Answer:
A. The biceps muscle contracts while the triceps muscle relaxes. Only then can the forearm move up to lift the mug.
Explanation:
Skeletal muscle is also known as voluntary muscle and it can be defined as a type of muscle connected with the skeleton by tendons, so as to form a mechanical system that enables the movement of the limbs and other body parts with respect to another.
Generally, skeletal muscles are only found in vertebrates.
The muscles in the arm work to lift a mug of coffee to your lips when the biceps muscle contracts while the triceps muscle relaxes.
Basically, the only way the forearm move up to lift a mug for example, is through the contraction of the biceps muscle and the relaxation of the triceps muscle.
This ultimately implies that, the elbow joint is straightened or bent due to the actions of the biceps muscle and triceps muscle against each other.
To bend the elbow and raise the forearm, the biceps muscle contracts and the triceps muscle relaxes.To straighten the elbow and lower the forearm, the biceps muscle relaxes and the triceps muscle contracts.Answer:A
Explanation:
got it right on plato
Cite another example of information technology companies pushing the boundaries of privacy issues; apologizing, and then pushing again once scandal dies down. As long as the controversy fades, is there anything unethical about such a strategy?
Answer:
Explanation:
Tech Social Media giant FB is one of those companies. Not long ago the ceo was brought to court to accusations that his company was selling user data. Turns out this is true and they are selling their users private data to companies all over the word. Once the news turned to something else, people focused on something new but the company still continues to sell it's users data the same as before. This is completely unethical as the information belongs to the user and they are not getting anything while the corporation is profiting.
Draw a sinusoidal signal and illustrate how quantization and sampling is handled by
using relevant grids.
A multipurpose transformer has a secondary coil with several points at which a voltage can be extracted, giving outputs of 5.60, 12.0, and 480 V. (a) The input voltage is 220 V to a primary coil of 230 turns. What are the numbers of turns in the parts of the secondary used to produce the output voltages
Answer:
Explanation:
A multipurpose transformer can act as step up as well as step down transformer according to the desired setting by a user.
When the voltage at the output is greater than the voltage at the input of the transformer then it acts as step-up transformer and vice-versa acting is a step down transformer.
Given that:
input (primary) voltage of the transformer, [tex]V_i=220~V[/tex]
no. of turns in the primary coil, [tex]N_i=230[/tex]
When the output voltage is 5.60 V:[tex]V_o=5.60~V[/tex]
[tex]\frac{N_i}{N_o} =\frac{V_i}{V_o}[/tex]
[tex]\frac{N_o}{230}=\frac{5.60}{220}[/tex]
[tex]N_o=5.85\approx 6[/tex] turns compensating the losses
When the output voltage is 12.0 V:[tex]V_o=12.0~V[/tex]
[tex]\frac{N_i}{N_o} =\frac{V_i}{V_o}[/tex]
[tex]\frac{N_o}{230}=\frac{12.0}{220}[/tex]
[tex]N_o=12.45\approx 13[/tex] turns compensating the losses
When the output voltage is 480 V:[tex]V_o=480~V[/tex]
[tex]\frac{N_i}{N_o} =\frac{V_i}{V_o}[/tex]
[tex]\frac{N_o}{230}=\frac{480}{220}[/tex]
[tex]N_o=501.8\approx 502[/tex] turns compensating the losses
1. Add:
(i) 5xy, -2xy, -11xy, 8xy
(iv) 3a - 2b + c, 5a + 8b -70
Answer:
(i) 0
(iv) 8a+6b+c-70
Explanation:
Hope this helps you
An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a lagging power factor of 0.77. Determine the size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging.
Answer:
[tex]Q=41.33 KVAR\ \\at\\\ 480 Vrms[/tex]
Explanation:
From the question we are told that:
Voltage [tex]V=480/0 \textdegree V[/tex]
Power [tex]P=120kW[/tex]
Initial Power factor [tex]p.f_1=0.77 lagging[/tex]
Final Power factor [tex]p.f_2=0.9 lagging[/tex]
Generally the equation for Reactive Power is mathematically given by
Q=P(tan \theta_2-tan \theta_1)
Since
[tex]p.f_1=0.77[/tex]
[tex]cos \theta_1 =0.77[/tex]
[tex]\theta_1=cos^{-1}0.77[/tex]
[tex]\theta_1=39.65 \textdegree[/tex]
And
[tex]p.f_2=0.9[/tex]
[tex]cos \theta_2 =0.9[/tex]
[tex]\theta_2=cos^{-1}0.9[/tex]
[tex]\theta_2=25.84 \textdegree[/tex]
Therefore
[tex]Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)[/tex]
[tex]Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)[/tex]
[tex]Q=-41.33VAR[/tex]
Therefore
The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is
[tex]Q=41.33 KVAR\ \\at\\\ 480 Vrms[/tex]
Imagine a cantilever beam fixed at one end with a mass = m and a length = L. If this beam is subject to an inertial force and a uniformly distributed load = w, what is the moment present at a length of L/4?
Answer:
jsow
hfhcffnbxhdhdhdhdhdhdddhdhdgdhdhdhdhdhdhhhdhdjsksmalalaksjdhfgrgubfghhhhhhh
Explanation:
j
grudbAn ideal neon sign transformer provides 9130 V at 51.0 mA with an input voltage of 240 V. Calculate the transformer's input power and current.
Answer:
Input power = 465.63 W
current = 1.94 A
Explanation:
we have the following data to answer this question
V = 9130
i = 0.051
the input power = VI
I = 51.0 mA = 0.051
= 9130 * 0.051
= 465.63 watts
the current = 465.63/240
= 1.94A
therefore the input power is 465.63 wwatts
while the current is 1.94A
the input power is the same thing as the output power.
A levee will be constructed to provide some flood protection for a residential area. The residences are willing to accept a one-in-five chance that the levee will be overtopped in the next 15 years. Assuming that the annual peak streamflow follows a lognormal distribution with a log10(Q[ft3/s]) mean and standard deviation of 1.835 and 0.65 respectively, what is the design flow in ft3/s?
Answer:
1709.07 ft^3/s
Explanation:
Annual peak streamflow = Log10(Q [ft^3/s] )
mean = 1.835
standard deviation = 0.65
Probability of levee been overtopped in the next 15 years = 1/5
Determine the design flow ins ft^3/s
P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2
∴ T = 67.72 years
Q₁₅ = 1 - 0.2 = 0.8
Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )
K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )
= 2.1504
back to equation 1
Zt = 1.835 + ( 2.1504 * 0.65 ) = 3.23276
hence:
Log₁₀ ( Qt(ft^3/s) ) = Zt = 3.23276
hence ; Qt = 10^3.23276
= 1709.07 ft^3/s
The following is a correlation for the average Nusselt number for natural convection over spherical surface. As can be seen in the above, the Nusselt number approaches 2 as Rayleigh number approaches zero. Prove that this situation corresponds to conduction heat transfer and in conduction heat transfer over sphere, the Nusselt number becomes 2. Hint: First step: Write an expression for heat transfer between two spherical shells that share the same center. Second step: Assume the outer spherical shell is infinitely large.
Answer:
Explanation:
[tex]r_2=[/tex]∞
[tex]q=4\pi kT_1(T_2-T_1)\\[/tex]
[tex]q=2\pi kD.[/tex]ΔT--------(1)
[tex]q=hA[/tex] ΔT[tex]=4\pi r_1^2(T_2_s-T_1_s)\\[/tex]
[tex]N_u=\frac{hD}{k} = 2+\frac{0.589 R_a^\frac{1}{4} }{[1+(\frac{0.046}{p_r}\frac{9}{16} )^\frac{4}{9} } ------(3)[/tex]
By equation (1) and (2)
[tex]2\pi kD.[/tex]ΔT=h.4[tex]\pi r_1^2[/tex]ΔT
[tex]2kD=hD^2\\\frac{hD}{k} =2\\N_u=\frac{hD}{k}=2\\[/tex]-------(4)
From equation (3) and (4)
So for sphere [tex]R_a[/tex]→0
Suppose a causal CT LTI system has bilateral Laplace transform H(s) 2s - 2 $2 + (10/3)s + 1 (8)
(a) Write the linear constant coefficient differential equation (LCCDE) relating a general input x(t) to its corresponding output y(t) of the system corresponding to this transfer function in equation (8).
(b) Suppose the input x(t) = e-tu(t). Find the output y(t). In part (c), the output signal can be expressed as y(t) = - e-(1/3)t u(t) + e-tu(t) e-3tu(t), - 019 Where a, b, and care positive integers. What are they? a = b = C=
Solution :
Given :
[tex]$H(S) =\frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]
Transfer function, [tex]$H(S) =\frac{Y(S)}{K(S)}= \frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]
[tex]$Y(S) \left(S^2+\frac{10}{3}S+1\right) = (2S-2) \times (S)$[/tex]
[tex]$S^2Y(S) + \frac{10}{3}(SY(S)) + Y(S) = 2(S \times (S)) - 2 \times (S)$[/tex]
Apply Inverse Laplace Transforms,
[tex]$\frac{d^2y(t)}{dt^2} + \frac{10}{3} \frac{dy(t)}{dt} + y(t)=2 \frac{dx(t)}{dt} - 2x(t)$[/tex]
The above equation represents the differential equation of transfer function.
Given : [tex]$x(t)=e^{-t} u(t) \Rightarrow X(S) = \frac{1}{S+1}$[/tex]
We have : [tex]$H(S) =\frac{Y(S)}{K(S)}= \frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]
[tex]$Y(S) = X(S) \times \frac{6S-6}{3S^2+10 S + 3} = \frac{6S-6}{(S+1)(3S+1)(S+3)}$[/tex]
[tex]$Y(S) = \frac{A}{S+1}+\frac{B}{3S+1} + \frac{C}{S+3}[/tex]
[tex]$A = Lt_{S \to -1} (S+1)Y(S)=\frac{6S-6}{(3S+1)(S+3)} = \frac{-6-6}{(-3+1)(-1+3)} = 3$[/tex]
[tex]$B = Lt_{S \to -1/3} (3S+1)Y(S)=\frac{6S-6}{(S+1)(S+3)} = \frac{-6/3-6}{(1/3+1)(-1/3+3)} = \frac{-9}{2}$[/tex]
[tex]$C = Lt_{S \to -3} (S+3)Y(S)=\frac{6S-6}{(S+1)(3S+1)} = \frac{-18-6}{(-3+1)(-9+1)} = \frac{-3}{2}$[/tex]
So,
[tex]$Y(S) = \frac{3}{S+1} - \frac{9/2}{3S+1} - \frac{3/2}{S+3}$[/tex]
[tex]$=\frac{3}{S+1} - \frac{3/2}{S+1/3} - \frac{3/2}{S+3}$[/tex]
Applying Inverse Laplace Transform,
[tex]$y(t) = 3e^{-t}u(t)-\frac{3}{2}e^{-t/3}u(t) - \frac{3}{2}e^{-3t} u(t)$[/tex]
[tex]$=\frac{-3}{2}e^{-\frac{1}{3}t}u(t) + \frac{3}{1}e^{-t}u(t)-\frac{3}{2}e^{-3t} u(t)$[/tex]
where, a = 2
b = 1
c= 2
a) Complete the following methods description using the correct tense for the verb in brackets. (This student is using passive voice rather than any human agents at the request of the instructor.) Student Lab Report Identical tensile test procedures were performed on all test specimens. Each of the metal specimens ____1____ [have] an indentation near the center to ensure that the fracture point would occur in this region. Tension tests ____2____ [conduct] as follows. Two pieces of reflective tape ____3____ [place] approximately 1 inch apart in the center of the specimen where the indentation 4 [locate]. The width and the thickness of the specimen at this location _____5_____ [measure] using a Vernier caliper. Then the specimen _____6____ [secure] in the MTS Load Frame. A laser extensometer _____7_____ [place] into position to measure the deformation of the specimen. The laser extensometer ______8_ __ [use] to measure the original distance between the pieces of reflective tape. The MTS ________9____ [set] to elongate the specimen one tenth of an inch every minute.
Answer:
Each of the metal specimens HAS an indentation near the center to ensure that the fracture point would occur in this region. Tension tests WERE CONDUCTED as follows. Two pieces of reflective tape WERE PLACED approximately 1 inch apart in the center of the specimen where the indentation 4 WAS LOCATED. The width and the thickness of the specimen at this location WAS MEASURED using a Vernier caliper. Then the specimen WAS SECURED in the MTS Load Frame. A laser extensometer WAS PLACED into position to measure the deformation of the specimen. The laser extensometer WAS USED to measure the original distance between the pieces of reflective tape. The MTS WAS SET to elongate the specimen one tenth of an inch every minute.
Water steam enters a turbine at a temperature of 400 o C and a pressure of 3 MPa. Water saturated vapor exhausts from the turbine at a pressure of 125 kPa. At steady state, the work output of the turbine is 530 kJ/kg. The surrounding air is at 20 o C. Neglect the changes in kinetic energy and potential energy. Determine (20 points) (a) the heat transfer from the turbine to the surroundings per unit mass flow rate, (b) the entropy generation during this process.
Answer:
a) -505.229 kJ/Kg
b) -1.724 kJ/kg
Explanation:
T1 = 400°C
P1 = 3 MPa
P2 = 125 kPa
work output = 530 kJ/kg
surrounding temperature = 20°C = 293 k
A) Calculate heat transfer from Turbine to surroundings
Q = h2 + w - h1
h ( enthalpy )
h1 = 3231.229 kj/kg
enthalpy at P2
h2 = hg = 2676 kj/kg
back to equation 1
Q = 2676 + 50 - 3231.229 = -505.229 kJ/Kg ( i.e. heat is lost )
b) Entropy generation
entropy generation = Δs ( surrounding ) + Δs(system)
= - 505.229 / 293 + 0
= -1.724 kJ/kg
A container can be made from steel [β = 36 × 10-6 (C°)-1] or lead [β = 87 × 10-6 (C°)-1]. A liquid is poured into the container, filling it to the brim. The liquid is either water [β = 207 × 10-6 (C°)-1] or ethyl alcohol [β = 1120 × 10-6 (C°)-1]. When the full container is heated, some liquid spills out. To keep the overflow to a minimum, from what should the container be made and what should the liquid be
Answer:
The container should be made of lead and the liquid should be water.
Explanation:
Since the volume of the container of liquid after expansion is V = V₀(1 + βΔθ) where V = initial volume, β = coefficient of volume expansion, Δθ = temperature change.
So, the volume change V₂ - V₁ where V₁ = volume of liquid and V₂ = volume of container
For steel, V₂ = V₀(1 + β₂Δθ) and V₁ = V₀(1 + β₁Δθ)
So, ΔV = V₀(1 + β₂Δθ) - V₀(1 + β₁Δθ) = V₀[1 + β₂Δθ - 1 - β₁Δθ] = V₀[β₂Δθ - β₁Δθ]
Since we want a minimum value for ΔV and V₀ and Δθ are the same, we need ΔV/V₀Δθ = β₂ - β₁ to be a minimum
where β₂ = coefficient of volume expansion of liquid and β₁ = coefficient of volume expansion of container.
So, trying each combination, with β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 36 × 10⁻⁶ (C°)⁻¹
β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 36 × 10⁻⁶ (C°)⁻¹ = 171 × 10⁻⁶ (C°)⁻¹
With β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 × 10⁻⁶ (C°)⁻¹
β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 87 × 10⁻⁶ (C°)⁻¹ = 120 × 10⁻⁶ (C°)⁻¹
With β₂ = 1120 × 10⁻⁶ (C°)⁻¹] and β₁ = 36 × 10⁻⁶ (C°)⁻¹
β₂ - β₁ = 1120 × 10⁻⁶ (C°)⁻¹ - 36 × 10⁻⁶ (C°)⁻¹ = 1084 × 10⁻⁶ (C°)⁻¹
With β₂ = 1120 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 × 10⁻⁶ (C°)⁻¹
β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 87 × 10⁻⁶ (C°)⁻¹ = 1033 × 10⁻⁶ (C°)⁻¹
The combination that gives the lowest value for β₂ - β₁ is β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 × 10⁻⁶ (C°)⁻¹
Since β₁ = 87 × 10⁻⁶ (C°)⁻¹ = coefficient of expansion for lead β₂ = 207 × 10⁻⁶ (C°)⁻¹] = coefficient of expansion for water, the container should be made of lead and the liquid should be water.
Consider two houses that are identical, except that the walls are built using bricks in one house, and wood in the other. The walls of the brick house are twice as thick. Which house do you think will be more energy efficient?
Answer:
Walls Built Using Bricks and Wood
The brick house is more energy-efficient than the one built with wood.
Explanation:
Because of their high thermal mass, which gives bricks the ability to absorb heat and release it over time, bricks remain more energy-efficient than other building materials, including wood. In summer, bricks leave your home cool. In winter, they make it warm. With these two advantages provided by bricks over other building materials, bricks are the most energy-efficient building material.
Lab 5A Problem Input two DWORD values from the keyboard. Determine which number is larger or if they are even. Your program should look like the following: First number larger Enter a number 12 Enter a number 10 12 is the larger number Press any key to close this window... Second number larger Enter a number 10 Enter a number 12 12 is the larger number Press any key to close this window... Numbers Equal Enter a number 12 Enter a number 12 Numbers are equal Press any key to close this window...
Answer:
Explanation:
#include<iostream>
using namespace std;
int main()
{
int n1,n2;
cout<<"Enter a number:"<<endl; //Entering first number
cin>>n1;
cout<<"Enter a number:"<<endl; //Entering second number
cin>>n2;
if(n1%2==0 && n1%2==0) //Checking whether the two number are even or not
{
if(n1>n2)
{
cout<<n1<<" is the larger number"<<endl;
}
else if(n1==n2)
{
cout<<"Numbers are equal"<<endl;
}
else
{
cout<<n2<<" is the larger number"<<endl;
}
}
else
{
cout<<"The number are not even"<<endl;
}
}
Explain by Research how a basic generator works ? using diagram
ow Pass Filter Design 0.0/5.0 points (graded) Determine the transfer function H(s) for a low pass filter with the following characteristics: a cutoff frequency of 100 kHz a stopband attenuation rate of 40 dB/decade. a nominal passband gain of 20 dB, which drops to 14 dB at the cutoff frequency Write the formula for H(s) that satisfies these requirements:
Answer:
H(s) = 20 / [ 1 + s / 10^5 ]^2
Explanation:
Given data:
cutoff frequency = 100 kHz
stopband attenuation rate = 40 dB/decade
nominal passband gain = 20 dB
new nominal passband gain at cutoff = 14 dB
Represent the transfer function H(s)
The attenuation rate show that there are two(2) poles
H(s) = k / [ 1 + s/Wc ]^2 ----- ( 1 )
where : Wc = 100 kHz = 10^5 Hz , K = 20 log k = 20 dB ∴ k = 20
Input values into equation 1
H(s) = 20 / [ 1 + s / 10^5 ]^2
If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phase
Answer:
The heat input from the combustion phase is 2000 watts.
Explanation:
The energy efficiency of the heat engine ([tex]\eta[/tex]), no unit, is defined by this formula:
[tex]\eta = \frac{\dot W}{\dot Q}[/tex] (1)
Where:
[tex]\dot Q[/tex] - Heat input, in watts.
[tex]\dot W[/tex] - Power output, in watts.
If we know that [tex]\dot W = 600\,W[/tex] and [tex]\eta = 0.3[/tex], then the heat input from the combustion phase is:
[tex]\eta = \frac{\dot W}{\dot Q}[/tex]
[tex]\dot Q = \frac{\dot W}{\eta}[/tex]
[tex]\dot Q = \frac{600\,W}{0.3}[/tex]
[tex]\dot Q = 2000\,W[/tex]
The heat input from the combustion phase is 2000 watts.
Use pseudocode. 1) Prompt for and input a saleswoman's sales for the month (in dollars) and her commission rate (percentage). Output her commission for that month. Note that you will need the following Variables: SalesAmount CommissionRate CommissionEarned
You will need the following formula: CommissionEarned= Sales Amount * (commissionrate/100)
Answer:
The pseudocode is as follows:
Input SalesAmount
Input CommissionRate
CommissionEarned= SalesAmount * (CommissionRate/100)
Print CommissionEarned
Explanation:
This gets input for SalesAmount
Input SalesAmount
This gets input for CommissionRate
Input CommissionRate
This calculates the CommissionEarned
CommissionEarned= SalesAmount * (CommissionRate/100)
This prints the calculated CommissionEarned
Print CommissionEarned
Given that the system function of a third order Butterworth type analog low-pass filter with a 3 dB cut-off frequency of 2 radian/second is:
2s HS = S2 + 0.2 s +1
Answer the following questions:
1. Use the bilinear transformation to obtain H(z). Use T=2 second.
2. Give H(w) for your filter.
3. Use MATLAB to give the magnitude spectrum.
4. Comment on the quality of the design.
5. With the aid of simple sketch graphs explain how frequency warping affects the frequency response of the digital filter.
6. Comment on the need for prewarping, i.e. give conditions when prewarping is needed.
answer
d just too the test
For a steel alloy it has been determined that a carburizing heat treatment of 3-h duration will raise the carbon concentration to 0.38 wt% at a point 2.6 mm from the surface. Estimate the time (in h) necessary to achieve the same concentration at a 6.1 mm position for an identical steel and at the same carburizing temperature.
Answer:
The right answer is "16.5 hrs".
Explanation:
Given values are:
[tex]x_1=2.6 \ mm[/tex]
[tex]t_1=3 \ hrs[/tex]
[tex]x_2=6.1 \ mm[/tex]
As we know,
⇒ [tex]\frac{x^2}{Dt}=constant[/tex]
or,
⇒ [tex]\frac{x_1^2}{t_1} =\frac{x_2^2}{t_2}[/tex]
⇒ [tex]t_2=(\frac{x_2}{x_1})^2\times t_1[/tex]
By putting the values, we get
[tex]=(\frac{6.1}{2.6} )^2\times 3[/tex]
[tex]=5.5\times 3[/tex]
[tex]=16.5 \ hrs[/tex]
A circular rod with a gage length of 3.1 m and a diameter of 3 cm is subjected to an axial load of 68 kN . If the modulus of elasticity is 200 GPa , what is the change in length
Answer:
1.49 mm
Explanation:
The modulus of elasticity, Y = stress/strain = σ/ε
σ = F/A where F = load = 68 kN = 68 × 10³ N and A = cross-sectional area of rod = πd²/4 where d = diameter of rod = 3 cm = 3 × 10⁻² m.
ε = ΔL/L where ΔL = change in length of the circular rod and L = length of circular rod = 3.1 ,
So, Y = σ/ε
Y = F/A ÷ ΔL/L
Y = FL/AΔL
making the change in length ΔL subject of the formula, we have
ΔL = FL/AY
substituting the value of A into the equation, we have
So, ΔL = FL/(πd²/4)Y
ΔL = 4FL/πd²Y
Since Y = 200 GPa = 200 × 10⁹ Pa
Substituting the values of the variables into the equation, we have
ΔL = 4FL/πd²Y
ΔL = 4 × 68 × 10³ N × ×3.1 m/[π(3 × 10⁻²m)² × 200 × 10⁹ Pa]
ΔL = 843.2 × 10³ Nm/[9π × 10⁻⁴m² × 200 × 10⁹ Pa]
ΔL = 843.2 × 10³ Nm/[1800π × 10⁵ N]
ΔL = 843.2 × 10³ Nm/5654.87 × 10⁵ N
ΔL = 0.149 × 10⁻² m
ΔL = 1.49 × 10⁻³ m
ΔL = 1.49 mm
The change in length of the circular rod is 1.49 mm
If there is a discrepancy between Chick-fil-A food safety requirements and local Health Department
regulations, what should Team Member do?
The following should be done by the team member:
It is important to follow both Chick-fil-A food safety requirements and local Health Department regulations. In the case when there is a discrepancy between the two, always follow the more stringent requirement. Any other appearance or grooming issue not covered in these materials may be addressed at the discretion of the Operator.
Learn more: brainly.com/question/17429689
6. When the engine stalls or the power unit fails, on a car with power
brakes, the service brake pedal will
A. Take about the same amount of pressure
B. Take more pressure to stop
C. Take less pressure to stop
D. Become locked in place and no longer help stop the car
Consider the equation y = 10^(4x). Which of the following statements is true?
A plot of log(y) vs. x would be linear with a slope of 4.
A plot of log(y) vs. log (x) would be linear with a slope of 10.
A plot of log(y) vs. x would be linear with a slope of 10.
A plot of y vs. log(x) would be linear with a slope of 4.
A plot of log(y) vs. log (x) would be linear with a slope of 4.
A plot of y vs. log(x) would be linear with a slope of 10.
Answer: Plot of [tex]\log y[/tex] vs [tex]x[/tex] would be linear with a slope of 4.
Explanation:
Given
Equation is [tex]y=10^{4x}[/tex]
Taking log both sides
[tex]\Rightarrow \log y=4x\log (10)\\\Rightarrow \log y=4x[/tex]
It resembles with linear equation [tex]y=mx+c[/tex]
Here, slope of [tex]\log y[/tex] vs [tex]x[/tex] is 4.
it is a small sharp and printed item for fine worker in trimming scallops clipping threads and cutting large eyelets
Answer:
embroidery scissor
Explanation:
is small, sharp and pointed good for fine work use trimming scallops,clipping threads,and cutting large eyelets.
hope this helps
Activity 1. Fill the blank with the correct answer. Write your answer on the blank. 1. ___________________ is a regular pattern of dots displayed on the screen which acts as a visual aid and also used to define the extent of your drawing. 2. Ortho is short for ___________________, which means either vertical or horizontal. 3. Tangent is a point where two _______________________ meet at just a single point. 4. If you want to create a new drawing, simply press ___________________ for the short cut key. 5. There are _______________Osnap that can help you performs your task easier.
Answer:
1. Drawing grid.
2. Orthogonal.
3. Geometries.
4. CTRL+N.
5. Thirteen (13).
Explanation:
CAD is an acronym for computer aided design and it is typically used for designing the graphical representation of a building plan. An example of a computer aided design (CAD) software is AutoCAD.
Some of the features of an AutoCAD software are;
1. Drawing grid: is a regular pattern of dots displayed on the screen of an AutoCAD software, which acts as a visual aid and it's also used to define the extent of a drawing.
2. Ortho is short or an abbreviation for orthogonal, which means either vertical or horizontal.
3. Tangent is a point where two geometries meet at just a single point.
4. If you want to create a new drawing, simply press CTRL+N for the short cut key.
5. There are thirteen object snaps (Osnap) that can help you perform your task on AutoCAD easily. The 13 object snaps (Osnap) are; Endpoint, Midpoint, Apparent intersect, Intersection, Quadrant, Extension, Tangent, Center, Insert, Perpendicular, Node, Parallel, and Nearest.
g The inside surface of a 17 mm inner diameter tube with a 2.4 mm thick wall indicates a temperature of 46 deg C. The outside temperature is 43 deg C. The tube is 5 m long. If the tube material has a conductivity of 0.15 W/m/K, estimate the heat transfer rate through the tube wall assuming SS 1D conduction. Indicate the direction of heat transfer with a or - sign ( meaning outward and vice versa). Express your answer using two significant digits in W.
Answer:
-50 W
Explanation:
The heat transfer rate Q = kA(T₂ - T₁)/d where k = thermal conductivity of material = 0.15 W/m-K, A = surface area of tube = πdL where d = diameter of tube = 17 mm = 0.017 m and L = length of tube = 5 m, T₁ = inside temperature = 46 °C, T₂ = outside temperature = 43 °C and d = thickness of tube = 2.4 mm = 0.0024 m
Since Q = kA(T₂ - T₁)/d ,
Q = kπdL(T₂ - T₁)/d
substituting the values of the variables into the equation, we have
Q = 0.15 W/m-K × π × 0.017 m × 5 m(43 °C - 46 °C )/0.0024 m
Q = 0.01275π Wm/K(-3 K )/0.0024 m
Q = -0.03825π Wm/0.0024 m
Q = -0.1202 Wm/0.0024 m
Q = -50.07 W
Q = -50 W
So, the heat transfer rate is -50 W meaning heat transfer out of the tube.
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Submit
Add a pair of radio buttons to your form, each nested in its own label element.
One should have the option of car and the other should have the option of bike.
Both should share the name attribute of “vehicle” to create a radio group
Make sure the radio buttons are nested with the form
Make sure that the name attributes appear after the type
Answer:
The code is as follows:
<form name = "myForm">
<div>
<input type="radio" name="vehicle" value="D0" id="D0"/>
<label for="D0">Car</label>
</div>
<div>
<input type="radio" name="vehicle" value="D1" id="D1"/>
<label for="D1">Bike</label>
</div>
</form>
Explanation:
This defines the first button
<input type="radio" name="vehicle" value="D0" id="D0"/>
<label for="D0">Car</label>
This defines the second button
<input type="radio" name="vehicle" value="D1" id="D1"/>
<label for="D1">Bike</label>
The code is self-explanatory, as it follows all the required details in the question
Alice and Bob both have RSA Public-Private key pairs: (PUA, PRA) and (PUB, PRB). They also have cryptographic functions E_AES / D_AES to encrypt / decrypt using AES; and E_RSA and D_RSA to encrypt / decrypt using RSA. Alice wants to sent a high resolution video of a large secret facility to Bob.
A. Show how Alice can securely and efficiently send the video to Bob. You are required to use the cryptographic functions above to get full credit;
B. Does your solutions assure confidentiality? How / Why not?
C. Does your solutions assure non-repudiation? How / Why not?
D. Does your solutions assure integrity? How / Why not?
E. Does your solutions assure replay attacks? How / Why not?
Solution :
B. yes, the given solution assures confidentiality. The sender Alice encrypting his messages with its own private key PRA which provides authentication. Sender Alice further encrypts his messages with the receiver's public key PUB provides confidentiality.
C. So the given solution provides non repudiation. Alice and Bob who are exchanging messages. In one case, Alice denies sending a messages to Bob that he claims to have received being able to counter Alice's denial is caused non repudiation of origin.
D. The given solution provides integrity. Because it provides authentication and have not been changed.
E. It does not provide replay attacks because it does not captures the traffic. The client does not receive the messages twice.
