An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2. The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2. (Hint: The police will not go against the law.) a) Find the total time required for the police car to overtake the automobile. (12 marks) b) Find the total distance travelled by the police car while overtaking the automobile. (2 marks) c) Find the speed of the police car at the time it overtakes the automobile

Answer:

Yes, the forging can be completed

Explanation:

Given h = 100 mm, ε = ㏑(450/100) = 1.504

[tex]Y_f = 230 \times 1.504^{0.15} = 244.52[/tex]

V = π·D²·L/4 = π × 50²×450/4 = 883,572.93 mm³

At h = 100 mm, A = V/h = 883,572.93 /100 = 8835.73 mm²

D = √(4·A/π) = 106.07 mm

[tex]K_f[/tex] = 1 + 0.4 × 0.1 × 106.07/100 = 1.042

F = 1.042 × 244.52 × 8835.73 = 2252199.386 N =2.25 MN

Hence the required force = 2.25 MN is less than the available force = 3 MN therefore, the forging can be completed.

Answer:

Blindspot Location

Explanation:

Just took the quiz

When you do a vehicle check, you do NOT need to keep an eye on Blind spot locations. The correct option is D.

A blind spot is the area of the road that can't be seen by looking forward through windscreen, or by rear-view and side-view mirrors.

While doing vehicle check, we need to check tire inflation, cleanliness of windows and mirrors along with the functioning indicator lights and headlights.

Blind spot locations does not need to be checked.

Thus, the correct option is D.

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Answer:

a. load factor = 0.893

shrinkage factor = 0.848

b. Number of Trucks loads = 113,585 Trucks loads

Explanation:

Here, we start by identifying the factors as given in the question.

γn = 2800 lb/cy

γloose = 2500 lb/cy

and γcompacted = 3300 lb/cy

a. Mathematically,

Load factor = γloose/γn = 2500/2800 = 0.893

Shrinkage factor = γn/γcompacted = 2800/3300 = 0.848

b. To find the number of trucks loads with a capacity of 5 lcy/truck, we use the mathematical formula as follows;

ρlcy = 5

Load factor × Shrinkage factor = ρloose/γn × γn/γcompacted = ρlcy/ρccy

0.893 × 0.848 = 5/ρccy

ρccy =5/(0.893 × 0.848) = 6.603

The number of truck loads = 750,000/6.603 = 113,584.7 which is approximately 113,585 trucks loads

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:

(i) The inductance of the inductor is = 43.43 mH (ii) the impedance of the circuit is = 16∠58.61° Ω (iii) the phase difference for current and the voltage applied is Q = 58.61°

Explanation:

Solution

Given that:

I= 5 A

V = 125V

Resistance R= Not known yet

Thus

To find the resistance we have the following formula which is shown below:

R = V/I

=125/15

R =8.333Ω

Now,

Voltage = 240

Frequency = 50Hz

Current (I) remain at = 15A

Z= not known (impedance)

so,

To find the impedance we have the formula which is shown below:

Z = V/I =240/15

Z= 16Ω⇒ Z = R + jXL

Z = 8.333 + jXL = 16

Thus

√8.333² + XL² = 16²

8.333² + XL² = 16²

XL² = 186.561

XL = 13.658Ω

Now

We find the inductance of the Inductor and the impedance of the circuit.

(i) In solving for the inductance of the inductor, a formula is applied here, which is shown below:

L =  XL/w

=13.658/ 2π * 50

=13.658/314.15 = 0.043 = 43.43 mH

Note: w= 2πf

(ii) For the impedance of the circuit we have the following:

z = 8.333 + j 13.658

z = 16∠58.61° Ω

(iii) The next step is to find the phase difference between the applied voltage and current.

Q =  this is the voltage across the inductor in a series of resonant circuit.

Q can also be called the applied voltage

Thus,

Q is described as an Impedance angle

Therefore, Q = 58.81°

Answer:

Pressure difference (ΔP) = 63,519.75 kpa

Explanation:

Given:

ρ = 1,000 kg/m³

Height of cylindrical container used (h) = 7m / 2 = 3.5m

Specific gravity (sg) = 0.85

Find:

Pressure difference (ΔP).

Computation:

⇒ Pressure difference (ΔP) = h g [ ρ(sg) + ρ]                ∵ [ g = 9.81]

⇒ Pressure difference (ΔP) = (3.5)(9.81) [ 1,000(0.85) + 1,000]

⇒ Pressure difference (ΔP) = 34.335 [8,50 + 1,000]

⇒ Pressure difference (ΔP) = 34.335 [1,850]

⇒ Pressure difference (ΔP) = 63,519.75 kpa

Answer:

The reason for their unusual properties of the greasy feel and low hardness is that the chemical bonds between the sheets is so weak that very low stresses can allow slip between the sheets.

Explanation:

Talc is a monoclinic mineral with a sheet structure similar to the micas and also has perfect cleavage that follows planes between the weakly bonded sheets.

Now, these sheets are held together only by van der Waals bonds and this allows them to slip past each other easily. Thus, this unique characteristic is responsible for talc's extreme softness, its greasy, soapy feel, and its value as a high-temperature lubricant.

While for graphite, it's carbon atoms are linked in a hexagonal network which forms sheets that are one atom thick. It's sheets are poorly connected and easily cleave or slide over one another when subjected to a small amount of force. Thus, gives graphite its very low hardness, its perfect cleavage, and its slippery feel.

So, we can conclude that the reason for their unusual properties is that the chemical bonds between the sheets is so weak that very low stresses can allow slip between the sheets; hence, the greasy feel and low hardness.

Answer:

[tex] T_A_C = 6.296 kN [/tex]

[tex] R = 10.06 kN [/tex]

Explanation:

Given:

[tex] T_A_B = 8.7 kN[/tex]

Required:

Find the tension TAC and magnitude R of this downward force.

First calculate [tex] \alpha, \beta, \gamma [/tex]

[tex] \alpha = tan^-^1 =\frac{40}{50} = 38. 36 [/tex]

[tex] \beta = tan^-^1 =\frac{50}{30} = 59.04 [/tex]

[tex] \gamma = 180 - 38.36 - 59.04 = 82.6 [/tex]

To Find tension in AC and magnitude R, use sine rule.

[tex] \frac{sin a}{T_A_C} = \frac{sin b}{T_A_B} = \frac{sin c}{R} [/tex]

Substitute values:

[tex]\frac{sin 38.36}{T_A_C} = \frac{sin 59.04}{8.7} = \frac{82.6}{R}[/tex]

Solve for T_A_C:

[tex] T_A_C = 8.7 * \frac{sin 38.36}{sin 59.04} = [/tex]

[tex] T_A_C = 8.7 * 0.724 = 6.296 kN [/tex]

Solve for R.

[tex] R = 8.7 * \frac{sin 82.6}{sin 59.04} = [/tex]

[tex] R = 8.7 * 1.156 [/tex]

R = 10.06 kN

Tension AC = 6.296kN

Magnitude,R = 10.06 kN

Answer:

None of these

Explanation:

There are different types of amplifiers, and each has different characteristics.

Therefore, the correct answer is "None of these "

Answer:

Develop social skills

Explanation:

Answer:

strengthen your college applications

Explanation:

Answer:

1) 2304 kPa

2) B. 200 N/m

Explanation:

The internal pressure of the of the tank  can be found from the following relations;

Resisting wall force F = p×(1/4·π·D²)

σ×A = p×(1/4·π·D²)

Where:

σ = Allowable stress of the tank

A = Area of the wall of the tank = π·D·t

t = Thickness of the tank = 15 mm. = 0.015 m

D = Diameter of the tank = 25 m

p = Maximum permissible internal pressure pressure

∴ σ×π·D·t = p×(1/4·π·D²)

p = 4×σ×t/D = 4 × 240 ×0.015/2.5 = 5.76 MPa

With a desired safety factor of 2.5, the permissible internal pressure = 5.76/2.5 = 2.304 MPa

2) The formula for average shear flow is given as follows;

[tex]q = \dfrac{T}{2 \times A_m}[/tex]

Where:

q = Average shear flow

T = Torque = 8 N·m

[tex]A_m[/tex] = Average area enclosed within tube

t = Thickness of tube = 6.35 mm = 0.00635 m

Side length of the square cross sectioned tube, s = 203 mm = 0.203 m

Average area enclosed within tube, [tex]A_m[/tex] = (s - t)² = (0.203 - 0.00635)² = 0.039 m²

[tex]\therefore q = \dfrac{8}{2 \times 0.039} = 206.9 \, N/m[/tex]

Hence the average shear flow is most nearly 200 N/m.

Following are the solution to the given question:

Calculating the allowable stress:

[tex]\to \sigma_{allow} = \frac{\sigma_y}{FS} \\\\[/tex]

              [tex]= \frac{240}{2.5} \\\\= 96\\\\[/tex]

Calculating the Thickness:

[tex]\to t =15\ mm = \frac{15\ }{1000}= 0.015\ m\\\\[/tex]

The stress in a spherical tank is defined as

[tex]\to \sigma = \frac{pD}{4t}\\\\\to 96 = \frac{p(25)}{4(0.015)}\\\\\to p = 0.2304\;\;MPa\\\\\to p = 230.4\;\;kPa\\\\\to p \approx 230\;\;kPa\\\\[/tex]

[tex]\bold{\to A= 203^2= 41209\ mm^2} \\\\[/tex]

Calculating the shear flow:

[tex]\to q=\frac{T}{2A}[/tex]

      [tex]=\frac{8}{2 \times 41209 \times 10^{-6}}\\\\=\frac{8}{0.082418}\\\\=97.066\\\\[/tex]

[tex]\to q=97 \approx 100 \ \frac{N}{m}\\[/tex]

Therefore, the final answer is "".

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Image of wheel is missing, so i attached it.

Answer:

ω = 14.95 rad/s

Explanation:

We are given;

Mass of wheel; m = 20kg

T = 20 N

k_o = 0.3 m

Since the wheel starts from rest, T1 = 0.

The mass moment of inertia of the wheel about point O is;

I_o = m(k_o)²

I_o = 20 * (0.3)²

I_o = 1.8 kg.m²

So, T2 = ½•I_o•ω²

T2 = ½ × 1.8 × ω²

T2 = 0.9ω²

Looking at the image of the wheel, it's clear that only T does the work.

Thus, distance is;

s_t = θr

Since 4 revolutions,

s_t = 4(2π) × 0.4

s_t = 3.2π

So, Energy expended = Force x Distance

Wt = T x s_t = 20 × 3.2π = 64π J

Using principle of work-energy, we have;

T1 + W = T2

Plugging in the relevant values, we have;

0 + 64π = 0.9ω²

0.9ω² = 64π

ω² = 64π/0.9

ω = √64π/0.9

ω = 14.95 rad/s

Answer:

(a) 12 seconds (b) t = 8.31 seconds (c) 10µ A (d) V = 20 V (e) V =0

Explanation:

Solution

Given that:

C = 6 µ which is = 6 * 10^ ⁻6

R = 2 MΩ, which is = 2 * 10^ 6

ε = 20 V

(a) When it is at the time constant we have the following:

λ = CR

= 6 * 10^ ⁻6 * 2 * 10^ 6

λ =12 seconds

(b) We solve for the half life of the circuit which is given below:

d₀ = d₀ [ 1- e ^ ⁺t/CR

d = decay mode]

d₀/2 =  d₀  1- e ^ ⁺t/12

2^⁻1 = e ^ ⁺t/12

Thus

t/12 ln 2

t = 12 * ln 2

t = 12 * 0.693

t = 8.31 seconds

(c) We find the current at t = 0

So,

I = d₀/dt

I = d₀/dt e ^ ⁺t/CR

= CE/CR e ^ ⁺t/CR

E/R e ^ ⁺t/CR

Thus,

at t = 0

I  E/R = 20/  2 * 10^ 6

= 10µ A

(d) We find the voltage across the capacitor at t = 0 which is shown below:

V = IR

= 10 * 10^ ⁻6 * 2 * 10^ 6

V = 20 V

(e)  We solve for he voltage across the resistor.

At t = 0

I = 0

V =0

Answer:

The average thickness of the blubber is 0.077 m

Explanation:

Here, we want to calculate the average thickness of the Walrus blubber.

We employ a mathematical formula to calculate this;

The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L

Where dQ is the change in amount of heat transferred

dT is the temperature gradient(change in temperature) i.e T2-T1

dQ/dT = 220 W

K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)

A is the surface area which is 2.23 m^2

T2 = 37.0 °C

T1 = -1.0 °C

L is ?

We can rewrite the equation in terms of L as follows;

L × dQ/dT = KA(T2-T1)

L = KA(T2-T1) ÷ dQ/dT

Imputing the values listed above;

L = (0.2 * 2.23)(37-(-1))/220

L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m

Answer:

Explanation:

The frequency components in the message signal are

f1 = 100Hz, f2 = 200Hz and f3 = 400Hz

When amplitude modulated with a carrier signal of frequency fc = 100kHz

Generates the following frequency components

Lower side band

[tex]100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\[/tex]

Carrier frequency 100kHz

Upper side band

[tex]100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz[/tex]

After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be

[tex]100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz[/tex]

At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be

[tex]100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz[/tex]

After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be

[tex]100k, 99.9k, 99.8k\ \ and \ \99.6kHz[/tex]

At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be

[tex]100.02 - 99.9k = 0.12k = 120Hz\\\\100.02 - 99.8k = 0.22k = 220Hz\\\\100.02 - 99.6k = 0.42k = 420Hz[/tex]

A) The frequency Components of the Detector Output are;

80 Hz, 120 Hz and 380 Hz

B) The frequency Components if only the lower sideband is transmitted are; 120 Hz, 220 Hz and 420 Hz

A) We are given the frequency components in the message signal as;

We are told that the carrier signal has a frequency; fc = 100kHz

Thus, the frequency components generated are;

Lower side band:

Upper side band:

We are told that the local oscillator now supplies a sinusoidal wave of frequency 100.02 kHz.

Thus, the output frequencies are;

100.02 kHz - 100.1 kHz = 80 Hz

100.02 kHz - 100.2 kHz = 180 Hz

100.02 kHz - 100.4 kHz = 380 Hz

B) Repeating the analysis assuming only the lower sideband is repeated gives us the frequencies as;

100.02 kHz - 99.9 kHz = 120 Hz

100.02 kHz - 99.8 kHz = 220 Hz

100.02 kHz - 99.6 kHz = 420 Hz

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Answer:

The dimension of the  square cross section is = 30mm * 30mm

Explanation:

Before proceeding with the calculations convert MPa to Kpsi

Sut ( ultimate strength ) = 770 MPa * 0.145 Kpsi/MPa = 111.65 Kpsi

the fatigue strength factor of Sut at 10^3 cycles for Se = Se' = 0.5 Sut

at 10^6 cycles" for 111.65 Kpsi  = f ( fatigue strength factor) = 0.83

To calculate the endurance limit  use  Se' = 0.5 Sut      since Sut < 1400 MPa

Se'( endurance limit ) = 0.5 * 770 = 385 Mpa

The surface condition modification factor

Ka = 57.7 ( Sut )^-0.718

Ka = 57.7 ( 770 ) ^-0.718

Ka = 0.488

Assuming the size modification factor (Kb) = 0.85 and also assuming all modifiers are equal to one

The endurance limit at the critical location of a machine part can be expressed as :

Se = Ka*Kb*Se'

Se = 0.488 * 0.85 * 385 = 160 MPa

Next:

Calculating the constants to find the number of cycles

α = [tex]\frac{(fSut)^2}{Se}[/tex]

α =[tex]\frac{(0.83*770)^2}{160}[/tex]  =  2553 MPa

b = [tex]-\frac{1}{3} log(\frac{fSut}{Se} )[/tex]

b = [tex]-\frac{1}{3} log (\frac{0.83*770}{160} )[/tex]  = -0.2005

Next :

calculating the fatigue strength using the relation

Sf = αN^b

N = number of cycles

Sf = 2553 ( 10^4) ^ -0.2005

Sf = 403 MPa

Calculate the maximum moment of the beam

M = 2000 * 0.6 = 1200 N-m

calculating the maximum stress in the beam

∝ = ∝max = [tex]\frac{Mc}{I}[/tex]

Where c = b/2 and   I = b(b^3) / 12

hence ∝max = [tex]\frac{6M}{b^3}[/tex]  =  6(1200) / b^3   =  7200/ b^3   Pa

note: b is in (meters)

The expression for the factor of safety is written as

n = [tex]\frac{Sf}{\alpha max }[/tex]

Sf = 403, n = 1.5 and ∝max = 7200 / b^3

= 1.5 = [tex]\frac{403(10^6 Pa/Mpa)}{7200 / B^3}[/tex]   making b subject of the formula in other to get the value of b

b = 0.0299 m * 10^3 mm/m

b = 29.9 mm therefore b ≈ 30 mm

since  the size factor  assumed is near the calculated size factor using this relation : de = 0.808 ( hb)^1/2

the dimension = 30 mm by 30 mm

Answer:

Check the v(t) signal referred to in the question and the solution to each part in the files attached

Explanation:

The detailed solutions of parts a to d are clearly expressed in the second file attached.

Answer:

a. [tex]\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}[/tex] which is constant therefore, n = constant

b. The temperature at the end of the process is 109.6°C

c. The work done by the balloon boundaries = 10.81 MJ

The work done on the surrounding atmospheric air = 10.6 MJ

Explanation:

p₁ = 100 kPa

T₁ = 27°C

D₁ = 10 m

v₂ = 1.2 × v₁

p ∝ α·D

α = Constant

[tex]v_1 = \dfrac{4}{3} \times \pi \times r^3[/tex]

[tex]\therefore v_1 = \dfrac{4}{3} \times \pi \times \left (\dfrac{10}{2} \right )^3 = 523.6 \ m^3[/tex]

v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³

Therefore, D₂ = 10.63 m

We check the following relation for a polytropic process;

[tex]\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}} \right )^{n} = \left (\dfrac{T_{1}}{T_{2}} \right )^{\dfrac{n}{n-1}}[/tex]

We have;

[tex]\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times \pi \times \left (\dfrac{D_2}{2} \right )^3}{\dfrac{4}{3} \times \pi \times \left (\dfrac{D_1}{2} \right )^3} \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}[/tex]

[tex]\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_2} }{ {D_1}} \right )^{3\times n} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}[/tex]

[tex]\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2 \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}[/tex]

[tex]log \left (\dfrac{D_{1}}{ D_{2}}\right ) = -3\times n \times log\left (\dfrac{ \left{D_1} }{ {D_2}} \right )[/tex]

n = -1/3

Therefore, the relation, pVⁿ = Constant

b. The temperature T₂ is found as follows;

[tex]\left (\dfrac{628.32 }{523.6} \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{1}{4}}[/tex]

T₂ = 300.15/0.784 = 382.75 K = 109.6°C

c. [tex]W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}[/tex]

[tex]p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}} \right )^{n} } = \dfrac{100\times 10^3}{ \left (1.2) \right ^{-\dfrac{1}{3} } }[/tex]

p₂ =  100000/0.941 = 106.265 kPa

[tex]W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3 \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J[/tex]

The work done by the balloon boundaries = 10.81 MJ

Work done against atmospheric pressure, Pₐ, is given by the relation;

Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J

The work done on the surrounding atmospheric air = 10.6 MJ

Answer:

a = 0.07m or 70mm

b = 0.205m or 205mm

Explanation:

Given the following data;

Modulus of rigidity, G = 14MPa=14000000Pa.

c = 80mm = 0.08m.

P = 46kN=46000N.

Shearing stress (r) in the rubber shouldn't exceed 1.4MPa=1400000Pa.

Deflection (d) of the plate is to be at least 7mm = 0.007m.

From shearing strain;

[

[tex]Modulus Of Elasticity, E = \frac{d}{a} =\frac{r}{G}[/tex]

Making a the subject formula;

[tex]a = \frac{Gd}{r}[/tex]

Substituting into the above formula;

[tex]a = \frac{14000000*0.007}{1400000}[/tex]

[tex]a = \frac{98000}{1400000}[/tex]

[tex]a = 0.07m or 70mm[/tex]

a = 0.07m or 70mm.

Also, shearing stress;

[tex]r = \frac{P}{2bc}[/tex]

Making b the subject formula;

[tex]b = \frac{P}{2cr}[/tex]

Substituting into the above equation;

[tex]b = \frac{46000}{2*0.08*1400000}[/tex]

[tex]b = \frac{46000}{224000}[/tex]

[tex]b = 0.205m or 205mm[/tex]

b = 0.205m or 205mm

Answer:

Explanation:

Given that,

The area of glass [tex]A_g[/tex] = [tex]0.11m^2[/tex]

The thickness of the glass [tex]t_g=4mm=4\times10^-^3m[/tex]

The area of the styrofoam [tex]A_s=11m^2[/tex]

The thickness of the styrofoam [tex]t_s=0.20m[/tex]

The thermal conductivity of the glass [tex]k_g=0.80J(s.m.C^o)[/tex]

The thermal conductivity of the styrofoam  [tex]k_s=0.010J(s.m.C^o)[/tex]

Inside and outside temperature difference is ΔT

The heat loss due to conduction in the window is

[tex]Q_g=\frac{k_gA_g\Delta T t}{t_g} \\\\=\frac{(0.8)(0.11)(\Delta T)t}{4.0\times 10^-^3}\\\\=(22\Delta Tt)j[/tex]

The heat loss due to conduction in the wall is

[tex]Q_s=\frac{k_sA_s\Delta T t}{t_g} \\\\=\frac{(0.010)(11)(\Delta T)t}{0.20}\\\\=(0.55\Delta Tt)j[/tex]

The net heat loss of the wall and the window is

[tex]Q=Q_g+Q_s\\\\=\frac{k_gA_g\Delta T t}{t_g}+\frac{k_sA_s\Delta T t}{t_g}\\\\=(22\Delta Tt)j +(0.55\Delta Tt)j \\\\=(22.55\Delta Tt)j[/tex]

The percentage of heat lost by the window is

[tex]=\frac{Q_g}{Q}\times 100\\\\=\frac{22\Delta T t}{22.55\Delta T t}\times 100\\\\=97.6 \%[/tex]

Answer:

Both technician A and technician B are correct

Explanation:

A planetary gearbox consists of a gearbox with the input shaft and the output shaft that is aligned to each other. It is used to transfer the largest torque in the compact form. A planetary gearbox has a compact size and low weight and it has high power density.

One planetary gear set can provide gear reduction, overdrive, and reverse. Also, most transmissions today use compound (multiple) planetary gears set.

So, both technician A and technician B are correct.

Answer:

Answer:

• it charges banks more interest

• it sells more securities

• it decreases the money supply

Explanation:

hope this help edge 21

Answer:

See Explanation

Explanation:

Required

- Pseudocode to determine the largest of 10 numbers

- C# program to determine the largest of 10 numbers

The pseudocode and program makes use of a 1 dimensional array to accept input for the 10 numbers;

The largest of the 10 numbers is then saved in variable Largest and printed afterwards.

Pseudocode (Number lines are used for indentation to illustrate the program flow)

1. Start:

2. Declare Number as 1 dimensional array of 10 integers

3. Initialize: counter = 0

4. Do:

4.1 Display “Enter Number ”+(counter + 1)

4.2 Accept input for Number[counter]

4.3 While counter < 10

5. Initialize: Largest = Number[0]

6. Loop: i = 0 to 10

6.1 if Largest < Number[i] Then

6.2 Largest = Number[i]

6.3 End Loop:

7. Display “The largest input is “+Largest

8. Stop

C# Program (Console)

Comments are used for explanatory purpose

using System;

namespace ConsoleApplication1

{

   class Program

   {

       static void Main(string[] args)

       {

           int[] Number = new int[10];  // Declare array of 10 elements

           //Accept Input

           int counter = 0;

           while(counter<10)

           {

               Console.WriteLine("Enter Number " + (counter + 1)+": ");

               string var = Console.ReadLine();

               Number[counter] = Convert.ToInt32(var);

               counter++;                  

           }

           //Initialize largest to first element of the array

           int Largest = Number[0];

           //Determine Largest

           for(int i=0;i<10;i++)

           {

               if(Largest < Number[i])

               {

                   Largest = Number[i];

               }

           }

           //Print Largest

           Console.WriteLine("The largest input is "+ Largest);

           Console.ReadLine();

       }

   }

}

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

[tex]n = \frac{S_{uts}}{\tau_{max}}[/tex]

Where:

[tex]n[/tex] - Safety factor, dimensionless.

[tex]S_{uts}[/tex] - Ultimate shear strength, measured in pascals.

[tex]\tau_{max}[/tex] - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: ([tex]n = 4.2[/tex], [tex]S_{uts} = 320\times 10^{6}\,Pa[/tex])

[tex]\tau_{max} = \frac{S_{uts}}{n}[/tex]

[tex]\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}[/tex]

[tex]\tau_{max} = 76.190\times 10^{6}\,Pa[/tex]

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

[tex]\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}[/tex]

Where:

[tex]\tau_{max}[/tex] - Maximum allowable shear stress, measured in pascals.

[tex]V[/tex] - Shear force, measured in kilonewtons.

[tex]A[/tex] - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

[tex]V = \frac{P}{5}[/tex]

[tex]V = \frac{450\,kN}{5}[/tex]

[tex]V = 90\,kN[/tex]

The minimum allowable cross section area is cleared in the shearing stress equation:

[tex]A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}[/tex]

If [tex]V = 90\,kN[/tex] and [tex]\tau_{max} = 76.190\times 10^{3}\,kPa[/tex], the minimum allowable cross section area is:

[tex]A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}[/tex]

[tex]A = 1.640\times 10^{-3}\,m^{2}[/tex]

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

[tex]A = \frac{\pi}{4}\cdot D^{2}[/tex]

The diameter is now cleared and computed:

[tex]D = \sqrt{\frac{4}{\pi}\cdot A}[/tex]

[tex]D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})[/tex]

[tex]D = 0.0457\,m[/tex]

[tex]D = 45.7\,mm[/tex]

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

We have that the minimum allowable bolt diameter is mathematically given as

From the question we are told

Generally the equation for the stress   is mathematically given as

[tex]\mu= 320/4.2 \\\\\mu= 76.190 N/mm^2[/tex]

Therefore

Force = Stress * area

Force = P/2

F= 425,000 N / 2 = 212,500 N

Hence area of each bolt is given as

212,500 = 76.190*( 5* area of each bolt)

area of each bolt = 557.815

Since

area of each bolt=\pi*d^2/4

\pi*d^2/4 = 557.815

For more information on diameter visit

https://brainly.com/question/8552546

Answer:

We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...

Explanation:

We can describe 15×-10 as an expression. we would describe 6×-2< 35 as an...

Answer:

Reduce sound transmission.

Explanation:

A caulking is a flexible material used to seal joints, cracks or gaps formed between building materials and pipes against leakage.

Caulking is recommended around the edges of partitions between apartments to reduce sound transmission.

Hence, in the event that an individual notices that air or sound is gaining entrance into their apartment, a caulking can be used to mitigate this noise or unwanted sound.

The caulking when applied to the gap or edges of partitions between apartments would create a tight seal and block the flow or entry of air, thereby reducing sound transmission.

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes  2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

Answer:

the overall elongation δ of the bar is  1.2337 mm

Explanation:

From the information given :

According to the principle of superposition being applied to the axial load P of the system; we have:

[tex]\delta = \delta_{AB} +\delta_{BC} + \delta_{CD}[/tex]    

where;

δ = overall elongation

[tex]\delta _{AB}[/tex] = elongation of bar AB

[tex]\delta _{BC}[/tex] = elongation of  bar BC

[tex]\delta _{CD} =[/tex]  elongation of bar CD]

If we replace; [tex]\dfrac{PL}{AE}[/tex] for  δ  and bt for area;

we have:

[tex]\delta = \dfrac{P_{AB}L_{AB}}{(b_{AB}t)E} +\dfrac{P_{BC}L_{BC}}{(b_{BC}t)E}+\dfrac{P_{CD}L_{CD}}{(b_{CD}t)E}[/tex]

where ;

P = load

L = length of the bar

A = area of the cross-section

E = young modulus of elasticity

Let once again replace:

P for [tex]P_{AB}, P_{BC} , P_{CD}[/tex]  (since load in all member of AB, BC and CD will remain the same )

[tex]\dfrac{L}{4}[/tex] for [tex]L_{AB}[/tex],  

[tex]\dfrac{L}{2}[/tex] for [tex]L_{BC}[/tex] and

[tex]\dfrac{L}{4}[/tex] for [tex]L_{CD}[/tex]

[tex]2\dfrac{b}{3}[/tex] for  [tex]b_{BC}[/tex]

b for  [tex]b_{CD}[/tex]

[tex]\delta = \dfrac{P (\dfrac{L}{4})}{btE}+ \dfrac{P (\dfrac{L}{2})}{2 \dfrac{b}{3}tE}+\dfrac{P (\dfrac{L}{4})}{btE}[/tex]

[tex]\delta = \dfrac{PL}{btE}[\dfrac{1}{4}+ \dfrac{1}{2}*\dfrac{3}{2}+ \dfrac{1}{4}][/tex]

[tex]\delta = \dfrac{5}{4}\dfrac{PL}{btE} --- \ (1)[/tex]

The stress in the central portion can be calculated as:

[tex]\sigma = \dfrac{P}{A}[/tex]

[tex]\sigma = \dfrac{P}{\dfrac{2}{3}bt}[/tex]

[tex]\sigma = \dfrac{3P}{2bt}[/tex]

So; Now:

[tex]\delta = \dfrac{5}{4}* \dfrac{2 * \sigma}{3}*\dfrac{L}{E}[/tex]

[tex]\delta= \dfrac{5}{4}* \dfrac{2 * 200}{3}*\dfrac{570}{77*10^3 \ MPa}[/tex]

δ = 1.2337 mm

Therefore, the overall elongation δ of the bar is  1.2337 mm