Conventional thermal irons and pressing combs are heated by:

The best practice in formatting spreadsheets include the following: D. Organize data in vertical columns with first row header.

In Computer technology, a spreadsheet can be defined as a type of computer document which comprises cells that are arranged in a tabulated format with rows and columns.

Additionally, a spreadsheet is typically used in various field to do the following on a data:

Generally speaking, a best practice that should be adopted by end users when formatting spreadsheets is making sure that the data are organized in vertical columns with first row header.

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Answer:

I think The Battle of Sabine Pass

Considering that the metal has a low specific heat, it could be a great material for cookware.

What are metals ?

Metals are substances that develop naturally beneath the Earth's surface. Most metals are shiny or lustrous. Because they are inorganic, metals are composed of materials that have never been alive.

In order to raise a body's temperature per unit mass, a certain amount of energy is needed, which is known as specific heat. It takes less energy to heat the cookware up when the specific heat is low. Cookware is made with the handles having a higher specific heat than the actual cooking portion, which is designed to have a low specific heat. Since the handles would take longer to get hot, the high specific heat at the handles ensures safety.

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Blank is the disparity between adjacent parts of a design to give it a striking overall look. ​

What is disparity?

A disparity is a marked and usually significant difference or dissimilarity.

disparity is a form of inequality because disparity contains the Latin dis, which means apart. The term is often used to describe any social or economic situation that is considered unfair. These include racial inequalities in hiring, health disparities between rich and poor, and income disparities between men and women. The adjectives differ (emphasized on the first syllable) are often used to emphasize sharp differences.

Blank is the disparity between adjacent parts of a design to give it a striking overall look. ​

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Answer:

for 13

Explanation:

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To find the answer to the question, we can use the formula of confidence interval which is given below:$$\bar{x} \pm z_{(α/2)}\frac{σ}{\sqrt{n}}$$Where,$\bar{x}$ is the sample mean. $z_{(α/2)}$ is the $Z$-score for the given level of confidence interval.$σ$ is the population standard deviation. $n$ is the sample size.

If the 98% confidence limits of the population mean are 73 and 80, then we can say that:$$\bar{x} \pm z_{(α/2)}\frac{σ}{\sqrt{n}} = (73, 80)$$Now, we need to find the 95% confidence limits. For this, we have to calculate the Z-score for the 95% confidence interval. The Z-score for 95% confidence interval is calculated as follows:$$Z_{(α/2)} = Z_{(0.025)} = 1.96$$Now, we can use this Z-score in the confidence interval formula as follows:$$\bar{x} \pm z_{(α/2)}\frac{σ}{\sqrt{n}} = (73, 80)$$$$\bar{x} \pm 1.96\frac{σ}{\sqrt{n}} = (73, 80)$$We do not know the value of $\bar{x}$ or $σ$ or $n$, so we can't solve the above equation directly.

However, we can use some logic to find out which of the options is correct. If we consider the option (b), i.e. the 95% confidence limits are 72 and 79, then we can say that the distance between the upper limit of the 98% confidence interval and the upper limit of the 95% confidence interval is 1 unit less than the distance between the lower limit of the 98% confidence interval and the lower limit of the 95% confidence interval. In other words, the difference between the upper limit and the lower limit of the 98% confidence interval is 7, and the difference between the upper limit and the lower limit of the 95% confidence interval is 7-1 = 6. This is possible only if the sample size is very small or the standard deviation of the population is very high. Both of these conditions are unlikely to be true in most cases. Therefore, we can eliminate option (b).

Similarly, we can eliminate option (d) as the difference between the upper and the lower limit of the 98% confidence interval is 5 and the difference between the upper and the lower limit of the 95% confidence interval is 5-1 = 4.This is also unlikely to happen in most cases.

Therefore, we can eliminate option (d) as well.

Now, if we consider option (a), i.e. the 95% confidence limits are 73 and 81, then we can say that the difference between the upper limit of the 98% confidence interval and the upper limit of the 95% confidence interval is 1 unit, and the difference between the lower limit of the 98% confidence interval and the lower limit of the 95% confidence interval is 1 unit as well. This is possible and likely to happen in most cases.

Therefore, we can select option (a) as the correct answer .Finally, the answer to the question is: Option (a), i.e. 73 and 81 could be the 95% confidence limits.

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Answer: Part tolerancing  

Answer:

4.30 gp

Explanation:

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Answer:

lol Jebus?

Explanation:

Answer:

addsa

sdadsa

Explanation:

dsads

Answer:

im not for sure

Explanation:

Complete question:

The acceleration of a particle traveling along a straight line is a = 1/4 s^1/2 m/s^2, where s is in meters. If v = 0, s = 1 m when t = 0, determine the particle’s velocity at s = 2 m.

Answer:

The particle’s velocity is 0.781 m/s.

Explanation:

Given;

acceleration of the particle, \(a = \frac{1}{4} s^{\frac{1}{2}} \ m/s^2\) \(= 0.25s^{0.5} \ m/s^2\)

Acceleration is given by;

\(a = \frac{dv}{dt}\\\\a = \frac{dv}{dt} *\frac{ds}{ds} = \frac{ds}{dt}* \frac{dv}{ds}\\\\a = v*\frac{dv}{ds} \\\\ads = vdv\\\\\int\limits^s_1 {a} \, ds = \int\limits^v_0 {v} \, dv\\\\ \int\limits^s_1 {0.25s^{0.5}} \, ds = \int\limits^v_0 {v} \, dv\\\\\frac{1}{6} (s^{1.5} -1^{1.5}) = \frac{v^2}{2} \\\\v^2 = \frac{2}{6} (s^{1.5} -1^{1.5})\\\\v^2 = \frac{1}{3} (s^{1.5} -1^{1.5})\\\\when \ s= 2 m\\\\v^2 = \frac{1}{3} (2^{1.5} -1^{1.5})\\\\v^2 = 0.6095\\\\v = \sqrt{0.6095}\\\\v = 0.781 \ m/s\)

Therefore, the particle’s velocity at s = 2 m, is 0.781 m/s.

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature \(T_1 = -16^0\ C\)

Quality \(x_1 = 0.2\)

Outlet:

Temperature \(T_2 = -16^0 C\)

Quality  \(x_2 = 1\)

The following data were obtained at saturation properties of R134a at the temperature of -16° C

\(v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\ v_g = 0.1247 \ m^3 /kg\)

\(v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\ v_1 = 0.0255 \ m^3/kg \\ \\ \\ v_2 = v_g = 0.1247 \ m^3/kg\)

\(m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6 \\ \\ \mathbf{m = 0.0534 \ kg/sec}\)

\(\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2 \\ \\ \rho_1v_1 = \rho_2v_2 \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}\)

Answer:

a. \(L_o\)  = 40 psf

b. L ≈ 30.80 psf

c. The uniformly distributed total load for the beam = 812.8 ft./lb

d. The alternate concentrated load is more critical to bending , shear and deflection

Explanation:

The given parameters of the beam the beam are;

The span of the beam = 26 ft.

The width of the tributary, b = 16 ft.

The dead load, D = 20 psf.

a. The basic floor live load is given as follows;

The uniform floor live load, = 40 psf

The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²

Therefore, the uniform live load, \(L_o\)  = 40 psf

b. The reduced floor live load, L in psf. is given as follows;

\(L = L_o \times \left ( 0.25 + \dfrac{15}{\sqrt{k_{LL} \cdot A_T} } \right)\)

For the school, \(K_{LL}\) = 2

Therefore, we have;

\(L = 40 \times \left ( 0.25 + \dfrac{15}{\sqrt{2 \times 416} } \right) = 30.80126 \ psf\)

The reduced floor live load, L ≈ 30.80 psf

c. The uniformly distributed total load for the beam, \(W_d\) = b × \(W_{D + L}\) =

∴  \(W_d\) =  = 16 × (20 + 30.80) ≈ 812.8 ft./lb

The uniformly distributed total load for the beam, \(W_d\) = 812.8 ft./lb

d. For the uniformly distributed load, we have;

\(V_{max}\) = 812.8 × 26/2 = 10566.4 lbs

\(M_{max}\) =  812.8 × 26²/8 = 68,681.6 ft-lbs

\(v_{max}\) = 5×812.8×26⁴/348/EI = 4,836,329.333/EI

For the alternate concentrated load, we have;

\(P_L\) = 1000 lb

\(W_{D}\) = 20 × 16 = 320 lb/ft.

\(V_{max}\) = 1,000 + 320 × 26/2 = 5,160 lbs

\(M_{max}\) =  1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs

\(v_{max}\) = 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI

Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load

Answer:

The main sections included in the bill of quantities are Form of Tender, Information, Requirements, Pricing schedule, Provisional sums, and Day works.

The correct option is(B). The height of the transverse metacentre above the centre of buoyancy is called metacentric height.

The height of the transverse metacentre above the centre of buoyancy is called metacentric height (GM).

The distance between the centre of gravity of a floating body and its metacentre is referred to as the metacentric radius.

GM determines the stability of a floating body.

Metacentric height (GM) is the term used to describe the distance between the center of gravity of a floating body and its metacentre.

It is the height of the transverse metacentre above the centre of buoyancy. When a ship is floating in calm water, it will come to a stable position after it is heeled because of the change in its centre of gravity and buoyancy.

The stability of the floating body is determined by the metacentric height (GM).When the GM value is high, the ship will have more stability, and it will be challenging to overturn it.

When the GM value is low, the ship will be less stable, and it will be easier to overturn. If the value of GM is zero, the ship will become unstable and will capsize.

Therefore, the value of GM is critical in the design of a ship and is calculated by naval architects to ensure the safety and stability of the vessel.

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Answer:

a). 139498.24 kg

b). 281.85 ohm

c). 10.2 ohm

Explanation:

Given :

Diameter, d = 22 m

Linear strain, \($\epsilon$\) = 3%

                        = 0.03

Young's modulus, E = 30 GPa

Gauge factor, k = 6.9

Gauge resistance, R = 340 Ω

a). Maximum truck weight

σ = Eε

σ = \($0.03 \times 30 \times 10^9$\)

\($\frac{P}{A} =0.03 \times 30 \times 10^9$\)

\($P = 0.03 \times 30 \times 10^9\times \frac{\pi}{4}\times (0.022)^2$\)

 = 342119.44 N

For the four sensors,

Maximum weight = 4 x P

                            =  4 x 342119.44

                            = 1368477.76 N

Therefore, weight in kg is \($m=\frac{W}{g}=\frac{1368477.76}{9.81}$\)

                     m = 139498.24 kg

b). Change in resistance

\(k=\frac{\Delta R/R}{\Delta L/L}\)

\($\Delta R = k. \epsilon R$\)    , since \($\epsilon= \Delta L/ L$\)

\($\Delta R = 6.9 \times 0.03 \times 340$\)

\($\Delta R = 70.38 $\) Ω

For 4 resistance of the sensors,

\($\Delta R = 70.38 \times 4 = 281.52$\) Ω

c). \($k=\frac{\Delta R/R}{\epsilon}$\)

If linear strain,

\($\frac{\Delta R}{R} \approx \frac{\Delta L}{L}$\)  , where k = 1

\($\Delta R = \frac{\Delta L}{L} \times R$\)

\($\Delta R = 0.03 \times 340$\)

\($\Delta R = 10.2 $\) Ω

Answer:

It is true that the sampling distribution of the proportions is approximately a normal distribution.

Explanation:

Solution

Given that:

N = 50

P= 0.8

Thus

p ^ =√p(1-p)/n

p ^=√0.8 (1-0.8)/50

p ^= 0.0566

Therefore,we conclude that the sampling distribution of the proportions is approximately a normal distribution.

Answer:

What specific fluid goes in the windshield wipers.

Distilled water

How much to put in fluid ounces?

There should be a tiny bit more than 3/4 of the way full.

Answer:

D would be correct because it maximizes it.

All of the above are acceptable means of making a steel column fire-resistant.

Spraying low-density fibers or cementitious compounds, also known as spray-applied fire-resistive materials, is the most common method of fireproofing.

Applying intumescent mastic/paint also known as intumescent paints protects steel members from fire. By creating a buffer between a steel element and the fire, intumescent coatings can expand by as much as 100 times their original thickness.

Encase the column in concrete is majorly utilized in large sections whereby steel is encased in. Based on the amount of concrete used, this requires more space than other options. Additionally, because it is not as visually appealing as others, it is utilized in settings where appearance is not the primary consideration.

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Answer:

C) Prospective Cohort study

Explanation:

prospective cohort study can be regarded as longitudinal cohort study that comes up with periods of time when group of individuals that are different in terms of some factors that are undergoing some study, so that how theses factors influence rates of some particular outcomes can be known.

Resolution-refutation is a proof strategy that helps in establishing that a given sentence is unsatisfiable. In other words, it shows that a given sentence cannot be true under any interpretation.

To prove the above statement using resolution-refutation, we need to follow the below

steps:

Step 1: Convert the given statement into Conjunctive Normal Form(CNF)

Step 2: Apply the resolution rule to the CNF formula until it can't be applied any further.

Step 1: Convert the given statement into CNFTo apply the resolution rule, we need to first convert the given statement into CNF form.For that, we need to use some of the following equivalences:1. De Morgan's Laws: ¬(P ∧ Q) ≡ ¬P ∨ ¬Q and ¬(P ∨ Q) ≡ ¬P ∧ ¬Q2. Distribution: P ∧ (Q ∨ R) ≡ (P ∧ Q) ∨ (P ∧ R) and P ∨ (Q ∧ R) ≡ (P ∨ Q) ∧ (P ∨ R)Using the above rules, we can convert the given statement as follows: ¬f(a) ∧ ¬g(a) ∧ ¬h(a) ∨ g(a) ∧ h(a)The above formula is in CNF form.

Step 2: Apply the resolution rule to the CNF formula until it can't be applied any further.

Now, we apply the resolution rule as follows: Clause 1: {¬f(a), ¬g(a), ¬h(a)}Clause 2: {g(a), h(a)}Resolve: {¬f(a), ¬g(a), ¬h(a), h(a)}Resolve: {¬f(a), ¬g(a)}Resolve: {¬f(a), ¬h(a)}Resolve: {¬f(a)}Hence, the proof is complete.

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Answer:

Heat rate. 10,535 kJ/kWh. Turbine speed. 7,700 rpm. Compressor pressure ratio. 14.0:1. Exhaust gas flow. 80.4 kg/s. Exhaust gas temperature. 543 deg C.

Explanation:

Answer:

Exit temperature = 32 °C

Explanation:

We are given;

Initial Pressure;P1 = 100 KPa

Cp =1000 J/kg.K = 1 KJ/kg.k

R = 500 J/kg.K = 0.5 Kj/Kg.k

Initial temperature;T1 = 27°C = 273 + 27K = 300 K

volume flow rate;V' = 15 m³/s

W = 130 Kw

Q = 80 Kw

Using ideal gas equation,

PV' = m'RT

Where m' is mass flow rate.

Thus;making m' the subject, we have;

m' = PV'/RT

So at inlet,

m' = P1•V1'/(R•T1)

m' = (100 × 15)/(0.5 × 300)

m' = 10 kg/s

From steady flow energy equation, we know that;

m'•h1 + Q = m'h2 + W

Dividing through by m', we have;

h1 + Q/m' = h2 + W/m'

h = Cp•T

Thus,

Cp•T1 + Q/m' = Cp•T2 + W/m'

Plugging in the relevant values, we have;

(1*300) - (80/10) = (1*T2) - (130/10)

Q and M negative because heat is being lost.

300 - 8 + 13 = T2

T2 = 305 K = 305 - 273 °C = 32 °C

13000 + 300 - 8000 = T2

Answer:

you might be facing some difficulty in observing this point of division between your question and me

Answer:

forced convection

Explanation:

When a fan, pump or suction device is used to facilitate convection, the result is forced convection. Everyday examples of this can be seen with air conditioning, central heating, a car radiator using fluid, or a convection oven.

To find the support reactions at points A and D, we can use the principles of static equilibrium. At point A, there is a vertical reaction force (Ra) and a horizontal reaction force (Ha). At point D, there is only a vertical reaction force (Rd).

Step 1: Calculate the vertical reaction force at point A (Ra)
Taking moments about point D:
- (12 lb/ft * 14 ft) + (50 lb * 7 ft) + (300 lb-ft) - (Ra * 14 ft)

= 0
- Solving for Ra, we get: Ra

= 150 lb

Step 2: Calculate the horizontal reaction force at point A (Ha)
Since there are no horizontal forces acting on the beam,

Ha = 0.

Step 3: Calculate the vertical reaction force at point D (Rd)
By using the principle of vertical equilibrium:
- Rd = (12 lb/ft * 14 ft) + (50 lb) - Ra
- Rd = (12 lb/ft * 14 ft) + (50 lb) - 150 lb
- Rd = 38 lb

To calculate the axial force (N), shear force (V), and bending moment (M) at mid-span of AB, we can use the equations for a simply supported beam with a uniformly distributed load.

Step 4: Calculate the axial force (N)
N = Ra + Rd
N = 150 lb + 38 lb
N = 188 lb

Step 5: Calculate the shear force (V)
V = (12 lb/ft * 7 ft) - (50 lb) - (300 lb-ft / 14 ft) * 7 ft
V = 84 lb - 50 lb - 150 lb
V = -116 lb

Step 6: Calculate the bending moment (M)
M = (12 lb/ft * 7 ft * (7 ft / 2)) - (50 lb * 7 ft) - (300 lb-ft)
M = 588 lb-ft - 350 lb-ft - 300 lb-ft
M = -62 lb-ft

bending moment (M) is -62 lb-ft.

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Answer:

The armature voltage required to run the motor at 1500 RPM is 143.75 V

The armature voltage required to run the motor at 100 RPM is 9.58 V

Explanation:

Given;

angular speed of the DC motor, ω = 1200 RPM

source voltage, V = 115 V

The armature voltage required to run the motor at 1500 RPM;

\(V_a = 115V \ \times \ \frac{1500 \ RPM}{1200 \ RPM} \\\\V_a = 143.75 \ V\)

The armature voltage required to run the motor at 100 RPM;

\(V_a = 115V \ \times \ \frac{100 \ RPM}{1200 \ RPM} \\\\V_a = 9.58 \ V\)