Answer:
The maximum load the rod can take before it starts to permanently elongate = = 6086.84 lbf or 27075.59 N
The maximum energy that the rod would store =2.536 ft·lbf or 3.439 J
Explanation:
We list out the variables to the question as follows
Rod diameter = 0.5 in
Length of rod = 5 in
Young's modulus of elasticity for the material = 15.5 Msi
The maximum load rthe rd can take before it starts to permanently elongate is the yield stress and it can vbe calclulated by applyng the 0.2% offset rule by taking the yeild strain to be equal to 2%
Thus Young's Modulus at yield point =[tex]\frac{Stress}{Strain}[/tex] so that the yield stess is
Yield stress = strain × Young's Modulus
= 2/100 × E = 0.002 × 15.5 Msi = 0.031 Msi = 31 ksi
The maximum load that the rod would take before it starts to permanently elongate = F = Stress×Area
where area = π·D²/4 = 0.196 in²
F = 31 kSi × 0.196 in² = 6086.84 lbf or 27075.59 N
To calculate the strain energy stored in the rod we apply the strain energy formula thus
U = V×σ²/2·E
To calculate the volume we have V = L ×π·D²/4 =5 in × 0.196 in² = 0.98 in³
then U = 0.98 × (31000)²/(2·15.5×10⁶)
= 30.43 in³·psi = 2.536 ft·lbf
or 3.439 J
The maximum energy that the rod would store = 2.536 ft·lbf
2. Volcanic islands that form over mantle plumes, such as the Hawaiian chain, are home to some of Earth’s largest volcanoes. However, several volcanoes on Mars are gigantic compared to any on Earth. What does this difference tell us about the role of plate motion in shaping the Martian surface?
The theory that explains this phenomenon is linked to the convergence of tectonic plates when there is immersion over the crazy oceanic ones. This movement could be referred to as an oceanic-oceanic convergence where the immersion of two ocean slabs occurs and one descends below another plate and initiates volcanic activity by the same mechanism that operates in all subduction zones.
In this way the movement of the plate on the Martian surface could be relatively much faster than the occurrence of the movement of the plate on Earth. Giant volcanoes form because the area of the most oceanic crust converges faster.
A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft), is subjected to an acceleration a = {6ti + 12t^2k} ft/s^2. Determine the particle’s position (x, y, z) when t = 2 s.
The particle's position at t = 2 s is (11 ft, 2 ft, 21 ft).
To determine the particle's position at t = 2 s, we need to integrate the acceleration function twice. First, we'll find the velocity function:
[tex]\[ \mathbf{v}(t) =\int \mathbf{a}(t) \, dt \][/tex]
[tex]\[ \int (6t\mathbf{i} + 12t^2 \mathbf{k}) \, dt \\= 3t^2 \mathbf{i} + 4t^3 \mathbf{k} + \mathbf{C} \][/tex]
where C₁ is the constant of integration.
Next, we'll find the position function:
[tex]\[ \mathbf{v}(t) = \int (3t^2 \mathbf{i} + 4t^3 \mathbf{k} + \mathbf{C}_1) \, dt \\= t^3 \mathbf{i} + t^4 \mathbf{k} + \mathbf{C}_1 t + \mathbf{C}_2 \][/tex]
where C₂ is the constant of integration.
Given the initial conditions, we know that at t = 0, the particle is at rest (v = 0) and located at point (3 ft, 2 ft, 5 ft). We can use these conditions to find the values of C₁ and C₂:
[tex]c_{1} = (0 i + 0 j + 0 k) - (3(0^{2} i + 4(0^{4} k) = 0[/tex]
[tex]c_{2}[/tex][tex]= (3 i + 2 j + 5 k)[/tex] - ([tex]i + 0^{4}k[/tex] )
[tex]= (3 i + 2 j + 5 k)[/tex]
Now we can find the position at t = 2 s:
= [tex]2^{3}[/tex][tex]i[/tex] + [tex]2^{4}[/tex][tex]k + (0)(2) + (3 i + 2 j + 5 k)[/tex]
[tex]= (8 i + 16 k) + (3 i + 2 j + 5 k)[/tex]
[tex]= (11 i + 2 j + 21 k)[/tex]
So the particle's position at t = 2 s is (11 ft, 2 ft, 21 ft).
a) the pre-exponential factor and activation energy for the hydrolysis of t-butyl chloride are 2.1xE16 sec-1 and 102 kJ mol-1, respectively. calculate the values of delta S and delta H at 286 K for the reaction.
b)the pre-exponential factor and activation energy for the gas-phase cycloaddition of maleic anyhyrdride and cycopentadiene are 5.9xE7 M-1S-1 and 51 kJ mol-1 respectively. calculate values of delta S and delta H at 293 K for the reaction.
Answer:
Delta S = 356.64J/mol and Delta H = 99620J/mol
Explanation:
The detailed steps and appropriate use of the arrehenius equation and necessary susbtitution were made as shown in the attached file.
Determine the values of the lumped parasitic capacitor and inductor in a 5 ft long two-wire transmission line with air dielectric (See Chapter 7 in the Stanley and Harrington textbook). Each wire has a diameter of 1 mm and the separation between the wires is 1 inch. Show your calculations and the equations used.
Answer: See attachment below
Explanation:
What is the energy change when the temperature of 15.0 grams of solid silver is decreased from 37.3 °C to 20.5 °C ?
Answer:
Q=58.716 W
Explanation:
Given that
mass ,m = 15 g
The decrease in the temperature ,ΔT = 37.3 - 20.5 °C
ΔT = 16.8°C
We know that specific heat of silver ,Cp = 0.233 J/g°C
The energy change of the silver is given as
Q = m Cp ΔT
Now by putting the values we get
[tex]Q= 15 \times 0.233\times 16.8\ W[/tex]
Q=58.716 W
Therefore the change in the energy will be 58.716 W.
Which of the following method header represents Function(true, 0, 2.6)? A. void Function( bool b, double d, int c ) B. void Function( bool b, double d, double e ) C. void Function( bool b, int c, double d ) D. b and c
Answer:
(d) b and c
Explanation:
Given;
Function (true, 0, 2.6)
From the above, the function Function receives three arguments;
=>First argument (i.e true), is of type boolean
=>Second argument (i.e 0), is of type int, float or double since integers are inherently floating point numbers. Therefore 0 can be int, float or double.
=>Third argument (i.e 2.6), is a floating point number and could be of type float or double.
Therefore the method header for the function Function could be any of the following;
i). void Function (bool b, double d, double e)
ii). void Function (bool b, int c, double d)
iii). void Function (bool b, int d, float e)
iv). void Function (bool b, float d, double e)
v). void Function (bool b, double d, float e)
vi). void Function (bool b, float d, float e)
Note:
i. The ordered list above has ii and v in bold form to show that they are part of the options specified in the question.
ii. The letters b, c, d and e specified in the method declaration variations do not matter. Since they are just dummy variables, they can be any letter.
Therefore b and c are a correct representation for the method header.
Calculate the molar heat capacity of a monatomic non-metallic solid at 500K which is characterized by an Einstein temperature of 300K. Express your result as a multiple of 3R.
Answer:
Explanation:
Given
Temperature of solid [tex]T=500\ K[/tex]
Einstein Temperature [tex]T_E=300\ K[/tex]
Heat Capacity in the Einstein model is given by
[tex]C_v=3R\left [ \frac{T_E}{T}\right ]^2\frac{e^{\frac{T_E}{T}}}{\left ( e^{\frac{T_E}{T}}-1\right )^2}[/tex]
[tex]e^{\frac{3}{5}}=1.822[/tex]
Substitute the values
[tex]C_v=3R\times (\frac{300}{500})^2\times (\frac{1.822}{(1.822-1)^2})[/tex]
[tex]C_v=3R\times \frac{9}{25}\times \frac{1.822}{(0.822)^2}[/tex]
[tex]C_v=0.97\times (3R)[/tex]
How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an integer.
The number of electrons that move past a fixed reference point in 2.55 ps when the current is 7.3 extmu A is approximately 116, after calculating the total charge and then dividing it by the charge of an electron.
Explanation:The number of electrons moving past a fixed reference point in a given time interval when the electric current is known can be calculated by first determining the total charge that passes through the reference point and then dividing that by the charge per electron. The total charge ( extit{Q}) that moves through a fixed point is the product of the current ( extit{I}) and the time interval ( extit{t}), which is expressed by the formula extit{Q} = extit{I} imes extit{t}. Given that the charge of an electron is approximately -1.60 imes 10^{-19} C (coulombs), one can calculate the number of electrons by dividing the total charge by the electronic charge.
In this case, the current is 7.3 extmu A (microamperes), which is equal to 7.3 imes 10^{-6} A (amperes), and the time interval is 2.55 ps (picoseconds), or 2.55 imes 10^{-12} s (seconds). Using the formula extit{Q} = extit{I} imes extit{t}, the total charge moved past the reference point is calculated as follows:
Q = 7.3 imes 10^{-6} A imes 2.55 imes 10^{-12} s = 1.8615 imes 10^{-17} C
Next, to find the number of electrons that move past the reference point, we divide the total charge by the charge of an electron:
Number of electrons = rac{Q}{e} = rac{1.8615 imes 10^{-17} C}{1.60 imes 10^{-19} C/e^-}
This yields approximately 116.34 electrons. Since the question asks for an integer value, we round this to 116 electrons as the number of electrons that move past a fixed reference point every 2.55 ps if the current is 7.3 extmu A.
To find the number of electrons moving past a fixed reference point with a given current, you can use the formula: Current (i) = Charge (q) / Time (t). By substituting the known values and calculating, you can determine the number of electrons passing the point at a specific time.
To find the number of electrons moving past a fixed reference point with given current:
Given: Current (i) = 7.3 μA = 7.3 x 10^-6 A
Formula: Current (i) = Charge (q) / Time (t)
Calculation: Number of electrons = (Current x Time) / Charge of an electron
Substitute values and calculate.
Write a linear equation expressing the number of t-shirts sold each month as a function of price per t-shirt. Please be explicit about what your chosen variables represent. Find the slope and intercepts of this equation (including their units). What real life meaning do these values have?
Answer:
slope of the equation = -0.006778 and the intercept = 27.11
Explanation:
The following details were also given ; They predict the monthly T shirts to be 1750 shirts, if you price them at $15.25 per shirt or 791 shirts if you price them at $21.75 per shirt.
The detailed steps and calculation is as shown in the attached file.
Compute the estimated actual endurance limit for SAE 4130 WQT 1300 steel bar with a rectangular cross section of 20.0 mm by 60 mm. It is to be machined and subjected to repeated and reversed bending stress. A reliability of 99% is desired
Answer:
estimated actual endurance strength = 183.22 MPa
Explanation:
given data
actual endurance limit for SAE 4130 WQT 1300
cross section = 20.0 mm by 60 mm
reliability = 99%
solution
we use here table for get some value for AISI 4130 WQT 1300 steel
Sut = 676 MPa
Sn = 260 MPa
and here
stress factor Cst = 1
and wrought steel Cm = 1
and reliability factor Cr is = 0.81
so here equivalent diameter will be
equivalent diameter = 0.808 [tex]\sqrt{bh}[/tex] ...........1
equivalent diameter = 0. 808 [tex]\sqrt{20*60}[/tex]
equivalent diameter = 28 mm
so here size factor will be = 0.87 by the table
so now we can get estimated actual endurance strength will be
actual endurance strength = Sn × Cm × Cst × Cr × Cs ...............2
put here value
actual endurance strength = 260 × 1 × 1 × 0.81 × 0.87
actual endurance strength = 183.22 MPa
g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively, from their initial infinite positions. A. Determine the electric potential due to the two charges at the origin. B. Determine the electric potential energy associated with the system, relative to their infinite initial positions.
Answer:
a) V =10¹¹*(1.5q₁ + 3q₂)
b) U = 1.34*10¹¹q₁q₂
Explanation:
Given
x₁ = 6 cm
y₁ = 0 cm
x₂ = 0 cm
y₂ = 3 cm
q₁ = unknown value in Coulomb
q₂ = unknown value in Coulomb
A) V₁ = Kq₁/r₁
where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m
V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁
V₂ = Kq₂/r₂
where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m
V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂
The electric potential due to the two charges at the origin is
V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)
B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows
U = Kq₁q₂/r₁₂
where
r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m
then
U = 9*10⁹q₁q₂/(3√5/100)
⇒ U = 1.34*10¹¹q₁q₂
The following questions present a twist on the scenario above to test your understanding.
Suppose another stone is thrown horizontally from the same building. If it strikes the ground 67 m away, find the following values.
(a) time of flight
s
(b) initial speed
m/s
(c) speed and angle with respect to the horizontal of the velocity vector at impact
m/s
Answer:
a) t = √(2*h/g)
b) v₀ = 67*√(g/(2*h))
c) ∅ = tan⁻¹(2*h/67)
Explanation:
Suppose that the height of the building is known (h), then we have
a) y = h = g*t²/2
then the time of flight is
t = √(2*h/g)
b) The initial speed (v₀) can be obtained as follows
x = v₀*t ⇒ v₀ = x / t
where
x = xmax = 67 m and t = √(2*h/g)
So, we have
v₀ = 67 / √(2*h/g) = 67*√(g/(2*h))
c) In order to get the speed and the angle with respect to the horizontal of the velocity vector at impact, we obtain vx and vy as follows
vx = v₀ = 67*√(g/(2*h))
and
vy = gt = g*√(2*h/g) ⇒ vy = √(2*h*g)
then we apply
v = √(vx² + vy²) = √((67*√(g/(2*h)))² + (√(2*h*g))²)
⇒ v = √(g*(4489 + 4*h²)/(2*h))
The angle is obtained as follows
tan ∅ = vy / vx ⇒ tan ∅ = √(2*h*g) / 67*√(g/(2*h))
⇒ tan ∅ = 2*h / 67 ⇒ ∅ = tan⁻¹(2*h/67)
A section of highway has a free-flow speed of 55 mph and a capacity of 3300 veh/hr. In a given hour, 2100 vehicles were counted at a specified point along this highway section. If the linear speed density relationship applies, what would you estimate the space-mean speed of the vehicles to be?
Answer:
Space mean speed = 44 mi/h
Explanation:
Using Greenshield's linear model
q = Uf ( D - [tex]D^{2}[/tex]/Dj )
qcap = capacity flow that gives Dcap
Dcap = Dj/2
qcap = Uf. Dj/4
Where
U = space mean speed
Uf = free flow speed
D = density
Dj = jam density
now,
Dj = 4 × 3300/55
= 240v/h
q = Dj ( U - [tex]U^{2}[/tex]/Uf)
2100 = 240 ( U - [tex]U^{2}[/tex]/55)
Solve for U
U = 44m/h
Consider the following two well-mixed, isothermal gas-phase batch reactors for the elementary and irreversible decomposition of A to B given by the following chemical equation: A k −−→ 2 B (1) In reactor 1, the reactor volume is held constant (reactor pressure therefore changes). In reactor 2, the reactor pressure is held constant (reactor volume therefore changes). Both reactors are charged with pure A at 2.0 atm, A and B are considered ideal gases, and k = 0.35 min−1 .
(a) What is the fractional decrease in the concentration of A in reactors 1 and 2 after five minutes?
(b) What is the total molar conversion of A in reactors 1 and 2 after five minutes?
Answer:
attached below
Explanation:
Maintaining an optimistic attitude, building self-efficacy, strengthening stress inoculation, experiencing secure personal relationships, and practicing spirituality or religiosity are factors to help individuals thrive in the face of _____.
Optimistic attitude, self-efficacy, stress inoculation, secure relationships, and spirituality are factors that help individuals cope with adversity. These elements foster hardiness and stress-related growth, allowing individuals to manage chronic and acute stressors more effectively and pursue personal development.
Explanation:Maintaining an optimistic attitude, building self-efficacy, strengthening stress inoculation, experiencing secure personal relationships, and practicing spirituality or religiosity are factors to help individuals thrive in the face of adversity.
Adversity can include both chronic stressors, such as long-term unemployment, and acute stressors, like illnesses or loss. These stressors require coping mechanisms and personal strengths for an individual to manage effectively. Optimism is a general tendency to expect positive outcomes and it contributes to less stress and happier dispositions. Self-efficacy, defined by Albert Bandura, is the belief in one's abilities to reach goals, which empowers a proactive response to stressors. The concept of hardiness, which relates to optimism and self-efficacy, describes those who are less affected by life's stressors and use effective coping strategies.
Furthermore, stress-related growth or thriving describes the ability of individuals to exceed previous performances and show increased strength in the face of adversity. The nurturing of secure personal relationships provides social support that can aid in increasing positive affect and reducing the number of physical symptoms during stressful times.
2) The magnetic field at a distance of 2 cm from a current carrying wire is 4 μT. What is the magnetic field at a distance of 4 cm from the wire?
Answer:
2 μT
Explanation:
Assuming that the diameter of the current carrying wire is very small compared with its length, we can treat it as an infinite wire.
For an infinite wire, the magnetic field, at a distance r from the wire, is given by Biot-Savart law equation, as follows:
[tex]B = \frac{\mu0*I}{2*\pi*r}[/tex]
where:
μ₀ = 4*π*10⁻⁷
I = current carried by the wire
r = distance from the wire.
So, B₁ can be calculated as follows:
[tex]B1 = \frac{\mu0*I}{2*\pi*0.02m} = 4e-6 T[/tex]
If we change the distance where we want to know the value of B, all other factors remain constant, we can write the following expression for the new value of B, B₂:
[tex]B2 = \frac{\mu0*I}{2*\pi*0.04m}[/tex]
Dividing B1 by B2, and simplifying common terms, we have:
[tex]\frac{B1}{B2} = \frac{0.04m}{0.02m} = 2[/tex]
⇒ [tex]B2 = \frac{B1}{2} = \frac{4e-6T}{2} = 2e-6 T[/tex]
⇒ B₂ = 2 μT
Calculate the link parameter and channel utilization efficiency (in error-free channels) for a system with the following parameters, if the simplex stop and wait protocol is used. a) Bitrate: R=12 Mbps, Modulation: BPSK, distance between the transmitter and receiver: d-6 miles, propagation velocity:v3 x 108 m/s and the length of each frame: L-500 bits. b) Symbolrate: Rs.10 Msps, Modulation: QPSK, distance between the transmitter and receiver: d=15 km, propagation velocity: v 3 x 108 m/s and the length of each frame: L- 256 bits.
Answer: a) 0.77 and 0.39 b) 3.9 and 0.20
Explanation:
We have to find two things here i.e link parameter and channel utilization efficiency.
a)
We are given
Bitrate= 12 Mbps,
Distance= 6 miles
Propagation Velocity= 3 x 10⁸ m/s
Length of frame= 500 bits
We know that The link parameter is related to propagation time and frame time
Where Propagation time= Bitrate x Distance= 12 Mbps x 6 miles
and Frame time= Propagation velocity x length of frame= 3 x 10⁸ x 500
So,
Link parameter= [tex]\frac{12*10^6*6*1609.34}{3*10^8*500}[/tex]
Link parameter= 0.77
Channel utilization efficiency= [tex]\frac{1}{1+2* link parameter}[/tex]
Channel utilization efficiency= 0.39
b)
We are given
Bitrate= 10 Mbps,
Distance= 15 km
Propagation Velocity= 3 x 10⁸ m/s
Length of frame= 256 bits
We know that The link parameter is related to propagation time and frame time, Here QPSK is used so bitrate is multiplied by 2,
Where Propagation time= Bitrate x Distance= 2 x 10 Mbps x 15 km
and Frame time= Propagation velocity x length of frame= 3 x 10⁸ x 256
So,
Link parameter= [tex]\frac{2*10*10^6*15*1000}{3*10^8*256}[/tex]
Link parameter= 3.9
Channel utilization efficiency= [tex]\frac{1}{1+2* link parameter}[/tex]
Channel utilization efficiency= 0.20
A water tank filled with water to a depth of 16 ft has in inspection cover (1 in. 3 1 in.) at its base, held in place by a plastic bracket. The bracket can hold a load of 9 lbf. Is the bracket strong enough? If it is, what would the water depth have to be to cause the bracket to break?
Answer:
[tex]F=6.88 [lbf][/tex]
The bracket is strong enough.
[tex]h=20.91 [ft][/tex]
Explanation:
Let's recall that the variation of the pressure respect to displacement in a liquid incomprehensible and static will be:
[tex]\frac{dP}{dy}=-\rho g[/tex]
If we take ρ (density) as a constant and solving this differential equation, we will have:
[tex]\Delta P=\rho gh[/tex]
P is the total pressureh is the heightNow, the pressure at the base will be:
[tex]P_{base}=\rho gh[/tex]
We use this equation knowing that we have atmospheric pressure on the outside of the tank.
The force on the inspection cover will be (A=1 in²):
[tex]F=P_{base}A=\rho ghA= 62.4 [lb/ft^{3}]*32.2 [ft/s^{2}]*16 [ft]*0.00689 [ft^{2}]=221.47 [\frac{lb*ft}{s^{2}}][/tex]
We know that 1 lbf = 32.17 (lb*ft)/s², so:
[tex]F=6.88 [lbf][/tex]
The statement says that the bracket can hold a load of 9 lbf, therefore the bracket is strong enough.
We can use the equation of the force to find the depth.
[tex]F=\rho ghA[/tex]
If we solve it for h we will have:
[tex]h=\frac{F}{\rho gA}[/tex]
F is the force that bracket can hold (9 lbf or 289.53 (lb*ft)/s²)A is the area (A=0.00689 ft²)[tex]h=\frac{289.53}{62.4*32.2*0.00689}=20.91 [ft][/tex]
I hope it helps you!
The passageway between the riser and the main cavity must have a small cross-sectional area to minimize waste of casting metal: True or False
Answer:
True
Explanation:
The passageway between the riser and the main cavity must have a small cross-sectional area to reduce waste of material in metal casting.
Define a function pyramid_volume with parameters base_length, base_width, and pyramid_height, that returns the volume of a pyramid with a rectangular base. Sample output with inputs: 4.5 2.1 3.0 Volume for 4.5, 2.1, 3.0 is: 9.45 Relevant geometry equations: Volume
Hi, you haven't provided the programing language in which you need the code, I'll just explain how to do it using Python, and you can apply a similar method for any programming language.
Answer:
1. def pyramid_volume(base_length, base_width, pyramid_height):
2. volume = base_length*base_width*pyramid_height/3
3. return(volume)
Explanation step by step:
In the first line of code, we define the function pyramid_volume and it's input parametersIn the second line, we perform operations with the input values to get the volume of the pyramid with a rectangular base, the formula is V = l*w*h/3In the last line of code, we return the volumeIn the image below you can see the result of calling the function with input 4.5, 2.1, 3.0.
The function 'pyramid_volume' can be used to compute the volume of a pyramid with a rectangular base. Its calculation uses the equation: V = (base_area * height) / 3, where base_area = base_length * base_width. Given the inputs 4.5, 2.1, and 3.0 for base_length, base_width, and pyramid_height respectively, the volume would be 9.45
Explanation:The function pyramid_volume is best defined in the context of Mathematics, particularly, geometric formulas. To compute the volume of a pyramid with a rectangular base, you can utilize the formula: V = (base_area * height) / 3. Here, base_area is equal to base_length times base_width. Hence, the function pyramid_volume could be defined as follows:
Volume = lambda base_length, base_width, pyramid_height: (base_length * base_width * pyramid_height) / 3.
Thus, if you input 4.5, 2.1, and 3.0 as your base_length, base_width, pyramid_height respectively into the function, it will return the volume of the pyramid, which in this case is 9.45.
Learn more about Volume Computation here:https://brainly.com/question/33636745
#SPJ3
A submarine dives at a constant speed pf 3 m/s. It is know that the water pressure increase at a rate of one atmospheric pressure (105 Pa) per 10 meter of depth. Find the time rate of pressure change measured by the submarine.
Answer:
dP/dt = 0.3003003003
Explanation:
speed of submarine = 3m/s
rate of change of pressure with depth = (105 Pa) per 10 meter of depth
to get time taken by the submarine to cover 10m ; t = 10/3 = 3.33s
time rate of change of pressure = dP/dt = 105 Pa/3.33
= 30030.03003or 0.3003003003
Write a static method called quadrant that takes as parameters a pair of real numbers representing an (x, y) point and that returns the quadrant number for that point. Recall that quadrants are numbered as integers from 1 to 4 with the upper-right quadrant numbered 1 and the subsequent quadrants numbered in a counter-clockwise fashion:
Full Question
Write a static method called quadrant that takes as parameters a pair of real numbers representing an (x, y) point and that returns the quadrant number for that point.
Recall that quadrants are numbered as integers from 1 to 4 with the upper-right quadrant numbered 1 and the subsequent quadrants numbered in a counter-clockwise fashion. Notice that the quadrant is determined by whether the x and y coordinates are positive or negative numbers. If a point falls on the x-axis or the y-axis, then the method should return 0.
Answer
public int quadrant(double x, double y) // Line 1
{
if(x > 0 && y > 0){ //Line 2
return 1;
}
if(x < 0 && y > 0){ // Line 3
return 2;
}
if(x < 0 && y < 0) {// Line 4
return 3;
}
if(x > 0 && y < 0) {// Line 5
return 4;
}
return 0;
}
Explanation:
Line 1 defines the static method.
The method is named quadrant and its of type integer.
Along with the method definition, two variables x and y of type double are also declared
Line 2 checks if x and y are greater than 0.
If yes then the program returns 1
Line 3 checks if x is less than 0 and y is greater than 0
If yes then the program returns 2
Line 4 checks if x and y are less than 0
If yes then the program returns 3
Line 5 checks is x is greater than 0 and y is less than 0
If yes then the program returns 4
An ant starts at one edge of a long strip of paper that is 34.2 cm wide. She travels at 1.3 cm/s at an angle of 61◦ with the long edge. How long will it take her to get across? Answer in units of s.
Answer:
t = 30.1 sec
Explanation:
If the ant is moving at a constant speed, the velocity vector will have the same magnitude at any point, and can be decomposed in two vectors, along directions perpendicular each other.
If we choose these directions coincident with the long edge of the paper, and the other perpendicular to it, the components of the velocity vector, along these axes, can be calculated as the projections of this vector along these axes.
We are only interested in the component of the velocity across the paper, that can be calculated as follows:
vₓ = v* sin θ, where v is the magnitude of the velocity, and θ the angle that forms v with the long edge.
We know that v= 1.3 cm/s, and θ = 61º, so we can find vₓ as follows:
vₓ = 1.3 cm/s * sin 61º = 1.3 cm/s * 0.875 = 1.14 cm/s
Applying the definition of average velocity, we can solve for t:
t =[tex]\frac{x}{vx}[/tex] = [tex]\frac{34.2 cm}{1.14 cm/s} =30.1 sec[/tex]
⇒ t = 30.1 sec
Please write the command(s) you should use to achieve the following tasks in GDB. 1. Show the value of variable "test" in hex format. 2. Show the top four bytes on your stack word-by-word, e.g. it should look something like this "0x0102 0x0304", NOT "0x01020304". 3. Check all register values. 4. Set a breakpoint at "ece.c" line 391
You can use a series of commands to accomplish various tasks in GDB: 'print /x test' to view a variable in hex format, 'x/4wx $sp' to read the top four bytes on your stack by word, 'info registers' to check all register values, and 'break ece.c:391' to set a breakpoint at a specific line.
Explanation:To achieve the tasks in GDB, you will need to use the following commands:
To show the value of variable "test" in hexadecimal format, use the command: 'print/x test'.Read the top four bytes on your stack word-by-word with the command: 'x/4wx $sp'.Check all register values with the command: 'info registers'.To set a breakpoint at "ece.c" line 391, use the command: 'break ece.c:391'.Learn more about GDB Commands here:
https://brainly.com/question/33364038
#SPJ2
Peter the postman became bored one night and, to break the monotony of the night shift, he carried out the following experiment with a row of mailboxes in the post office. These mailboxes were numbered 1 through 150, and beginning with mailbox 2, he opened the doors of all the even-numbered mailboxes, leaving the others closed. Next, beginning with mailbox 3, he went to every third mail box, opening its door if it were closed, and closing it if it were open. Then he repeated this procedure with every fourth mailbox, then every fifth mailbox, and so on. When he finished, he was surprised at the distribution of closed mailboxes. Write a program to determine which mailboxes these were.
Answer:
//This Program is written in C++
// Comments are used for explanatory purpose
#include <iostream>
using namespace std;
enum mailbox{open, close};
int box[149];
void closeAllBoxes();
void OpenClose();
void printAll();
int main()
{
closeAllBoxes();
OpenClose();
printAll();
return 0;
}
void closeAllBoxes()
{
for (int i = 0; i < 150; i++) //Iterate through from 0 to 149 which literarily means 1 to 150
{
box[i] = close; //Close all boxes
}
}
void OpenClose()
{
for(int i = 2; i < 150; i++) {
for(int j = i; j < 150; j += i) {
if (box[j] == close) //Open box if box is closed
box[j] = open;
else
box[j] = close; // Close box if box is opened
}
}
// At the end of this test, all boxes would be closed
}
void printAll()
{
for (int x = 0; x < 150; x++) //use this to test
{
if (box[x] = 1)
{
cout << "Mailbox #" << x+1 << " is closed" << endl;
// Print all close boxes
}
}
}
Explanation:
You buy a 75-W lightbulb in Europe, where electricity is delivered to homes at 240V. If you use the lightbulb in the United States at 120 V (assume its resistance does not change), how bright will it be relative to 75-W 129-V bulbs?
Answer:
The bulb will be [tex]\frac{1}{4}[/tex] times as bright as it is in Europe.
Explanation:
Data provided in the question:
Power of bulb in Europe = 75 W
Voltage provided in Europe = 240 V
Voltage provided in United Stated = 120 V
Now,
We know,
Power = Voltage² ÷ Resistance
Therefore,
75 = 240² ÷ Resistance
or
Resistance = 240² ÷ 75
or
Resistance = 768 Ω
Therefore,
Power in United States = Voltage² ÷ Resistance
= 120² ÷ 768
= 18.75 W
Therefore,
Ratio of powers
[tex]\frac{\text{Power in united states}}{\text{Power in Europe}}=\frac{18.75}{75}[/tex]
= [tex]\frac{1}{4}[/tex]
Hence,
The bulb will be [tex]\frac{1}{4}[/tex] times as bright as it is in Europe.
Define the following concepts in your own words: (a) stiffness, (b) strength, (c) ductility, (d) yielding, (e) toughness, and (f) strain hardening.
Answer:
Please find detail answer given below
Explanation:
(a) Stiffness
It is defined as the property of a material that is rigid and difficult to bend. The example of stiffness is the rubber band. If an elastic band is stretched with two fingers, the stiffness is less and the flexibility is greater. Similarly, if we use the rubber band assembly and stretch it with two fingers, the stiffness will be more rigid and the flexibility will be less.
(b) Strength
Force measures how much stress can be applied to an element before it is permanently deformed or fractured.
(c) ductility
Ductility is a measure of a metal's ability to resist tensile stress: any force that separates the two ends of an object. The tug of war game provides a good example of the tension of tension that is applied to a rope. Ductility is the plastic deformation that occurs in the metal as a result of such types of deformation. The term "ductile" literally means that a metallic substance is capable of stretching on a thin wire without weakening or becoming more fragile in the process.
(d) Yielding
The performance of a material is explained as the tension at which a material begins to deform irreversibly. Before the elastic limit, the material will deform elastically, which means that it will return to its original shape when the applied tension is eliminated (that is, without permanent and visible change in the shape of the material).
(e) toughness
The ability of a metal to deform plastically and to absorb energy in the process before the fracture is called resistance. The emphasis of this definition should be placed on the ability to absorb energy before the fracture.
(f) strain hardening
The process of strain hardening is at what time when metal is strained beyond the yield point. An intensifying pressure is required to produce extra plastic deformation and the metal apparently becomes stronger and more difficult to deform.
Consider the generic state space model *(t)= Ax(t) + Bu(t) y(t) = Cx(t)+Du(t). Use the definition of Al to show that x(t)=1*[4=) Bu(t)dt is the state response under input u(t).
Answer:
Explanation:
[tex]\dot x (t) - ax(t)=bu(t)\\\\e^{-at}\dot x = e^{-at}ax = \frac{d}{dt} (e^{-at})=e^{-at}bu\\\\\int\limits^t_0 \frac{d}{d\tau}(e^{-a\tau}x(\tau))d\tau=e^{-at}x(t)-x(0)=\int\limits^t_0 e^{-a\tau}bu(\tau))d\tau\\\\x(t)=e^{at}x(0)+ \int\limits^t_0 e^{-a(t-\tau)}bu(\tau))d\tau[/tex]
Similarly,
[tex]e^{-At}\dot x(t) -e^{-At}Ax(t) = \frac{d}{dt} (e^{-At}x(t))=e^{-At}Bu(t)\\\\\int\limits^t_0 \frac{d}{d\tau}(e^{-A\tau}x(\tau))d\tau=e^{-At}x(t)-e^{-A.0}x(0)=\int\limits^t_0 e^{-A\tau}Bu(\tau))d\tau\\\\since\:\:\:e^{-A.0}=I \:\:\: and\:\:\:[e^{-At}]^{-1}=e^{At}\\\\x(t)=e^{At}x(0)+ e^{At}\int\limits^t_0 e^{-A\tau}Bu(\tau))d\tau\\\\x(t)=e^{At}x(0)+ \int\limits^t_0 e^{A(t-\tau)}Bu(\tau))d\tau[/tex]
For initial condition x(0) = 0
[tex]x(t)= \int\limits^t_0 e^{A(t-\tau)}Bu(\tau))d\tau[/tex]
A signal is band-limited between 2 kHz and 2.1 kHz. If the signal is sampled at 1750 Hz, the baseband signal will lie between?a. 00 and 650 Hzb. 750 and 2000 Hzc. 100 and 110 Hzd. 250 and 350 Hz
Answer:
Explanation: 100 and 110 Hz
Here fL=2kHz, fH=2.1kHz, B=0.1kHz, fL=2kHz,
n≤fH/(fH-fL)
n≤(2.1*1000)/((2.1-2)*1000)
n≤20
2fH/n ≤ fs
(2*2.1*1000)/20 ≤ fs
210≤ fs
The velocity components u and v in a two-dimensional flow field are given by: u = 4yt ft./s, v = 4xt ft./s, where t is time. What is the time rate of change of the velocity vector V (i.e., the acceleration vector) for a fluid particle at x = 1 ft. and y = 1 ft. at time t = 1 second?
Answer:
vec(a) = 16 i + 16 j
mag(a) = 22.63 ft/s^2
Explanation:
Given,
- The two components of velocity are given for fluid flow:
u = 4*y ft/s
v = 4*x ft/s
Find:
What is the time rate of change of the velocity vector V (i.e., the acceleration vector) for a fluid particle at x = 1 ft. and y = 1 ft. at time t = 1 second?
Solution:
- The rate of change of velocity is given to be acceleration. We will take derivative of each components of velocity with respect to time t:
a_x = du / dt
a_x = 4*dy/dt
a_y = dv/dt
a_y = 4*dx/dt
- The expressions dx/dt is the velocity component u and dy/dt is the velocity component v:
a_x = 4*(4*y) = 16y
a_y = 4*(4*x) = 16x
- The acceleration vector can be expressed by:
vec(a) = 16y i + 16x j
- Evaluate vector (a) at x = 1 and y = 1:
vec(a) = 16*1 i + 16*1 j = 16 i + 16 j
- The magnitude of acceleration is given by:
mag(a) = sqrt ( a^2_x + a^2_y )
mag(a) = sqrt ( 16^2 + 16^2 )
mag(a) = 22.63 ft/s^2