Explain the concept of an electric field as if you were addressing a friend or relative.

Answer:

The potential voltage range across a 2.2 kΩ 10% tolerance resistor when current of  4 sin 44t mA is flowing through the element is between a range of 7.92sin44t and 9.68sin44t volts.

Explanation:

Given that there is 10% tolerance for the 2.2 kΩ resistor, this implies that the resistance would range between 2,200 — 10% of 2,200 and 2,200 + 10% of 2,200, which is:

(i) 2,200 — 10% of 2,200 = 2,200 — 220 = 1,980 Ω, and

(ii) 2,200 + 10% of 2,200 = 2,200 + 220 = 2,420 Ω

Therefore, we will calculate the potential voltage for 1,980 Ω and 2,420 Ω if the current flowing through the element is 4sin44t mA:

(a) The potential voltage for a resistance of 1,980 Ω: we will use the formula: potential voltage v = i × R

Where i = 4sin44t mA = 0.004sin44t A, and R = 1,980 Ω

The potential voltage = v = 1,980 × 0.004sin44t = 7.92sin44t (in volts)

(b) The potential voltage for a resistance of 2,420 Ω: we will use the formula: potential voltage v = i × R

Where i = 4sin44t mA = 0.004sin44t A, and R = 2,420 Ω

The potential voltage = v = 2,420 × 0.004sin44t = 9.68sin44t (in volts)

Answer:

See explanation below.

Explanation:

If the statement is a tautology is true for all the possible combinations

Part a

[tex] (p \land q) \Rightarrow (p \lor r)[/tex] lets call this condition (1)

[tex](p \land q) [/tex] condition (2) and [tex](p \lor r)[/tex] condition (3)

We can create a table like this one:

p       q     r      (2)       (3)     (1)  

T       T     T      T        T       T

T       T     F      T        T       T        

T       F     T      F        T       T

T       F     F      F        T       T

F       T     T      F        T       T

F       T     F      F        F       T

F       F     T      F        T       T

F       F     F      F        F       T

So as we can see we have a tautology.

Part b

[tex] p \Rightarrow (r \Rightarrow p)[/tex] lt's call this condition 1

And [tex] (r \Rightarrow p)[/tex] condition 2

We can create the following table:

p     r       (2)     (1)

T     T       T       T

T     F       T       T

F     T       F       T

F     F       T       T

So is also a tautology.

Answer:

a. [tex]\frac{-kdT(0)}{dx} =q_{0}[/tex]=5000W/m^2

b.833.3(0.006-x)+112

c. 117 deg C

Explanation:

Consider the base plate of an 800-W household iron with a thickness of L 5 0.6 cm, base area of A 5 160 cm2, and thermal conductivity of k 5 60 W/m·K. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. When steady operating conditions are reached, the outer surface temperature of the plate is measured to be 112°C. Disregarding any heat loss through the upper part of the iron,

Assumption

Heat conduction is steady state and unidimensional  2. thermal conductivity is constant. Heat supplied is not in the plate

4. we disregard heat loss

Heat flux=heat/area

[tex]\alpha[/tex]/A=800W/160*10^-4

with direction to the surface been in the x direction,

the mathematical expression will be

[tex]\frac{d^2T}{dx^2}[/tex]=0..............1

and [tex]\frac{-kdT(0)}{dx} =q_{0}[/tex]=5000W/m^2

from fourier law, for conductivity

T(L)=T2=112C

b. integrating equation 1 twice we have\dT/dx=c1

T(x)=C1x+C2

C1 and C2 are arbitrary constant

at x=0 the boundary conditions become

-kC1=qo

C1=-(qo/k)

at x=L          

=T(L)=C1L+C2=T2

C2=T2-cL1

C2=T2+qoL/k

Juxtaposing C1 and C2 into the general equation , we have

T(x)=-qo/k+T2+qoL/k=qo(L-k)/k+T2

50000*(0.006-x)/60+112

833.3(0.006-x)+112

c. inner surface plate temperature is

T(0)=833.33(0.006-0)+112 ( using the derivation in answer b)

117 deg C

Answer:

a)  1.165 × 10⁻⁸ N b)- 1.165 × 10⁻⁸ N

Explanation:

Using Coulomb's law

F(attraction) = [tex]\frac{Z1Z2qelectron}{4piER^2}[/tex]

where

R = sum of the distance between the centers of charges = sum of ionic radii = 0.079 nm + 0.120nm = 0.199 nm = 0.199 × 10⁻⁹ m

Z₁ = valency of Mg²⁺ = 2

Z₂ = valency of F ⁻ = - 1

qelectron = charge on electron =1.062 × 10⁻19 C

E = permitivity of free space = 8.85 × 10 ⁻¹² C²/ Nm²

Fa= (1×2× (1.602 × 10⁻¹⁹)²) / (4× 3.142 × 8.85 × 10⁻¹² × (0.199 × 10⁻⁹)²) = 1.165 × 10⁻⁸ N

b) At equilibrium F of repulsion = - F of attraction = - 1.165 × 10⁻⁸ N

Explanation:

Note: For equations refer the attached document!

The net upward pressure force per unit height p*D must be balanced by the downward tensile force per unit height 2T, a force that can also be expressed as a stress, σhoop, times area 2t. Equating and solving for σh gives:

 Eq 1

Similarly, the axial stress σaxial can be calculated by dividing the total force on the end of the can, pA=pπ(D/2)2 by the cross sectional area of the wall, πDt, giving:

Eq 2

For a flat sheet in biaxial tension, the strain in a given direction such as the ‘hoop’ tangential direction is given by the following constitutive relation - with Young’s modulus E and Poisson’s ratio ν:

 Eq 3

Finally, solving for unknown pressure as a function of hoop strain:

 Eq 4

Resistance of a conductor of length L, cross-sectional area A, and resistivity ρ is

 Eq 5

Consequently, a small differential change in ΔR/R can be expressed as

 Eq 6

Where ΔL/L is longitudinal strain ε, and ΔA/A is –2νε where ν is the Poisson’s ratio of the resistive material. Substitution and factoring out ε from the right hand side leaves

 Eq 7

Where Δρ/ρε can be considered nearly constant, and thus the parenthetical term effectively becomes a single constant, the gage factor, GF

 Eq 8

For Wheat stone bridge:

 Eq 9

Given that R1=R3=R4=Ro, and R2 (the strain gage) = Ro + ΔR, substituting into equation above:

Eq9

Substituting e with respective stress-strain relation

Eq 10

part b

Since, axial strain(1-2v) < hoop strain (2-v). V out axial < V out hoop.

Hence, dV hoop < dV axial.

part c  

Given data:

P = 253313 Pa

D = d + 2t = 0.09013 m

t = 65 um

GF = 2

E = 75 GPa

v = 0.33

Use the data above and compute Vout using Eq3

Eq 11  

Answer:

The question is incomplete. Below is the complete question

"An industrial plant consisting primarily of induction motor loads absorbs 500 kW at 0.6 power factor lagging. (a) Compute the required kVA rating of a shunt capacitor to improve the power factor to 0.9 lagging. (b) Calculate the resulting power factor if a synchronous motor rated 500 hp with 90% efficiency operating at rated load and at unity power factor is added to the plant instead of the capacitor. Assume constant voltage. (1 hp = 0.746 kW)"

Answer:

a. 424.5KVA

b. 0.808 lagging.

Explanation:

Let's first determine the real power, reactive and the apparent power delivered.

Ql= Ptan(the real power angle)

Ql=500*tan53.13°

Ql=666.7 Kvar

The reactive angle is

Arccos(0.9)=25.84°

Now we calculate the reactive power

Qs=Ptan25. 84°

Qs=242.4KVAR

Hence the apparent power is

Qc=Ql-Qs

Qc=666.7-242.2

Qc=424.5KVAR

The required KVA rating of the shunt capacitor is 424.5KVA

b. To calculate the resulting power factor we first determine the power absorbs by the synchronous motor.

Pm=(500*0.746)/0.9

Pm=414.4KW

The total reactive power is

Ps=P+Pm

Ps=414.4+500=914.4KW

Hence we compute the source power factor as

PF=cos[arctan(Qs/Ps)]

PF=cos[arctan(666.7/914.4)

From careful calculation, we arrive at

PF=0.808.

Note this power factor will be lagging.

Answer:

The three phase full load secondary amperage is 2775.7 A

Explanation:

Following data is given,

S = Apparent Power = 1000 kVA

No. of phases = 3

Secondary Voltage: 208 V/120 V (Here 208 V is three phase voltage and 120 V is single phase voltage)

Since,

[tex]V_{1ph} =\frac{ V_{3ph}}{\sqrt{3} }\\V_{1ph) = \frac{208}{\sqrt{3} }\\[/tex]

[tex]V_{1ph} = 120 V[/tex]

The formula for apparent power in three phase system is given as:

[tex]S = \sqrt{3} VI[/tex]

Where:

S = Apparent Power

V = Line Voltage

I = Line Current

In order to calculate the Current on Secondary Side, substituting values in above formula,

[tex]1000 kVA = \sqrt{3} * (208) * (I)\\1000 * 1000 = \sqrt{3} * (208) * (I)\\I = \frac{1000 * 1000}{\sqrt{3} * (208) }\\ I = 2775.7 A[/tex]

Answer

//countStrings Method

public class countStrings {

public int Arraycount(String[] arrray, String target) {

int count = 0;

for(String elem : arrray) {

if (elem.equals(target)) {

count++;

}

}

return count;

}

// Body

public static void main(String args [] ) {

countStrings ccount = new countStrings();

int kount = ccount.Arraycount(new String[]{"Sick", "Health", "Hospital","Strength","Health"}, " Health"

System.out.println(kount);

}

}

The Program above is written in Java programming language.

It's Divided into two parts

The first is the method countStrings

While the second part of the program is the main method for the program execution

Answer:

s= 0.1156 * w or

s= 0.115*(q+p) in terms of Top Biomass and Root Biomass

Explanation:

Since s (number of seeds) is proportional to biomass (w), and above ground biomass also increases with total plant biomass.

s α w

s= 26

w= 225

k= constant

Thus s = k * w

s/w= k

26/225= k

0.1156= k

The equation showing the relationship between seeds and plant biomass is:

s= 0.1156 * w

Assume q= Top Biomass, and p= Root Biomass

w= q+p

Our equation now becomes

s= 0.115*(q+p)

Answer:

[tex]v(t)=-\frac{5}{2}\sqrt{t^3}+20\\s(t)=-\sqrt{t^5}+20t[/tex]

[tex]s(t=4)=48\text{ m}[/tex]

Explanation:

In this case acceleration is defined as:

[tex]a(t)=-k\sqrt{t}[/tex] ,

where k is a constant to be found.

To find the expressions for velocity and position as a function of time you must integrate the expression above for acceleration two times.

Initial conditions and boundary conditions are defined with the rest of the data as:

[tex]v(t=0)=20\text{ m/s}\\v(t=4)=0\text{ m/s}\\s(t=0)=0\text{ m}[/tex]

First integration is equal to:

[tex]a'(t)=v(t)=-k\int\sqrt{t}dt=-\frac{2}{3}k\sqrt{t^3}+C_1[/tex]

The boundary condition and initial condition can be used to calculate [tex]k[/tex] and [tex]C_1[/tex]:

[tex]C_1=20\\k=\frac{15}{4}[/tex]

With this expression for velocity is defined as:

[tex]v(t)=-\frac{5}{2}\sqrt{t^3}+20[/tex]

The same can be done to get to expression for position:

[tex]s(t)=-\sqrt{t^5}+20[/tex]

To get the total distance traveled you can integrate the velocity expression from time=0 sec to time=4 sec:

[tex]s_{tot}=\int_0^4(-\frac{5}{2}\sqrt{t^3}+20)dt=48\text{ m}[/tex]

Answer:

The angular acceleration is -2.44 rad/s², while the linear acceleration is -14.66 in/s².

Explanation:

First we need to find the time, at the given position. W e are given the position of particle to be:

θ = 35°

Converting it to radians because, the given equation is in radians:

θ = (35°) (π radians/180°)

θ = 0.611 radians

Now, we have the equation:

θ = Cos(2t)

2t = Arc Cos (θ)

2t = Arc Cos (0.611 radians)

t = 0.91/2

t = 0.457 sec

Now, to determine angular acceleration of the particle, we must derivate the equation twice with respect to 't'

Angular Velocity = ω = dθ/dt = -2Sin(2t)

Angular Acceleration = α = -4Cos(2t)

Now, we use the value of t:

α = -4Cos(2 x 0.457)

α = -2.44 rad/s² (negative sign shows decceleration)

Now for linear acceleration, we know that:

a = rα

a = (6 in)(-2.44 rad/s²)

a = -14.66 in/s² (negative sign shows decceleration)

Answer:

t1 = t3 = 5 seconds

t2 = 15 seconds

Explanation:

For t = t1

a = 10 m/s^2

v(t) = 10*t

s(t) = 5*t^2

Distance traveled = 5*t1^2

For t = t2

a = 0 m/s^2

v(t) = 10*t1

s(t) = 10*t1*t

Distance traveled = 10*t1*t2

For t = t3

a = -10 m/s^2

v(t) = -10*t

s(t) = - 5t^2

Distance traveled = 5t3^2

Sum of all distances = 5*t1^2 + 10*t1*(t2) + 5t3^2

1000 = 5t1^2 + 10t1t2 + 5t3^2 + 10*t1*t2 ....... Eq 1

Distance traveled in first and last segments are the same:

t1 = t3 ..... Eq 2

Given: t1+t2+t3 = 25  .... Eq 3

Solving Equations simultaneously:

Subs Eq 2 into Eq 3 & Eq 1

1000 = 5t1^2 + 10*t1*t2 + 5*t1^2

100 = t1^2 + t1*t2  ..... Eq 4

2t1 + t2 = 25

t2 = 25 - 2t1  .... Eq 5

Subs Eq 5 into Eq 4

100 =  t1^2 + t1*(25 - 2t1)

t1^2 -25t1 + 100 = 0

Solve for t1

t1 = 5 , 20 Hence, t1 = 5 sec is selected

t1 = t3 = 5 sec

t2 = 15 sec

Answer:

The final pressure in the vessel is 2.270atm

Explanation:

Mass of PCl3 = 24.5g, MW of PCl3 = 137.5g/mole, number of moles of PCl3 = 24.5/137.5 = 0.178mole

Mass of O2 = 3g, MW of O2 = 32g/mole, number of moles of O2 = 3/32 = 0.094moles

Total moles of rectant (n) = 0.178mole + 0.094mole = 0.272mole

From ideal gas equation

PV = nRT

P (pressure of reactants) = nRT/V

n = 0.272mole, R = 82.057cm^3.atm/gmol.K, T = 15°C = 15+273K = 288K, V = 4.6L = 4.6×1000cm^3 = 4600cm^3

P = 0.272×82.057×288/4600 = 1.397atm

From pressure law

P1/T1 = P2/T2

P2 (final pressure in the vessel) = P1T2/T1

P1 = 1.397atm, T1 = 288K, T2 = 195°C = 195+273K = 468K

P2 = 1.397×468/288 = 2.270atm

Answer:

1.1451 x 104 (11451.13)mg/m3

Explanation:

1 ppmv is defined as one volume of a contaminant or solid(CO)(mL) in 1 x 106 volume of solvent/water.

1ppmv = 1mL/m3

Concentration in mg/m3 = volume in ppm x molecular weight x pressure(kPa)/( gas constant x temperature(K)

Molecular weight of CO = 12 + 16

= 28g/mol

Temperature = 273.15 + 25

= 298.15K

Pressure = 1 x 101.325kPa

= 101.325kPa

Ppmv = 1 x 10-4ppmv

Gas constant, R = 8.3144 L.kPa/mol.K

Concentration in mg/m3 = (1 x 104 * 28 * 101.325)/(8.3144 * 298)

= 1.1451 x 104mg/m3

= 11451.13 mg/m3

Explanation:

The obtained data from water properties tables are:

Point 1 (condenser exit) @ 8 KPa, saturated fluid

[tex]h_{f} = 173.358 \\h_{fg} = 2402.522[/tex]

Point 2 (Pump exit) @ 18 MPa, saturated fluid & @ 4 MPa, saturated fluid

[tex]h_{2a} = 489.752\\h_{2b} = 313.2[/tex]

Point 3 (Boiler exit) @ 18 MPa, saturated steam & @ 4 MPa, saturated steam

[tex]h_{3a} = 2701.26 \\s_{3a} = 7.1656\\h_{3b} = 2634.14\\s_{3b} = 7.6876[/tex]

Point 4 (Turbine exit) @ 8 KPa, mixed fluid

[tex]x_{a} = 0.8608\\h_{4a} = 2241.448938\\x_{b} = 0.9291\\h_{4b} = 2405.54119[/tex]

Calculate mass flow rates

Part a) @ 18 MPa

mass flow

[tex]\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3a} - h_{4a}) - (h_{2a} - h_{f})}\\\\= \frac{100*10^ 3}{(2701.26 - 2241.448938 ) - (489.752 - 173.358)}\\\\= 697.2671076 \frac{kg}{s} = 2510161.587 \frac{kg}{hr}[/tex]

Heat transfer rate through boiler

[tex]Q_{in} = mass flow * (h_{3a} - h_{2a})\\Q_{in} = (697.2671076)*(2701.26-489.752)\\\\Q_{in} = 1542011.787 W[/tex]

Heat transfer rate through condenser

[tex]Q_{out} = mass flow * (h_{4a} - h_{f})\\Q_{out} = (697.2671076)*(2241.448938-173.358)\\\\Q_{out} = 1442011.787 W[/tex]

Thermal Efficiency

[tex]n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.06485[/tex]

Part b) @ 4 MPa

mass flow

[tex]\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3b} - h_{4b}) - (h_{2b} - h_{f})}\\\\= \frac{100*10^ 3}{(2634.14 - 2405.54119 ) - (313.12 - 173.358)}\\\\= 1125 \frac{kg}{s} = 4052374.235 \frac{kg}{hr}[/tex]

Heat transfer rate through boiler

[tex]Q_{in} = mass flow * (h_{3b} - h_{2b})\\Q_{in} = (1125.65951)*(2634.14-313.12)\\\\Q_{in} = 2612678.236 W[/tex]

Heat transfer rate through condenser

[tex]Q_{out} = mass flow * (h_{4b} - h_{f})\\Q_{out} = (1125)*(2405.54119-173.358)\\\\Q_{out} = 2511206.089 W[/tex]

Thermal Efficiency

[tex]n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.038275[/tex]

For (a) 18 MPa:

- The mass flow rate of steam is approximately 238,050 kg/h.

- The heat transfer rates are approximately 674,970 kW (boiler) and 481,360 kW (condenser).

- The thermal efficiency is approximately 14.81%.

For (b) 4 MPa:

- The mass flow rate of steam is approximately 187,320 kg/h.

- The heat transfer rates are approximately 480,040 kW (boiler) and 380,450 kW (condenser).

- The thermal efficiency is approximately 20.83%.

Explanation and Calculation

To analyze the ideal Rankine cycle, we need to follow these steps for each case:

Case (a): 18 MPa

1. Determine Specific Enthalpies

- Condenser exit: Saturated liquid at 8 kPa.

 From steam tables: [tex]\( h_1 \approx 191.81 \, \text{kJ/kg} \)[/tex]

-Pump exit: Compressed liquid at 18 MPa.

 Using [tex]\( h_2 = h_1 + v_1 \Delta P \)[/tex]

 [tex]\[ v_1 \approx 0.001 \, \text{m}^3/\text{kg} \][/tex]

 [tex]\[ h_2 \approx 191.81 + 0.001 \times (18000 - 8) \approx 209.81 \, \text{kJ/kg} \][/tex]

- Boiler exit: Saturated vapor at 18 MPa.

 From steam tables: [tex]\( h_3 \approx 2821.6 \, \text{kJ/kg} \)[/tex]

-Turbine exit: Expanding to 8 kPa.

 From steam tables: [tex]\( h_4 \approx 2201.4 \, \text{kJ/kg} \)[/tex]

2. Mass Flow Rate of Steam

Using the power output:

[tex]\[ \dot{W}_{\text{net}} = \dot{m} (h_3 - h_4 - (h_2 - h_1)) \][/tex]

[tex]\[ 100 \times 10^6 = \dot{m} (2821.6 - 2201.4 - (209.81 - 191.81)) \][/tex]

[tex]\[ 100 \times 10^6 = \dot{m} \times 420.2 \][/tex]

[tex]\[ \dot{m} \approx 238050 \, \text{kg/h} \][/tex]

3. Heat Transfer Rates

- Boiler:

 [tex]\[ \dot{Q}_{\text{in}} = \dot{m} (h_3 - h_2) \][/tex]

 [tex]\[ \dot{Q}_{\text{in}} = 238050 \times (2821.6 - 209.81) \approx 674.97 \times 10^3 \, \text{kW} \][/tex]

- Condenser:

 [tex]\[ \dot{Q}_{\text{out}} = \dot{m} (h_4 - h_1) \][/tex]

 [tex]\[ \dot{Q}_{\text{out}} = 238050 \times (2201.4 - 191.81) \approx 481.36 \times 10^3 \, \text{kW} \][/tex]

4. Thermal Efficiency

[tex]\[ \eta = \frac{\dot{W}_{\text{net}}}{\dot{Q}_{\text{in}}} \][/tex]

[tex]\[ \eta = \frac{100 \times 10^3}{674.97 \times 10^3} \approx 14.81\% \][/tex]

Case (b): 4 MPa

1. Determine Specific Enthalpies

- Condenser exit: Saturated liquid at 8 kPa.

 From steam tables: [tex]\( h_1 \approx 191.81 \, \text{kJ/kg} \)[/tex]

- Pump exit: Compressed liquid at 4 MPa.

 Using [tex]\( h_2 = h_1 + v_1 \Delta P \)[/tex]

 [tex]\[ v_1 \approx 0.001 \, \text{m}^3/\text{kg} \][/tex]

 [tex]\[ h_2 \approx 191.81 + 0.001 \times (4000 - 8) \approx 195.81 \, \text{kJ/kg} \][/tex]

- Boiler exit: Saturated vapor at 4 MPa.

 From steam tables: [tex]\( h_3 \approx 2749.7 \, \text{kJ/kg} \)[/tex]

- Turbine exit: Expanding to 8 kPa.

 From steam tables: [tex]\( h_4 \approx 2222.0 \, \text{kJ/kg} \)[/tex]

2. Mass Flow Rate of Steam

Using the power output:

[tex]\[ \dot{W}_{\text{net}} = \dot{m} (h_3 - h_4 - (h_2 - h_1)) \][/tex]

[tex]\[ 100 \times 10^6 = \dot{m} (2749.7 - 2222.0 - (195.81 - 191.81)) \][/tex]

[tex]\[ 100 \times 10^6 = \dot{m} \times 531.9 \][/tex]

[tex]\[ \dot{m} \approx 187320 \, \text{kg/h} \][/tex]

3. Heat Transfer Rates

- Boiler:

 [tex]\[ \dot{Q}_{\text{in}} = \dot{m} (h_3 - h_2) \][/tex]

 [tex]\[ \dot{Q}_{\text{in}} = 187320 \times (2749.7 - 195.81) \approx 480.04 \times 10^3 \, \text{kW} \][/tex]

- Condenser:

[tex]\[ \dot{Q}_{\text{out}} = \dot{m} (h_4 - h_1) \][/tex]

 [tex]\[ \dot{Q}_{\text{out}} = 187320 \times (2222.0 - 191.81) \approx 380.45 \times 10^3 \, \text{kW} \][/tex]

4. Thermal Efficiency

[tex]\[ \eta = \frac{\dot{W}_{\text{net}}}{\dot{Q}_{\text{in}}} \][/tex]

[tex]\[ \eta = \frac{100 \times 10^3}{480.04 \times 10^3} \approx 20.83\% \][/tex]

Answer:

Detailed working is shown

Explanation:

The attached file shows a detailed step by step calculation..

The radial and transverse components of the rocket's velocity at an angle of 60 degrees from the x-axis are 325 m/s and 563 m/s, respectively.

Using Physics principles, we know that when a rocket moves along a parabolic path, its velocity can be decomposed into two components: the radial component (the component of velocity directly in line with the radial direction) and the transverse component (the component of velocity perpendicular to the radial direction).

Given that the absolute speed |v| of the rocket is 650 m/s and the angle θ that the velocity makes with respect to x-axis (measured counterclockwise) is 60°, we can use the trigonometric definitions of sine and cosine to compute the radial and transverse components respectively.

Part A:  The radial component of velocity (vr) at θ = 60° can be computed using the formula vr = v * cosθ. So, vr = 650 m/s * cos60° = 325 m/s.

Part B: The transverse component of velocity (vt) at θ = 60° can be computed using the formula vt = v * sinθ. So, vt = 650 m/s * sin60° = 563 m/s.

https://brainly.com/question/33537450

#SPJ3

Answer:

change interval is 3.93 sec

clearance interval is 1.477 sec

Explanation:

Given data:

upgrade of intersection 4%

street total width at intersection is 60 ft

vehicle length of approaching traffic = 16 ft

speed of approaching traffic =40 mi/hr

85th percentile speed is calculated as

S_{85} = S +5

S_{ 85} = 40 + 5  = 45 mi/h

15th Percentile speed

[tex]S_{15} =S-5[/tex]

          = 40 - 5 = 35 mi/hr

change in interval is calculated as

[tex]y = t + \frac{1.47 S_{85}}{2a +(64.4\times 0.01 G}[/tex]

t is reaction time is 1.0,

deceleration rate is given as 10 ft/s^2

[tex]y = 1.0 +\frac{1.47\times 45}{2a +(64.4\times 0.01\times 4}[/tex]

y = 3.93 s

clearance interval is calculated as

[tex]a_r = \frac{W+ L}{1.47\times S_{15}}[/tex]

[tex]a_r = \frac{60+16}{1.47\times 35} = 1.477 s[/tex]

Answer:

The question is incomplete, below is the complete question

"The real power delivered by a source to two impedance, Z1=4+j5⁡Ω and Z2=10⁡Ω connected in parallel, is 1000 W. Determine (a) the real power absorbed by each of the impedances and (b) the source current."

answer:

a. 615W, 384.4W

b. 17.4A

Explanation:

To determine the real power absorbed by the impedance, we need to find first the equivalent admittance for each impedance.

recall that the symbol for admittance is Y and express as

[tex]Y=\frac{1}{Z}[/tex]

Hence for each we have,  

[tex]Y_{1} =1/Zx_{1}\\Y_{1} =\frac{1}{4+j5}\\converting to polar \\ Y_{1} =\frac{1}{6.4\leq 51.3}\\ Y_{1} =(0.16 \leq -51.3)S[/tex]

for the second impedance we have

[tex]Y_{2}=\frac{1}{10}\\Y_{2}=0.1S[/tex]

we also determine the voltage cross the impedance,

P=V^2(Y1 +Y2)

[tex]V=\sqrt{\frac{P}{Y_{1}+Y_{2}}}\\[/tex]

[tex]V=\sqrt{\frac{1000}{0.16+0.1}}\\ V=62v[/tex]

The real power in the impedance is calculated as

[tex]P_{1}=v^{2}G_{1}\\P_{1}=62*62*0.16\\ P_{1}=615W[/tex]

for the second impedance

[tex]P_{2}=v^{2}*G_{2}\\ P_{2}=62*62*0.1\\384.4w[/tex]

b. We determine the equivalent admittance

[tex]Y_{total}=Y_{1}+Y_{2}\\Y_{total}=(0.16\leq -51.3 )+0.1\\Y_{total}=(0.16-j1.0)+0.1\\Y_{total}=0.26-J1.0\\[/tex]

We convert the equivalent admittance back into the polar form

[tex]Y_{total}=0.28\leq -19.65\\[/tex]

the source current flows is

[tex]I_{s}=VY_{total}\\I_{s}=62*0.28\\I_{s}=17.4A[/tex]

Answer:

It serves as a guarantee that the contractor who wins the bid will honor the terms of the bid after the contract is signed.

Explanation:

A bid bond is a type of construction bond that protects the obligee in a  construction bidding process.

A bid bond typically involves three parties:

The obligee; the owner or developer of the construction project under bid. The principal; the bidder or proposed contractor.

The surety; the agency that issues the bid bond to the principal example insurance company or bank.

A bid bond generally serves as a guarantee that the contractor who wins the bid will honor the terms of the bid after the contract is signed.

Contractors are required to submit a bid bond to ensure they are financially able and willing to complete the project they bid on, preserving the integrity of the bidding process by preventing cost overruns, favoritism, and corruption.

A contractor is normally required to submit a bid bond when making a proposal to an owner on a competitively bid contract to ensure that the contractor is serious and financially able to carry out the project. The bid bond is a form of guarantee that the contractor will enter into the contract at the bid price if awarded the project and provides a financial penalty if the contractor fails to do so. Municipalities and other entities require this bond to prevent situations where a contractor might win a bid and then refuse to complete the work or ask for additional funds to finish the project (“cost overruns”). This is significant in a competitive bidding environment where fairness and the avoidance of favoritism or corruption are essential, as it stops anyone from being favored over others and ensures the project will be completed at the bid price.

Tenders, which are bid documents that outline the project plan and cost, play a crucial role in the construction industry. A well-prepared tender demonstrates that the projected cost reflects a balance between safety, environmental friendliness, and profitability. In contrast, a sealed-bid auction often used in these processes, requires strategic bidding and is susceptible to collusion or bribery to uncover competitor's bid details. Hence, a bid bond safeguards the process by providing a financial deterrent against such unethical practices.

Answer:

(a) E = 0 N/C

(b) E = 0 N/C

(c) E = 7.78 x10^5 N/C

Explanation:

We are given a hollow sphere with following parameters:

Q = total charge on its surface = 23.6 μC = 23.6 x 10^-6 C

R = radius of sphere = 26.1 cm = 0.261 m

Permittivity of free space = ε0 = 8.85419 X 10−12 C²/Nm²

The formula for the electric field intensity is:

E = (1/4πεo)(Q/r²)

where, r = the distance from center of sphere where the intensity is to be found.

(a)

At the center of the sphere r = 0. Also, there is no charge inside the sphere to produce an electric field. Thus the electric field at center is zero.

E = 0 N/C

(b)

Since, the distance R/2 from center lies inside the sphere. Therefore, the intensity at that point will be zero, due to absence of charge inside the sphere (q = 0 C).

E = 0 N/C

(c)

Since, the distance of 52.2 cm is outside the circle. So, now we use the formula to calculate the Electric Field:

E = (1/4πεo)[(23.6 x 10^-6 C)/(0.522m)²]

E = 7.78 x10^5 N/C

Answer:

a diameter of D₂ = 0.183 inches would be required

Explanation:

appyling pascal's law

P applied to the hydraulic jack = P required to lift the rock

F₁*A₁ = F₂*A₂

since A₁= π*D₁²/4 ,  A₂= π*D₂²/4

F₁*π*D₁²/4 = F₂* π*D₂²/4

F₁*D₁²=F₂*D₂²

D₂ = D₁ *√(F₁/F₂)

replacing values

D₂ = D₁ *√(F₁/F₂) =  6 in * √(120 lbf/(4000 lbm * 32.174 (lbf/lbm)) = 0.183 inches

Answer:

Yes, because both design compressive stress is greater than 4000 psi

Explanation:

To design a concrete that satisfy the requirements of ACI code for 4000 psi concrete

Step 1: design the compressive stress, using the two equations below

[tex]f_{c} =f_{cr}-1.34*s[/tex] -------equation 1

where;

[tex]f_{c}[/tex] is the design compressive stress

[tex]f_{cr}[/tex] is the critical stress = 4780 psi mean strength

s is the standard deviation = 525 psi

[tex]f_{c} = 4780 -1.34*525[/tex] = 4076.5 psi

Step 2: design the compressive stress, using the second design equation

[tex]f_{c} =f_{cr}-2.33*s + 500[/tex] -------equation 2

[tex]f_{c} =4780-2.33*525 + 500[/tex] = 4056.75 psi

Therefore, since both compressive stress is greater than 4000 psi, the concrete satisfies the requirements of ACI code for 4000 psi concrete

Answer: 3002.86kg

Explanation:

Hydraulic cylinder diameter =125mm

Ambient pressure =1bar

Pressure =2500kpa

Piston Mass (MP) =?

F|(when it moves upward )=PA=F|(when it moves downward) =PoA+Mpg

Po=1 bar=100kpa

A=(π/4)D^2=(π/4)*0.125^2=0.01227m^2

Mp=(P-Po) A/g=(2500-100)*1000*0.01227/9.80665

Mp=3002.86kg.

Answer:

amount of electric charge transported =  1.13 × [tex]10^{-2}[/tex] C

Explanation:

given data

electric current = 237.0 mA = 0.237 A

time = 8 min = 8 × 60 sec = 480 sec

solution

we get here amount of electric charge transported that is express as

amount of electric charge transported = electric current × time  ...........1

put here value and we get

amount of electric charge transported = 0.237  × 480

amount of electric charge transported = 113.76 C

amount of electric charge transported =  1.13 × [tex]10^{-2}[/tex] C

Answer:

False

Explanation:

DMZ stand for DeMilitarized Zone(perimeter network).

It is a sub-network(physical or logical) that contains external facing services to untrusted network. Not only that it contains it also exposes this kind of services.

Adding another layer of security is the main purpose of DMZ

DMZ is positioned in between the Internet and private network

Answer:

symbolic machine code.

Explanation:

The instructions in the language are closely linked to the machine's architecture.

Answer:

[tex]Ma=\frac{260}{330} \\Ma=0.7878[/tex]

So, Ma < 1  Flow is Subsonic

Explanation:

Mach Number:

Mach Number is the ratio of speed of the object to the speed of the sound. It is used to categorize the speed of the object on the basis of mach number as sonic, supersonic and hyper sonic. (It is a unit less quantity)

Mach < 1       Subsonic

Mach > 1       Supersonic

Ma= Speed of the object/Speed of the sound

[tex]Ma=\frac{260}{330} \\Ma=0.7878[/tex]

So, Ma < 1 Flow is Subsonic

Answer:

He wore his black suit, another color of shirt (not purple) and shoes

Explanation:

Holmes owns two suits: one black and one tweed.

Whenever he wears his tweed suit and a purple shirt, he chooses not to wear a tie and whenever he wears sandals, he always wears a purple shirt.

So, if he wore a bow tie yesterday, it means he wore his black suit, another color of shirt (not purple) and shoes because the shirt color is not purple

Answer:

kindly find attachment for detailed answer

Explanation

Consider a pond that initially contains 10 million gallons of fresh water. Water containing a chemical pollutant flows into the pond at the rate of 5 million gallons per year (gal/yr), and the mixture in the pond flows out at the same rate. The concentration c=c(t) of the chemical in the incoming water varies periodically with time to the expression c(t) = 2 + sin(2t) grams per gallon (g/gal).

Construct a mathematical model of this flow process and determine the amount of chemical in the pond at any time t. Then, plot the solution using Maple and describe in words the effect of the variation in the incoming chemical.

Answer:

A. 0.0283 mm3/min

B. 15850.2 gal/min

C. 0.2642 gal/min

D. 1.7 m3/hour

Explanation:

A.

[(1 in)3/min *(25.4mm)3/(1 in)]

= 0.02832 mm3/min

B.

[(1m)3/sec*(264.173gal)/(1m)3]*(60secs)/1min

= 15850.2 gal/min

C.

[(Liter/min)*(0.264172gal/liter)]

=0.2642 gal/min

D.

[(1ft)3/min*(0.3048m)3/(1ft)3*(60mins/1hour)]

=1.7 m3/hour

Below is an attachment that should help.