 cm. Determine the mass flow rate of the refrigerant, in kg/s, and the velocity at the exit, in m/s.
Answer:
mass flow rate = 0.0534 kg/sec
velocity at exit = 29.34 m/sec
Explanation:
From the information given:
Inlet:
Temperature [tex]T_1 = -16^0\ C[/tex]
Quality [tex]x_1 = 0.2[/tex]
Outlet:
Temperature [tex]T_2 = -16^0 C[/tex]
Quality [tex]x_2 = 1[/tex]
The following data were obtained at saturation properties of R134a at the temperature of -16° C
[tex]v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\ v_g = 0.1247 \ m^3 /kg[/tex]
[tex]v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\ v_1 = 0.0255 \ m^3/kg \\ \\ \\ v_2 = v_g = 0.1247 \ m^3/kg[/tex]
[tex]m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6 \\ \\ \mathbf{m = 0.0534 \ kg/sec}[/tex]
[tex]\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2 \\ \\ \rho_1v_1 = \rho_2v_2 \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}[/tex]
Which statements describe the motion of car A and car B? Check all that apply. Car A and car B are both moving toward the origin. Car A and car B are moving in opposite directions. Car A is moving faster than car B. Car A and car B started at the same location. Car A and car B are moving toward each other until they cross over.
Answer:
car a is moving faster than the car b
Answer:
B: Car A and car B are moving in opposite directions.
C: Car A is moving faster than car B.
E: Car A and car B are moving toward each other until they cross over.
Explanation:
I just did the assignment on EDGE2020 and it's 200% correct!
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The distribution of SAT scores of all college-bound seniors taking the SAT in 2014 was approximately normal with mean μ=1497 and standard deviation σ=322. A certain test-retake preparation course is designed for students whose SAT scores are in the lower 25%, percent of those who take the test in a given year. What is the maximum SAT score in 2014 that meets the course requirements?
Answer:
1279
Explanation:
We have the mean u = 1497
Standard deviation sd = 322
We find the x distribution using 25%
P(Z<z) = 0.25
Z = -0.675
From here we use the formula for z score
X = z(sd) + u
X = -0.675*322 + 1497
X = -217.35 + 1497
X = 1279.6
Which is approximately 1279
So we conclude that the maximum sat scores in year 2014that meets with the requirements of this course is 1279
Answer:
1831
Explanation:
state the parallelogram law of forces
Answer:
The law of parallelogram of forces states that if two vectors acting on a particle at the same time be represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point their resultant vector is represented in magnitude and direction by the diagonal of the parallelogram drawn from .
Identify parts of the E-Cig that constitute voltage, current, and resistance. Discuss the role each plays in the E-Cig and typical values for each including units.
Discuss the electrical dangers of an E-Cig. Give specific examples.
There are many electrical safety rules. Pick one, and discuss its application on a small system, such as the E-Cig.
Answer: c
Explanation:
Define hermetic compressor
Answer:
Hermetic compressors are ideal for small refrigeration systems, where continuous maintenance cannot be ensured.
In 1951, a small approach embankment was constructed for a highway bridge over a river south of Los Angeles. The embankment was underlain by 5 ft of organic clay. Records of the settlement rate indicate that 90% of the consolidation settlements occurred in the first 4.5 years after construction. A new bridge over the river is now planned for a site a few hundred yards from the old bridge. The approach embankment to the new bridge will be underlain by 20 ft of the same organic clay found at the old bridge site. Estimate the time required to achieve an average degree of consolidation of 90% under the new embankment. Assume single drainage from the organic clay at both sites..
Answer:
72 years
Explanation:
The degree of consideration is the same for both bridges = 90%
Height of first highway bridge( d1 ) = 5 ft
Time to consolidation ( t1 )= 4.5 years
Height of second bridge ( d2 ) = 20 ft
Time to consolidation ( t2 ) = ?
we will apply this relation below
Tv = Cv * t / d^v
Tv = constant
for a single drainage condition : t ∝ d^v hence; d = H
∴ [tex]\frac{t_{2} }{t_{1} } = (\frac{d_{2} }{d_{1} })[/tex]^2
t2 = t1 ( d2/d1 )^2
= 4.5 ( 20 / 5 )^2
= 72 years
Barries of effective
communication?
Answer: barries
Explanation:
Consider a convergent-
How may the desire for a perfect lawn be related to environmental pollution? A. Lawns have little to do with environmental pollution. B. Perfect lawns require excess use of manual labor. C. Manicured lawns are subject to increased runoff. D. Lawns absorb pesticides and fertilizers and these chemicals leach out of the lawn and run into creeks, streams and eventually, rivers, with rain or watering, eventually reaching the ocean.
Answer: is b
Explanation:
Fig_Q5
6. A steel rod is stressed by a tension force of 250 N. It is found that the rod has length of 45
m and diameter of 1.5 mm. If the modulus of elasticity of the steel rod is assumed as 2 x 105
MPa, determine the strain of the steel rod due to the applied force.
Answer:
The strain of the steel rod due to the applied force is 41.93
Explanation:
Modulus of elasticity is equal to stress divided by strain.
And stress is equal to force divided by area
Surface area of cylindrical rod
[tex]2\pi r (r+h)[/tex]
Substituting the given values we get -
[tex]2 *3.14 * \frac{1.5}{1000} * 45 (45 + \frac{1.5}{1000}) = 19.07[/tex]
[tex]2 * 10 ^5 = \frac{250}{19.07 * S=(\frac{\Delta L}{L} )}[/tex]
Hence, strain is equal to
Strain = 41.93
g Design of a spindle present in an existing design needs to be reviewed for use under new loading needs. It is currently designed to withstand combined torsional and compressive loading. The design team needs to determine what the maximum torque it can withstand before failing if a compressive, axial load of 15 kips is present. The ultimate compressive and tensile strengths of the material are 135 ksi and 40 ksi, respectively. Determine the torsional shear stress that will just cause failure using a non-conservative failure theory.
Answer: its c
Explanation:
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 412 MPa (59760 psi) is applied if the original length is 480 mm (18.90 in.)? Assume a value of 0.22 for the strain-hardening exponent, n.
Answer:
the elongation of the metal alloy is 21.998 mm
Explanation:
Given the data in the question;
K = σT/ (εT)ⁿ
given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,
strain-hardening exponent n = 0.22
we substitute
K = 345 / [tex]0.02^{0.22[/tex]
K = 815.8165 Mpa
next, we determine the true strain
(εT) = (σT/ K)^1/n
given that σT = 412 MPa
we substitute
(εT) = (412 / 815.8165 )^(1/0.22)
(εT) = 0.04481 mm
Now, we calculate the instantaneous length
[tex]l_i[/tex] = [tex]l_0e^{ET[/tex]
given that [tex]l_0[/tex] = 480 mm
we substitute
[tex]l_i[/tex] =[tex]480mm[/tex] × [tex]e^{0.04481[/tex]
[tex]l_i[/tex] = 501.998 mm
Now we find the elongation;
Elongation = [tex]l_i - l_0[/tex]
we substitute
Elongation = 501.998 mm - 480 mm
Elongation = 21.998 mm
Therefore, the elongation of the metal alloy is 21.998 mm
Let xa(t)be an analog signal with bandwidth B=3kHz. We wishto use an ????=2m–pointDFT to compute the spectrum ofthe signal with a resolution less than or equal to 50 Hz.
Determine
(a) the minimum sampling rate,
(b) the minimum number of required samples, and
(c) the minimumlength of the analog signal record(in seconds).
Answer:
a) the minimum sampling rate is 6 kHz
b) the minimum numbers of required samples are 120
c) the minimum length of the analog signal is 0.02 s
Explanation:
Given the data in the question;
(a) the minimum sampling rate;
band width of analog signal xₐ(t) is;
bandwidth B = 3kHz
Now, according to sampling theorem, minimum sampling rate F[tex]_s[/tex] must be twice the bandwidth of the signal.
so
F[tex]_s[/tex] = 2B
F[tex]_s[/tex] = 2( 3 kHz )
F[tex]_s[/tex] = 6 kHz
Therefore, the minimum sampling rate is 6 kHz
(b) the minimum number of required samples;
Let L represent the minimum number of samples required,
given that; required resolution of the spectrum of the signal is less than or equal to 50 Hz
F[tex]_s[/tex]/L ≤ 50
L ≥ F[tex]_s[/tex]/50
L ≥ ( 6 × 1000 Hz ) / 50
L ≥ 6000 / 50
L ≥ 120
Therefore, the minimum numbers of required samples are 120
(c) the minimum length of the analog signal record(in seconds).
minimum number of samples required is 120
T = L / F[tex]_s[/tex]
T = 120 / ( 6 × 1000 Hz )
T = 120 / 6000
T = 0.02 s
Therefore, the minimum length of the analog signal is 0.02 s
calculate the radius of a circular orbit for which the period is 1 day
Answer:
(T²/D³)sys1 = (T²/D³)sys2
sys1 = earth-moon
sys2 = earth-sat
(27.33day)²/(3.8e8m)³ = (1day)²/D³
D = cbrt(7.3e22m³) = 4.2e7 m
Explanation:
exchange in capacity whilst a satellite tv for pc differences altitude How plenty paintings could be executed to flow the satellite tv for pc into yet another around orbit it is bigger above the outdoors of the Earth? satellite tv for pc exchange in capacity with top Assuming the satellite tv for pc is to be boosted to a clean top r? Gravitational capacity capacity (to center of earth) new orbit(2) has a greater robust PE than old one(a million), so exchange is helpful PE = G m?m?/r earth GM = 3.98e14 ?PE = (GM)(m)(a million/r? – a million/r?) KE additionally differences. Get speed at each and every top. New orbit(2) has decrease speed, so exchange is damaging v = ?(GM/R) V? = ?(GM/r?) V? = ?(GM/r?) ?KE = –½m(V?² – V?²) ?KE = –½mGM(a million/r? – a million/r?) including the two ?E = (GM)(m)(a million/r? – a million/r?)– ½mGM(a million/r? – a million/r?) ?E = ½mGM(a million/r? – a million/r?) ?E = ½(3.98e14)(7500) [a million/(0.5e7) –a million/(3.3e7) ] ?E = ½(3.98e14)(7500)(1e-7) [a million/(0.5) –a million/(3.3) ] ?E = ½(3.98e7)(7500) [2 – 0.303 ] ?E = ½(3.98e7)(7500)(a million.70) ?E = 2.04e11 Joules edit, corrected .
The radius of a circular orbit will be "[tex]\frac{V}{2 \pi} \ km[/tex]".
According to the question,
The orbit period of satellite,Time = 1 day
Total distance will be equal to the orbit's circumference, thenDistance = [tex]2 \pi r[/tex]
Let,
The velocity be "V km/day".As we know,
→ [tex]Distance = Velocity\times time[/tex]
By substituting the values, we get
→ [tex]2 \pi r = V\times 1[/tex]
→ [tex]r = \frac{V}{2 \pi} \ km[/tex]
Thus the above is the right answer.
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can someone please help me with this
I've an exams tomorrow
Answer:
I am in Eight Grade
Explanation:
In the production of soybean oil, dried and flaked soybeans are brought in contact with a solvent (often hexane) that extracts the oil and leaves behind the residual solids and a small amount of oil.
a. Draw flow diagram of the process, labeling the two feed streams (beans and solvent) and the leaving streams (solids and extract).
b. The soybeans contain 18.5 wt% oil and the remainder insoluble solids, and the hexane is fed at a rate corresponding to 2.0 kg hexane per kg beans. The residual solids leaving the extraction unit contain 35.0 wt% hexane, all of the non-oil solids that entered with beans, and 1.0% of the oil that entered the beans. For a feed rate of 1000 kg/h of dried flaked soybeans, calculate mass flow rates of extract and residual solids and the composition of extract.
Answer: its c
Explanation: