The volume of air in one beach ball is 288π in 3. What is the volume of air in a ball whose radius is 3 inches greater than the first beach ball?

To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are

[tex]V = 690V[/tex]

[tex]P_{real} = 2.3MW[/tex]

Real power in 3 phase

[tex]P_{real} = 3V_{ph}I_{ph} Cos\theta[/tex]

Now the Phase Voltage is,

[tex]V_{ph} = \frac{V}{\sqrt{3}}[/tex]

[tex]V_{ph} = \frac{690}{\sqrt{3}}[/tex]

[tex]V_{ph} = 398.37V[/tex]

The current phase would be,

[tex]P_{real} = 3V_{ph}I_{ph} Cos\theta[/tex]

Rearranging,

[tex]I_{ph}=\frac{P_{real}}{3V_{ph}Cos\theta}[/tex]

Replacing,

[tex]I_{ph}=\frac{2.3MW}{3( 398.37V)(0.85)}[/tex]

[tex]I_{ph}= 2.26kA/phase[/tex]

Therefore the current per phase is 2.26kA

Answer:

a.The current is cut in half

b.the current stays the same

c.the current doubles

d.the current becomes zero

Explanation:

The current doubles

Answer:

SaaS

Explanation:

Software as a service (SaaS) is also called software on demand, it involves a third party that centrally hosts the software and provides it to the end user.

All aspects of hosting is handled by the third party: application, data, runtime, middleware, operating system, server, virtualization, storage and networking are all handled by the provider.

This is an ideal software service for Fictional corp, as there will be no need to hire additional IT staff to maintain the new CRM software.

In the process of analyzing a thermodynamic system it is important to identify what system is being worked on and the processes and properties if the system

The magnitude of the force acting on the shaft, is approximately 1,336.5 N

The reason the value for the force magnitude acting on the shaft is correct is as follows:

The known parameters are:

The cross-sectional area of the shaft, Aₐ = 0.8 cm²

The required gas pressure in the cylinder, P = 3 bar

The mass of the piston, m₁ = 24.5 kg

The mass of the shaft, m₂ = 0.5 kg

The diameter of the piston, D = 10 cm

The atmospheric pressure, Pₐ = 1 bar

Required:

The magnitude of the force F acting on the shaft

Solution:

The force due to the gas in the cylinder, [tex]\mathbf{F_{gas}}[/tex], is given as follows;

[tex]F_{gas}[/tex] = 3 bar × π × (10 cm)²/4 = 2,359.19449 N

The force due to the atmosphere, [tex]\mathbf{F_{atm}}[/tex], is given as follows;

[tex]F_{atm}[/tex] = 1 bar × ((π × (10 cm)²/4) -  0.8 cm²) ≈ 777.4 N

The force due to the piston and shaft, [tex]\mathbf{F_{ps}}[/tex], is given as follows;

[tex]F_{ps}[/tex] = (24.5 kg + 0.5 kg) × 9.81 m/s² = 245.25 N

The magnitude of the force acting on the shaft, F = [tex]F_{gas}[/tex] - ([tex]\mathbf{F_{atm}}[/tex] + [tex]\mathbf{F_{ps}}[/tex])

∴ F = 2,359.19449 N - (777.4 N + 245.25 N) ≈ 1,336.5449 N

The magnitude of the force acting on the shaft, F ≈ 1,336.5 N

Learn more about forces due to pressure here:

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Answer:

The case with replacement has higher entropy

Explanation:

The complete question is given:

'Drawing with and without replacement. An urn contains r red,  w white  and b black balls. Which has higher entropy, drawing k ≥ 2 balls from the urn with  replacement or without replacement?'

Solution:

- n drawing is the same irrespective of whether there is replacement or not.

-  X to denotes drawing from an urn with r red balls,  w white balls and b black balls. So, n = b + r +  w.

We have:

                                      p_X(cr) = r / n

                                      p_X(cw) = w / n

                                      p_X(cb) = b / n

- Now, if  Xi is the ith drawing with replacement then Xi are independent and p_Xi (x) = pX(x) for x e ( cr , cb , cw ).

- Now, let  Yi be the ith drawing with replacement where Yi are not independent p_Yi (x) = pX(x) for x ∈ X.

- To see this, note  Y1 =  X and assume it is true for  Yi and consider  Yi+1:

    p_Y(i+1) (cr) = p(Y(i+1),Yi).(cr, cr) + p(Y(i+1),Yi).(cr, cw) + p(Y(i+1),Yi).(cr, cb)

= pY(i+1)|Yi  (cr|cr)*pYi  (cr) +pY(i+1)|Yi  (cr|cw)*pYi (cw) + pY(i+1)|Yi  (cr|cb)*pYi (cb)

= r*( r - 1 )/n*(n-1) + w*r/n*(n-1) + b*r/n*(n-1) = r / n =  p_X(cr)

- This means, using the chain rule and the conditioning theore m:

H(Y1, Y2, . . . , Yn) =  H(Y1) +  H(Y2|Y1) +  H(Y3|Y2, Y1) + ... H(Yn|Yn−1, . . . , Y1)

=< SUM H(Yi) = n*H(X) =  H(X1, X2, . . . , Xn)

- with equality if and only if the  Yi were independent:

                          H(Y1, Y2, . . . , Yn) < H(X1, X2, . . . , Xn)

Answer: The case with replacement has higher entropy

Answer:

1121.7 × 10³⁰ photons per second

Explanation:

Data provided in the question:

Power transmitted by the AM radio,P = 550 kW = 550 × 10³ W

Frequency of AM radio, f = 740 kHz = 740 × 10³ Hz

Now,

P = [tex]\frac{NE}{t}[/tex]

here,

N is the number of photons

t is the time

E = energy = hf

h = plank's constant = 6.626 × 10⁻³⁴ m² kg / s

Thus,

P = [tex]\frac{NE}{t}[/tex] = [tex]\frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}[/tex]          [t = 1 s for per second]

or

550 × 10³ = [tex]\frac{N\times(6.626\times10^{-34}\times740\times10^{3})}{1}[/tex]

or

550 = N × 4903.24 × 10⁻³⁴

or

N = 0.11217 × 10³⁴ = 1121.7 × 10³⁰ photons per second

The number of photons that are emitted by this AM radio transmitter is equal to [tex]1.12 \times 10^{33}\;photons.[/tex]

Given the following data:

Scientific data:

In order to determine the number of photons that are being emitted by this AM radio transmitter, we would solve for the quantity of energy it consumes by using Planck-Einstein's equation.

Mathematically, the Planck-Einstein relation is given by the formula:

[tex]E = hf[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]E = 6.626 \times 10^{-34}\times 740 \times 10^3\\\\E = 4.903 \times 10^{-28}\;Joules.[/tex]

For the number of photons:

[tex]n=\frac{Power}{Energy} \\\\n=\frac{550 \times 10^3}{4.903 \times 10^{-28}} \\\\n=1.12 \times 10^{33}\;photons.[/tex]

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Answer:

The PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.

Explanation:

From table 3.5 of Traffic Engineering by Mannering

R_v=835

R=835+(10ft/2)= 840 ft.

Now T is given as

T=R tan(Δ/2)

Here Δ is the central angle of curve given as 35°

So

T=R tan(Δ/2)

T=840 x tan(35/2)

T=840 x tan(17.5)

T=264.85

Now

STA PC=482+72-(2+64.85)=480+07.15

Also L is given as

L=(π/180)RΔ

Here R is the radius calculated as 840 ft, Δ is the angle given as 35°.

L=(π/180)RΔ

L=(π/180)x840 x35

L=512.87 ft

STA PT=480+07.15+5+12.87=485+20.02

Now Ms is the minimum distance which is given as

[tex]M_s=R_v(1-cos(\frac{90 \times SSD}{\pi Rv}))\\[/tex]

Here R_v is given as 835

SSD for 50 mi/hr is given as 425 ft from table 3.1 of Traffic Engineering by Mannering

So Ms is

[tex]M_s=R_v(1-cos(\frac{90 \times SSD}{\pi Rv}))\\M_s=835(1-cos(\frac{90 \times 425}{\pi 835}))\\M_s=26.92 ft[/tex]

Now for the clearance from the inside lane

Ms=Ms-lane length

Ms=26.92-5= 21.92 ft.

So the PT station is at 485+20.02 and 21.92 ft are to be cleared from the lane's shoulder to provide adequate stopping sight distance.

Answer:

a) the two-way concrete joist framing system

Explanation:

A waffle slab is also known as ribbed slab, it is a slab which as waffle like appearance with holes beneath. It is adopted in construction projects that has long length, length more than 12m. The waffle slab is rigid, therefore it is used in building that needs minimal vibration.

Answer:

Due to differences in the nature of the adhesive force and cohesive forces in the molecules of the individual substances and the glass tube.

Explanation:

To understand why this is so, a deep understanding of adhesive and cohesive force is required.

First, what is adhesive force?

Note: Adhesive and cohesive forces are best discussed under macroscopic level.

Adhesive force is the Intermolecular forces that exist between atoms of different molecules e g the forces explain when unlike charges stick together.

Cohesive force on the other hand is the Intermolecular force that exist between atoms of the same molecules e.g the force between hydrogen bonding.

Hence to explain the case scenario above, the adhesive force between the water molecules and the glass molecules is higher than the cohesive force between the water molecules. Hence the high rise.

For the case of the Hexane, the cohesive forces between the molecules hexane is far greater then the adhesive force between the glass molecules and the hexane molecules.

Answer:

View Image

Explanation:

You didn't provide me a picture of the opamp.

I'm gonna assume that this is an ideal opamp, therefore the input impedance can be assumed to be ∞ . This basically implies that...

Since no current is going into the + and - side of the opamp, then

i₁ = i₂

Since V₊ is connected to ground (0V) then V₋ must also be 0V.

V₊ = V₋  = 0

Use whatever method you want to solve for v_out and v_in then divide them. There's so many different ways of solving this circuit.

You didn't give me what the input voltage was so I can't give you the entire answer. I'll just give you the equations needed to plug in your values to get your answers.

Answer:

Detailed Answer is given below,

Explanation:

(a) Elastic strain

Elastic deformation is defined as the limit for the deformation values up to which the object will bounce and return to the original shape when the load is removed.

When the object is subjected to an external load, it deforms. If the load exceeds the load corresponding to the elastic limit of the object, then, after removing the load, the object cannot return to its original geometric specification.

(b) Plastic strain

Plastic Strain is a deformation that cannot be recovered after eliminating the deforming force. In other words, if the applied tension is greater than the elastic limit of the material, there will be a permanent deformation.

(c) Creep strain

The deformation of the material at a higher temperature is much more than at the normal temperature known as creep.

Subsequently, the deformation produced in heated material is called creep deformation. Creep deformation is a function of temperature.

(d) Tensile viscosity

Viscosity is a main parameter when measuring flux fluids, such as liquids, semi-solids, gases and even solids. Brookfield deals with liquids and semi-solids. Viscosity measurements are made together with the quality and efficiency of the product. Any person involved in the characterization of the flow, in research or development, quality control or transfer of fluids, at one time or another is involved with some kind of viscosity measurement.

(e) Recovery

Recovery is a process whereby deformed grains can reduce their stored energy by eliminating or reorganizing defects in their crystalline structure. These defects, mainly dislocations, are introduced by plastic deformation of the material and act to increase the elastic limit of a material.

(f) Relaxation

Stress relaxation occurs in polymers when they remain tense for long periods of time.

These alloys have very good resistance to stress relaxation and, therefore, are used as spring materials.

Stress relaxation is a gradual reduction of stress over time in constant tension.

Complete question:

The heat capacity of solid lead oxide is given by Cp,m=44.35+1.47×10⁻³T/K in units of J K−1 mol−1.

Calculate the change in enthalpy of 1.94 mol of PbO(s) if it is cooled from 732 K to 234 K at constant pressure.

Answer:

The change in enthalpy of PbO(s) is -39.488 x10³J

Explanation:

Given:

Initial temperature of PbO(s) (T₁) = 732 K

Final temperature of PbO(s) (T₂) = 234 K

[tex]\delta H = n\int\limits^{T_2}_{T_1} {C_p m} \, dT[/tex]

where;

Cp,m is the specific heat capacity of PbO(s)

[tex]\delta H = 1.94 molX\int\limits^{234}_{732} {[44.35 +1.47 X10^{-3}\frac{T}{K} ]} \,d (\frac{T}{K})[/tex]

[tex]\delta H = 1.94 molX {[44.35 (234-732) +1.47 X10^{-3}(\frac{234^2 -732^2}{2}) ]}[/tex]

     = 1.94mol [(-19957.5)+(-396.9)]

     = -38717.55 J -769.986J

     = -39487.536 J

ΔH = -39.488 x10³J

Therefore, the change in enthalpy of PbO(s) is -39.488 x10³J

Answer:

Power of the motor that kept the wheel rotating at 12 rev/hr despite friction is 342.79W.

Explanation:

Pls refer to the attached file. The explanation is long to pen down here.

Answer:

z=60.32°, i=0.32°, Beam Radiation = 1097.2 W/m²,  Id = 94.2 W/m², Ir=14.1W/m², total radiation = 1205.4 W/m², Local time=1:21PM

Explanation:

A. Zenith Angle:

As we know that,

Zenith angle=z=90⁰-α=L(latitude)=29.68⁰

Another way to do it is to find α first,

At solar time hour angle is 0⁰. So, solar altitude becomes equal to latitude which could be written as

sinα=cosL

α=sin⁻¹(cosL)=sin⁻¹(cos29.68⁰)=60.32°

B. Angle of incidence:

angle of incidence= cosi=sin(α+β)=sin(60.32°+32°)=sin92.32°

i=cos⁻¹(sin92.32°)=0.32°

C. Beam Radiation:

First we need to calculate extra terrestrial radiations

Iext.=1353[1+0.034cos(360n/365)]

where n=264

=1345 W/m²

Now,

Beam Radiation=CIext⁻ⁿ

where n=0.1/sin60.32°

Beam Radiation = 1097.2 W/m²

D. Diffude Radiation:

difuse radiation = Id = 0.0921ₙcos²(β/2)

where β=30°

Id = 94.2 W/m²

E. Reflected Radiations:

Ir=pIn(sinα+0.092)sin²(β/2)

= (0.2)(1097.1)(sin60.32+0.092)sin²(30/2)

= 14.1W/m²

F. Total Radiation:

total radiation = beam radiation + diffuse radiation + reflected raddiation

= 1205.4 W/m²

G. Local Time:

LST= ST-ET-(lₓ-l(local))4min/₀

     = 12:00-7.9min-(75°-82.27°)4min/₀

     =12:21PM

Local time

LDT=LST+=12:21+1:00=1:21PM

Answer:

Explanation:

A- Specific gravity and Absorption Test: Specific gravity is a measure of a material’s density as compared to the density of water at 73.4°F (23°C). Therefore, by definition, water at a temperature of 73.4°F (23°C) has a specific gravity of 1. Absorption is also determined by the same test procedure and it is a measure of the amount of water that an aggregate can absorb into its pore structure.

Specific gravity is used in a number of applications including Superpave mix design, deleterious particle identification and separation and material property change identification while

B- Soundness Test : This determines an aggregate's resistance to disintegration by weathering and in particular, freeze-thaw cycles. Aggregates that are durable (resistant to weathering) are less likely to degrade in the field and cause premature HMA pavement distress and potentially failure.It is used to identify the excess amount of lime in cement.

C - Sieve analysis Test: is a practice or procedure used to assess the particle size distribution (also called gradation) of a granular material by allowing the material to pass through a series of sieves of progressively smaller mesh size and weighing the amount of material that is stopped by each sieve as a fraction of the whole mass. This test is used to describe the properties of the aggregate and to see if it is appropriate for various civil engineering purposes such as selecting the appropriate aggregate for concrete mixes and asphalt mixes as well as sizing of water production well screens.

Answer:

The cathode ray oscilloscope (CRO) consists of a set of blocks. Those are vertical amplifier, delay line, trip circuit, time base generator, horizontal amplifier, cathode ray tube (CRT) and power supply. The CRO block diagram is shown in attached figure.

The function of each CRO block is mentioned below,

Vertical amplifier amplifies the input signal, which will be displayed on the CRT screen.

Delay line provides a certain amount of delay to the signal, which is obtained at the output of the vertical amplifier. This delayed signal is then applied to the CRT vertical deflection plates.

Trigger circuit produces a trigger signal to synchronize the horizontal and vertical deviations of the electron beam.

Time base generator produces a sawtooth signal, which is useful for horizontal deviation of the electron beam.

Horizontal amplifier amplifies the sawtooth signal and then connects it to the CRT horizontal deflection plates.

Power supply produces high and low voltages. The high negative voltage and the low positive voltage apply to CRT and other circuits respectively.

Cathode ray tube (CRT)

it is the main important block of CRO and consists mainly of four parts. Those are electronic guns, vertical deflection plates, horizontal deflection plates and fluorescent display.

The electron beam, which is produced by an electron gun, is deflected both vertically and horizontally by a pair of vertical deflection plates and a pair of horizontal deflection plates, respectively. Finally, the deflected beam will appear as a point on the fluorescent screen.

In this way, CRO will display the input signal applied on the CRT screen. So, we can analyze the signals in the time domain using CRO.

Explanation:

The oscilloscopes which is widely used for analysis purpose of circuits is divided into four main groups: the horizontal and vertical controls, the input controls and the activation controls.

Found in the front panel section marked Horizontal, the oscilloscope's horizontal controls allow users to adjust the horizontal scale of the screen. This section includes the control of the horizontal delay (displacement), as well as the control that indicates the time per division on the x-axis. The first control allows users to scan through a time range, while the latter allows users to approach a particular time range by decreasing the time per division.

Meanwhile, the oscilloscope's vertical controls are usually found in a section specifically marked as Vertical. The controls found in this section allow users to adjust the vertical appearance of the screen and include the control that indicates the number of volts per division on the axis and the grid of the screen. Also in this section is the control of the vertical displacement of the waveform, which translates the waveform up or down on the screen.

Signal activation helps provide a usable and stable display and allows users to synchronize the oscilloscope acquisition in the waveform of interest. The oscilloscope trigger controls allow users to choose the vertical trigger level, as well as the desired trigger capability. Common types of activation include fault activation, edge activation and pulse width activation.

Useful for identifying random errors or failures, the activation of faults allows users to fire at a pulse or event whose width is less than or greater than a specific period of time. This activation mode allows users to capture errors or technical problems that do not occur very frequently, which makes them very difficult to see.

The most famous trigger mode, edge tripping occurs when the voltage exceeds a set threshold value. This mode allows users to choose between shooting on a falling or rising edge.

Although pulse width activation is comparable to fault activation when users search for pulse width, it is, however, more general since it allows users to fire pulses of specified width. Users can also select the polarity of the pulses to be activated and set the horizontal position of the trigger. This allows users to see what happened during pre-shot or post-shot.

The input panels of an oscilloscope usually include two or four analog channels. They are usually numbered and have a button associated with each channel that allows users to activate and deactivate them. This section may also include a selection that allows users to specify the DC or AC coupling. Selecting the DC coupling implies that the entire signal will be input. The AC pairing, on the other hand, blocks the DC component and focuses the waveform around zero volts. Operators can also identify the probe impedance of the channels through a selection button. In adding, the input panels permit users to select the type of sampling to be used.

Answer:

attached below

Explanation:

Answer:

A) while (num >= 0)

Explanation:

To understand why we need to focus on the module and division operation inside the loop. num % 10 divide the number by ten and take its remainder to then add this remainder to sum, the important here is that we are adding up the number in reverse order and wee need to repeat this process until we get the first number (1%10 = 1), therefore, num need to be one to compute the last operation.

A) this is the correct option because num = 1 > 0 and the last operation will be performed, and after the last operation, num = 1 will be divided by 10 resulting in 0 and 0 is not greater than 0, therefore, the cycle end and the result will be printed.

B) This can not be the option because this way the program will never ends -> 0%10 = 0 and num = 0/10 = 0

C) This can not be the option because num = 1 > 1 will produce an early end of the loop printing an incomplete result

D) The same problem than C

E) There is a point, before the operations finish, where sum > num, this will produce an early end of the loop, printing an incomplete result

Loops are program statements that are used to carry out repetitive and iterative operations

The missing loop header is (a) while (num > 0)

To calculate the sum of the digits of variable num, the following must be set to be true

The loop header must be set to keep repeating the loop operations as long as the value of variable num is more than 0

To achieve this, we make use of while loop,

And the loop condition (as described above) would be num > 0

Hence, the true option is (a) while (num > 0)

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Answer:

The motor inertia is 7.958 X 10⁻² kg.m²

Explanation:

To determine the motor inertia, the following formula applies.

Neglecting the damping effect,

[tex]T = \frac{J}{9.55}.\frac{\delta n}{\delta T}[/tex]

Where;

T is the constant torque applied to the motor = 5Nm

J is the motor inertia = ?

δn is the change in angular speed of the motor = 1800 r/min

δT is change in time of the unloaded motor from rest = 3 sec

[tex]J = \frac{9.55* \delta T* T}{\delta n}[/tex]

[tex]J = \frac{9.55* 3* 5}{1800}[/tex] = 0.07958 kg.m² = 7.958 X 10⁻² kg.m²

Therefore, the motor inertia is 7.958 X 10⁻² kg.m²

Answer:

Interface temperature = -7.27C

Rate of heat transfer = 10.91W/m2

Explanation:

The steps and the application of fourier's law of heat conduction is as shown in the attached file.

For the given composite wall, the temperature at the interface between the two layers is 17.275 °C and the rate of heat transfer through the wall is -0.0109 W/m².

To determine the temperature at the interface between the two layers, we can use the formula for 1-D steady-state heat conduction:

q = (T2 - T1) / ((L1 / k1) + (L2 / k2))

where q is the heat transfer rate per unit area, T1 and T2 are the temperatures of the inner and outer surfaces, L1 and L2 are the thicknesses of the insulation and siding layers, and k1 and k2 are the thermal conductivities of the insulation and siding layers, respectively.

Plugging in the given values:

q = (-10 - 20) / ((125 / 0.05) + (25 / 0.10)) = -30 / (2500 + 250) = -30 / 2750 = -0.0109 W/m².

The negative sign indicates heat transfer from the siding to the insulation.

To find the temperature at the interface, we can use the formula:

T_interface = T1 + (q * (L1 / k1))

Plugging in the values:

T_interface = 20 + (-0.0109 * (125 / 0.05)) = 20 - 2.725 = 17.275 °C.

Therefore, the temperature at the interface between the two layers is 17.275 °C and the rate of heat transfer through the wall is -0.0109 W/m².

Answer:

Explanation:

The concept of differential in rate of change is applied to solve the problem.

First is to compare the equation given V=ai+bxj  , where u = a and v = bx.

the detailed steps and appropriate integration and substitution is as shown in the attached file.

The streamline equation and the coordinates of the point are

A) [tex]x^2-4y+16[/tex]

B) (4, 12)

C) (3, 11.25)

Given that, [tex]V=ai+bxj[/tex], [tex]a=\frac{2m}{s}[/tex] and [tex]b=1s^{-1}[/tex]

Putting the value, we get

[tex]V=2i+1.xj[/tex]

From this equation, we get

[tex]u=2\frac{m}{s}[/tex] and [tex]v=x\frac{m}{s}[/tex]

Obtain the equation for the streamline passing through the point (2,5), streamline equation

[tex]\frac{dx}{u}=\frac{dy}{v}[/tex]

Putting the value, we get

[tex]\frac{dx}{2}=\frac{dy}{x}[/tex]

rearrange it and do integration:

[tex]\int xdx=\int 2dy[/tex]

[tex]\frac{x^2}{2}=2y+c -------(i)[/tex]

C is the integration constant.

[tex]\frac{2^2}{2}=(2\times5)+C[/tex]

[tex]C=-8[/tex]

Put the value of C in equation (i), we get

[tex]\frac{x^2}{2}=2y-8[/tex]

[tex]y=\frac{x^2}{4}+4[/tex]

[tex]x^2-4y+16=0[/tex]

b) [tex]u=2\frac{m}{s}[/tex]

[tex]\frac{dx}{dt}=2[/tex]

Rearrange and integrate

[tex]\int dx=\int 2dt[/tex]

[tex]x=2t+C_1[/tex]

Put x=0 and t=0

[tex]0=0+C_1[/tex]

[tex]C_1=0[/tex]

Put the value of constant we get

[tex]x=2t[/tex]

at [tex]t=2s[/tex]

[tex]x=2\times2= 4 --------(2)[/tex]

Now, [tex]v=x \frac{m}{s}[/tex]

[tex]\frac{dy}{dt}=x[/tex]

Rearrange and integrate:

[tex]\int dy=\int xdt[/tex]

[tex]y=xt+C_2[/tex]

Putting the value of x, y and t we get the value of constant.

[tex]4(0\times0)+C_2[/tex]

[tex]C_2=4[/tex]

Putting the value of constant we get

[tex]y=xt+4[/tex]

Put the value of [tex]t=2[/tex] and [tex]x=4[/tex] we get

[tex]y=(4\times2)+4[/tex]

[tex]= 8+4[/tex]

[tex]y= 12 -------(3)[/tex]

From (2) and (3)

we get the coordinates of the particle that passed through point [tex](0,4)[/tex] at [tex]t=0[/tex].

After [tex]t=2s[/tex] as [tex](4, 12)[/tex]

c) first we have to find the integration constant through initial condition then put the value of time to get the coordinates.

From above we have:

(1) [tex]x=2t+C_1[/tex] and

(2) [tex]y=xt+C_2[/tex]

Given [tex]t=2s[/tex] and [tex]P(1, 4.25)[/tex]

Putting the value in first equation, we get

[tex]1=(2\times2)+C_1[/tex]

[tex]C_1 =-3[/tex]

Put the value of time [tex]t=3s[/tex] we get

[tex]x=(2\times3)-3=3 ---------(a)[/tex]

Now, putting the value in second equation  we get

[tex]4.25=(1\times2)+C_2[/tex]

[tex]C_2 =2.25[/tex]

Then [tex]y=xt+2.25[/tex]

Put the value of time [tex]t=3s[/tex] we get

[tex]y=(3\times3)+2.25=11.25 -------(b)[/tex]

From (a) and (b) we get the coordinate of the point after [tex]t=3s[/tex] is [tex](3, 11.25)[/tex].

Therefore, streamline equation and the coordinates of the point are

A) [tex]x^2-4y+16[/tex]

B) (4, 12)

C) (3, 11.25)

Learn more about the streamline passing through the point here:

https://brainly.com/question/34111016.

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Properties of Carpenter's hammer possess

Explanation:

1.The head of a carpenter's hammer should possess the impact resistance, so that the chips do not peel off the striking face while working.

2.The hammer head should also be very hard, so that it does not deform while driving or eradicate any nails in wood.

3.Carpenter's hammer is used to impact smaller areas of an object.It can drive nails in the wood,can crush  the rock and shape the metal.It is not suitable for heavy work.

How hammer head is manufactured :

1.Hammer head is produced by metal forging process.

2.In this process metal is heated and this molten metal is placed in the cavities said to be dies.

3.One die is fixed and another die is movable.Ram forces the two dies under the forces which gives the metal desired shape.

4.The third process is repeated for several times.

Answer:

A. 36 minutes

B. 120 minutes

C.

i. 144 minutes

ii. 984 minutes

D. Car 1 is 1.67km ahead of Cat 2 when Car 2 passed the toll B.

E. 98.33km

Explanation

A.

Given

dAb = 10km

dBc = 10km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 0.2 hours

= 24 minutes + 0.2 * 60 minutes

= 24 minutes + 12 minutes

= 36 minutes

B.

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 10km + 10km = 20km

So, Propagation delay = 20km/100km/hr

Propagation delay = 0.2 hour

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 0.2 hours

= 3 minutes + 0.2 * 60 minutes

= 3 minutes + 12 minutes

= 15 minutes

Total End delay for 8 cars = 8 * 15 = 120 minutes

C.

Given

dAb = 100km

dBc = 100km

Propagation Speed = 100km/hr

Delay time = 1 minute

Numbers of cars = 8

Number of tolls = 3

i. Cars travelling together

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

Translation delay = delay time* numbers of tolls * numbers of cars

Transitional delay = 1 * 3 * 8

Transitional delay = 24 minutes

Total End delay = 24 minutes + 2 hours

= 24 minutes + 2 * 60 minutes

= 24 minutes + 120 minutes

= 144 minutes

ii. Cars travelling separately

Total End to End delay = Propagation delay + Transition delay

Calculating Propagation Delay

Propagation delay = Total Distance/Propagation speed

Total distance = 100km + 100km = 200km

So, Propagation delay = 200km/100km/hr

Propagation delay = 2 hours

Translation delay = delay time* numbers of tolls ------ Cars traveling separately

Transitional delay = 1 * 3

Transitional delay = 3 minutes

Total End delay for one car = 3 minutes + 2 hours

= 3 minutes + 2 * 60 minutes

= 3 minutes + 120 minutes

= 123 minutes

Total End delay for 8 cars = 8 * 123 = 984 minutes

D.

Distance = 100km

Time = 1 min/car

Car 1 is 1 minute ahead of car 2 --- at toll A and B

If car 1 leaves toll B after 10 minutes then cat 2 leaves after 11 minutes

Time delay = 11 - 10 = 1 minute

Distance = time * speed

= 1 minute * 100km/hr

= 1 hr/60 * 100 km/hr

= 100/60

= 1.67km

E.

Given

Distance = 100km

Distance behind = 1.67

Maximum value of dBc = 100km - 1.67km = 98.33km

The maximum distance that can be reached is 98.33km

Answer: First pond= 8.0 mg/L

second pond = 4.2 mg/L

Explanation:

Cn/Co = [ 1/ [ 1 + (k*t/n)]]

for the first pond

where Co into the first pond is 20mg/L,

Cn = 20*[ 1/ [ 1+ ((1 x 5)/0.3)]]

Cn = 8 mg/L

into the second pond we calculate the BOD leaving the pond of volume 3million liters

we have Cn = 4.2 mg/L

Answer:

The maximum temperature of the cycle is 1065⁰R

Explanation:

The maximum temperature in degree Rankin can be obtained using the formula below;

[tex]\frac{T_2}{T_1} =[\frac{V_1}{V_2}]^{1.4-1}[/tex]

Where;

T₂ is the  maximum temperature of the cycle

T₁ is the initial temperature of the cycle = 520 deg R = 520 ⁰R

V₁/V₂ is the compression ratio = 6

[tex]T_2 = T_1(\frac{V_1}{V_2})^{0.4}[/tex]

[tex]T_2=T_1(6)^{0.4}[/tex]

[tex]T_2=520^0R(6)^{0.4}[/tex]

T₂ = 1064.96 ⁰R

Therefore, the maximum temperature of the cycle is 1065⁰R

Answer:

The question is incomplete, below is the complete question "The electric potential difference between the ground and a cloud in a particular thunderstorm is [tex]3.0*10^{9}V[/tex]. What is the magnitude of the change in the electric potential energy of an electron that moves between the ground and the cloud?"

Answer:

[tex]U=3.0*10^{9}eV[/tex]

Explanation:

data given,

Potential difference,V=3.0*10^9V

charge on an electron, q=1e.

Recall that the relationship between potential difference (v), charge(Q) and the potential energy(U) is expressed as

[tex]U=qV[/tex]

from the question, we asked to determine potential energy given the charge and the potential difference.

Hence if we substitute values into the equation, we arrive at

[tex]U=qV\\U=3.0*10^{9}*1e\\U=3.0*10^{9}eV[/tex]

Hence the magnitude of the change in the electric potential energy of an electron that moves between the ground and the cloud is [tex]3.0*10^{9}eV[/tex]

Answer:

[tex]\eta=0.303[/tex]

Explanation:

Given that

Temperature of the hot reservoir ,T₁ = 970 K

Temperature of the cold reservoir ,T₂ = 480 K

We know that only Carnot engine is the ideal engine which gives us the maximum power.The efficiency of Carnot engine is given as

[tex]\eta_{max}=1-\dfrac{T_2}{T_1}[/tex]

[tex]\eta_{max}=1-\dfrac{480}{970}[/tex]

[tex]\eta_{max}=0.505[/tex]

Therefore the efficiency of the given engine will be

[tex]\eta=\dfrac{3}{5}\eta_{max}[/tex]

[tex]\eta=\dfrac{3}{5}\times 0.505[/tex]

[tex]\eta=0.303[/tex]

Answer:

not provide the sprinklers, then the type of construction will be Type II B under IBC

Explanation:

given data

3 story office building = 20,000 square feet per floor

solution

we know when sprinkler is provide in high rise building to resist fire

and it provide in building as floor area exceed allowable permissible area of  building as IBC

so IBC for Type II B allowable area =  19000 square feet per floor

and type III B allowable area =  23000 square feet per floor

so when we design the building by type III B construction, the sprinklers require to provide

but not provide the sprinklers, then the type of construction will be Type II B under IBC

Answer: BULK RELATIVE DENSITY.

WHY?

BULK RELATIVE DENSITY GIVES A BETTER UNDERSTANDING OF THE QUALITY OF THE MATERIAL.

Explanation:Bulk relative density is a type of Specific gravity which is often used in determining the volume occupied by the aggregates in various mixtures containing aggregate including Portland cement concrete, bituminous concrete etc this mixtures are proportioned based on an absolute volume basis. Bulk relative density is considered because of its ability to give a better understanding of the materials which makes up the mixture.

Answer:

The answers to the question are

Explanation:

The number of employees in the firm = 62

Number of first-aid cases = 7

Number of medical treatment injuries = 3

Injury resulting restricted work activity = 1

Illness resulting in one week loss work day = 1

Illness resulting in loss of 6 weeks of work =1

Incident resulting in fatality = 1

Total hours worked = 40×62×50  = 124000 Hrs

Where 200,000 is the number of hours worked by Full time employees numbering 100 that  work for 40 Hours a week for 50 weeks in a year

Number of recordable incident =  Those incident that results in lost work days, death, restricted ability to work, or transfer to another task or more severe injury treatment beyond first aid

Therefore number recordabe incident = 7

Recordable Incident Rate is calculated by

IR = (Number of recordable incident Cases X 200,000)÷(Number of Employee labor hours worked)

= (7×200000)/124000= 11.29

The Lost Workday Rate is given by

Lost Workday Rate = (Total number of days lost to injury or illness)÷( Cumulative number of hours employees worked) × 200000

Total number of days lost to injury or illness = 42 days

Lost Workday Rate = (42/124000) × 200000 = 67.74

LWDI = (Number of incident resulting in lost workdays and restricted activity) ×200000 /(Total number of hours worked by all employees in one year)

LWDI = LWD×200000/ EH = 4×200000/124000 = 6.45

LWDI  = 6.45