A motorcycle starts from rest with an initial acceleration of 3 m/s^2, and the acceleration then changes with the distance s as shown. Determine the velocity v of the motorcycle when s=200 m. At this point, also determine the value of the derivative dv/ds.

Answer:

(a)This portion of the problem asks that we calculate, for a Pb-Mg alloy, the mass of lead in 7.5 kg of thesolidphase at 300C just below the solubility limit.From Figure 9.20, the solubility limit for thephase at

Explanation:

Answer:

stagflation

Explanation:

it was used in the article

A word which was first used during the 1970s economic crisis is stagflation.

In the 1970s, an energy crisis took place in the United States of America due to the oil embargo that was imposed on it by OPEC. This oil embargo was imposed on the United States of America by the Organization of Petroleum Exporting Countries (OPEC) in 1973 because of its role in the Arab-Israeli War.

Consequently, the economy of the United States of America experienced stagflation in the following ways:

Read more on stagflation here: https://brainly.com/question/25505087

Answer:

0.21

Explanation:

This would have been a fairly easy one, except for that the first part of the question is missing, and as such, I'd assume a value.

We need to use chord, so, I'm assuming the length of the chord to be 0.045 m

The Area is given by the formula

Area = span * chord

Area = 0.245 * 0.045

Area = 0.011 m²

This area gotten, is what we then divide the glider weight by to get our answer.

Lee = area / weight

Lee = 0.011 / 0.0523

Lee = 0.21

Therefore, using the values of the chord I'd assumed, the Lee of the same wing is 0.21

Answer: 0.2108

Explanation:got it correct

Answer:

In the case of the production Line, we know that,

No of gasoline cans = 5

probability that 1st can is out of tolerance = 0.03

probability that 2nd can is out of tolerance = 0.03

.

.

probability that the 5th can is out of tolerance = 0.03

Therefore,

probability of 1st can out of tolerance + probability of 1st can not out of tolerance = 1

Probability of 1st can not out of tolerance = 1 -- 0.03 = 0.97

probability of 2nd can not out of tolerance = 0.97

.

.

probability of 5th can not out of tolerance = 0.97

Question A:

Probability that they are all out of tolerance

= P(1st can out of tolerance) * P(2nd can out of tolerance) * P(3rd can out of tolerance) * P(4th can out of tolerance) * P(5th can out of tolerance)  

= (0.03 ) * (0.03) * (0.03) * (0.03) * (0.03) =  2.43 E⁻⁸   (2.43 ˣ 10⁻⁸)

Question B:

Probability that exactly two are out of tolerance

= P(1st can is out of tolerance) * P(2nd can is out of tolerance) * P(3rd can is not out of tolerance) * P(4th can is not out of tolerance) * P(5th can is not out of tolerance)

= (0.03) * (0.03) * (0.97) * (0.97) * (0.97) = 0.0008214057

Explanation:

Answer:

The ammonia is still in the gas phase

Explanation:

Given that 1 bar is approximately = 1 atm

From;

PV=nRT

P= 310 atm

V= the unknown

n= 1

R = 0.082atm LK-1mol-1

T = 92oC + 273 = 365 K

V= nRT/P

V= 1 * 0.082 * 365/310

V = 0.0965 L = 96.5 mL

molar mass of NH3 = 17 g/mol

Molar density of NH3 = 17g/ 96.5mL = 0.176g/mL

The ammonia is still in the gas phase

Answer:

the answer would be d its d

Answer:

Pretty sure the answer is "C"

Explanation:

"The media doesn't affect me at all because I'm smart enough to know the difference between right and wrong."

Answer:

a) Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

b) principal stress :

б1 = ( бx + бy ) / 2  - [tex]\sqrt{}[/tex]( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

Explanation:

Combined normal stress and shear stress  sketches attached below

The terms in the sketch are :

бx = tensile stress in x direction

бy =  tensile stress in y direction

Txy = y component of shear stress acting on the perpendicular plane to x axis

бn = Normal stress acting on the inclined plane EF

Tn = shear stress acting on the inclined plane EF

A) Normal and shear stresses on inclined plane

Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

B) principal and maximum shear stresses

principal stress :

б1 = ( бx + бy ) / 2  - [tex]\sqrt{}[/tex]( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

Answer: attached below is the power triangles

7.13589 MVAR

Explanation:

Power ( P1 ) = 10 MW

power factor ( cos ∅ ) = 0.6 lagging

New power factor = 0.85

Calculate the reactive power of a capacitor to be connected in parallel

Cos ∅ = 0.6

therefore ∅ = 53.13°

S = P1 / cos ∅ = 16.67 MVA

Q1 = S ( sin ∅ ) = 13.33 MVAR  ( reactive power before capacitor was connected in parallel )

note : the connection of a capacitor in parallel will cause a change in power factor and reactive power while the active power will be unchanged i.e. p1 = p2

cos ∅2 = 0.85 ( new power factor )

hence ∅2 =  31.78°

Qsh ( reactive power when power factor is raised to 0.85 )

= P1 ( tan∅1 - tan∅2 )

= 10 ( 1.333 - 0.6197 )

= 7.13589 MVAR

Answer:

284.4 m/s

Explanation:

At the inlet of the nozzle P =1 atm.

Temperature T = 250° C

Velocity of the steam at the inlet V_1 = 30 m/s

Change in enthalpy Δh = 40 KJ/kg

let V_2 be the final velocity

then

[tex]V_2 =\sqrt{2\Delta h+V_1^2} \\=\sqrt{2\times40+30^2}\\= 284.4 m/s[/tex]

hooooooooooooooooooooooooooooooooooooooooooooooooooooooe    

Answer:

0.2

Explanation:

Since the span and chord of the rectangular wing is missing, due to it being from the other question, permit me to improvise, or assume them. While you go ahead and substitute the ones from your question to it, as it's both basically the same method.

Let the span of the rectangular wing be 0.225 m

Let the chord of the rectangular wing be 0.045 m.

Then, the area of any rectangular chord is

A = chord * span

A = 0.045 * 0.225

A = 0.010 m²

And using the weight of the glider given to us from the question, we can find the LER for the wing.

LER = Area / weight.

LER = 0.010 / 0.05

LER = 0.2.

Therefore, using the values of the rectangular wing I adopted, and the weight of the glider given, we can see that the LER of the glider is 0.2

Please mark brainliest...

Answer: 0.2025

Explanation: I got it correct

Answer:

The answer for the question is true

Explanation:

If you get a virus or get hacked you will still have it saved

Answer:

The hardenability increases with increasing austenite grain size, because the grain boundary area is decreasing. This means that the sites for the nucleation of ferrite and pearlite are being reduced in number, with the result that these transformations are slowed down, and the hardenability is therefore increased.

Answer:

β = [tex]\frac{c}{\sqrt{km} }[/tex] =  0.7071 ≈ 1 ( damping condition )

closed-form expression for the response is attached below

Explanation:

Given :  x + 2x + 2x = 0   for Xo = 0 mm and Vo = 1 mm/s

computing a solution :

M = 1,

c = 2,

k = 2,

Wn = [tex]\sqrt{\frac{k}{m} }[/tex]  = [tex]\sqrt{2}[/tex]  

next we determine the damping condition using the damping formula

β = [tex]\frac{c}{\sqrt{km} }[/tex] =  0.7071 ≈ 1

from the condition above it can be said that the damping condition indicates underdamping

attached below is the closed form expression for the response

Answer:the critical radius for the homogeneous nucleating ice is  0.986 nm

Explanation:

Using the formulae below to calculate the  critical radius for homogeneous nucleation,

We Have that

Critical radius ( r *) =  [2γ Tm/ΔHf]{ 1/ Tm- T]

where γ = surface free energy =25 × 10^-3 J/m^2

Tm= solidification temperature at equilibrium =273K

Hf= latent heat of fusion = -3.1 × 10^8 J/m3

Temperature , T = -44.6°C

Critical radius ( r *) = (2 X 25 × 10^-3 J/m^2 x 273)/ (-3.1 × 10^8 J/m3 ) X ( 1/ 273 - ( -44.6 +273)

4.40x 10^-8X ( 1/273-228.4)

4.40x 10^-8  X 1/44.6

4.40x 10^-8  x 0.0224

9.86x 10 ^-10m =0.986 x 10 ^-9m = 0.986nm

Answer: the mole fraction of D in the mixture is 0.2222

Explanation:

Given that;

mixture analysis shows 2 kmol A, 1 kmol B, 4 kmol C and some unknown kmol of D was present.

4A+3B→2C+D

As from reaction stoichiometry, for every 2 kmol of C produced, kmol of D produced = 1 kmol

so, for 4 kmol C, kmol of D produced = 4/2 × 1 kmol = 2 kmol

Now our mixture has 2 kmol A, 1 kmol B, 4 kmol C and also 2 kmol of D

so, total moles in mixture, we have (2 + 1 + 4 + 2) kmol = 9 kmol

mole fraction of D in mixture  will be;

( Kmol of D) / (total moles in mixture) = 2 / 9 = 0.2222

Therefore the mole fraction of D in the mixture is 0.2222

Answer:people tend to do this when they are in a different environment they lose something or just have something going on in their life

Explanation:

Answer: Option D) 298 g/mol  is the correct answer

Explanation:

Given that;

Mass of sample m = 13.7 g

pressure P = 2.01 atm

Volume V = 0.750 L

Temperature T = 399 K

Now taking a look at the ideal gas equation

PV = nRT

we solve for n

n = PV/RT

now we substitute

n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K )

= 1.5075 / 32.7579

= 0.04601 mol

we know that

molar mass of the compound = mass / moles

so

Molar Mass = 13.7 g / 0.04601 mol

= 297.7 g/mol  ≈ 298 g/mol

Therefore Option D) 298 g/mol  is the correct answer

Answer:

the motor input power is 19.42 KW

the Reactive power is 17.65 KVAR

Current is 31.56 A

Explanation:

Given that;

V = 480V

h.p = 25 hp

p.f = 0.74 lagging

n_motor = 96%

so output = 25hp

and we know that;

1hp = 746 watt

watt = hp × 1hp

so output in watt = 25 × 746 = 18650 Watt = 18.65 KW

n_motor = (output / input) × 100

96 = 1865 / Input

96Input = 1865

Input = 1865 / 96

Input = 19.42 KW

Therefore the motor input power is 19.42 KW

P = √( 3 × V × I × cos∅)

19.42 = √( 3 ×480 × I × 0.74)

I = 31.56 A

Therefore Current is 31.56 A

Q = √( 3 × V × I × sin∅)

we know that

cos∅ = 0.74

so ∅ = cos⁻¹(0.74) = 42.26

so we substitute

Q = √( 3 × 480 × 31.56 × sin(42.26))

 = 17.65 KVAR

Therefore the Reactive power is 17.65 KVAR

Answer:

Building technology is building technology such as coding an app or a website

Building engineering is making computers or cars or phones

Explanation:

Explanation:

D. B. C. A. E. Is this a good idea

Answer:

There are several similarities between the nucleus and a liquid droplet.

Explanation:

A droplet of liquid simply is is very small or tiny drop of liquid. It is also considered as a tiny column of liquid that is surrounded by surfaces that have zero shear stress.

A nucleus on the other hand is an assembly between protons and neutrons. The latter is electrically charged whilst the former is positively charged. The number of protons present in an element is very crucial to the qualities of an element.

The main similarities between a nucleus and a liquid droplet are:

1. a nucleus consists of a large amount of neutrons and protons in the same volume as would a liquid which contains large numbers of molecules in the same volume;

2. both the nucleus and the droplet are similar for their homogeneity in electric charge and density;

3. the molecules exert the same amount for forces towards one another as would the nuclear forces in the nucleons.

4. both of them cannot be compressed

5. both molecules and nucleus are can be subject to nuclear fission which simply mean the breaking apart into smaller units (in the case of the nucleus) or the breaking apart into smaller droplets in the case of the liquid molecule.

6. There are two types of phenomena which occurs in both the liquid droplet and the nucleus which are similar to one another. They are:

Evaporation (in the case of the liquid molecule) and reaction emission (in the case of the nucleus). In evaporation, particles are lost, in Atomic transmutation, particles are lost as well.

B)  the forces which determine the stability of droplets are surface tension and gravitation. The smaller the area, the stronger the surface tension available to keep the drops from going out of shape.

Cheers

Answer:

A. 7199.55 volts

B. 120A

Explanation:

In this question we have the

line voltage = VLL = 12470volts

Phase current = Iph = 120 amps

A.)

We are to calculate the line-to-neutral/phase voltage here

VLL = √3VL-N

VL-N = VLL/√3

VL-N = 12470/√3

This gives a line to neutral phase/voltage of 7199.55 volts.

B.

We are to calculate the line current here:

In this connection, the line current and the phase current are equal

ILL = Iph = 120A

Answer:

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

Explanation:

Given that;

h_lam(x)= 1.74 W/m^1.5. Kx^-0.5

h_turb(x)= 3.98 W/m^1.8 Kx^-0.2

conditions for plates of length L = 0.1 m and 1 m

Now

Average heat transfer coefficient is expressed as;

h⁻ = 1/L ₀∫^L hxdx

so for Laminar flow

h_lam(x)= 1.74 . Kx^-0.5  W/m^1.5

from the expression

h⁻_lam = 1/L ₀∫^L 1.74 . Kx^-0.5   dx

= 1.74k / L { [x^(-0.5+1)] / [-0.5 + 1 ]}₀^L

= 1.74k/L = [ (x^0.5)/0.5)]⁰^L

= 1.74K × L^0.5 / L × 0.5

h⁻_lam= 3.48KL^-0.5

For turbulent flow

h_turb(x)= 3.98. Kx^-0.2 W/m^1.8

form the expression

1/L ₀∫^L 3.98 . Kx^-0.2   dx

= 3.98k / L { [x^(-0.2+1)] / [-0.2 + 1 ]}₀^L

= (3.98K/L) × (L^0.8 / 0.8)

h⁻_turb = 4.975KL^-0.2

Now at L = 0.1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(0.1)^-0.5  W/m^1.5

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(0.1)^-0.2

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48KL^-0.5  =  3.48K(1)^-0.5  W/m^1.5

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975KL^-0.2 = 4.975K(1)^-0.2

h⁻_turb = 4.975K   W/m^1.8

Therefore

At L = 0.1 m

h⁻_lam = 11.004K   W/m^1.5

h⁻_turb = 7.8848K   W/m^1.8

At L = 1 m

h⁻_lam = 3.48K   W/m^1.5

h⁻_turb = 4.975K   W/m^1.8

Answer: the air flow rate a is 90 kg/sec; Option b) 90 kg/sec is the correct answer

Explanation:

Given that;

product of combustor flow rate m = 100 kg/s

air-fuel = 9

Airflow rate = ?

⇒We know that in the combustor, air fuel are mixed and then ignited,

⇒air fuel products are exited at the combustor

let air and fuel be a and b respectively

⇒ a + b = 100 kg/sec ----- let this be equation 1

now

⇒ air / fuel = 9

a / b = 9

a = 9b -----------let this be equation 2

now input a = 9b in equation 1

9b + b = 100 kg/sec

10b = 100 kg/sec

b = 10 kg/sec

we know that

a = 9b

so a = 9 × 10 = 90 kg/sec

Therefore the air flow rate a is 90 kg/sec

Answer: 1.137*10^7 Btu/h.

Explanation:

Given data:

Efficiency of the plant = 4.5percent

Net power output of the plant = 150kw

Solution:

The required collection rate

QH = W/n

= 150/0.045 * 0.94782/ 1 /60 */60 Btu/h.

= 3333.333 *3412.152Btu/h.

= 11373840 Btu/h

= 1.137*10^7 Btu/h.

Answer:Total Voltage = 14V

Explanation: it is possible that a circuit  can contain more than one source of electromotive force which can cause flow of current in the same or opposite direction . When the  connection to  voltage sources  allows for current  from the voltage sources to flow in  same direction,it is termed  Series aiding  Thus, the  Total/effective voltage in a series aiding circuit is  computed as the sum of series aiding voltages .

 Here we have the series aiding voltages to be 5V and 9V ,

therefore,

Total Voltage = 5V + 9V

= 14V

Answer:

155 KJ

Explanation:

The total enthalpy is given by

ΔH=ΔU + PV

Where;

ΔH = enthalpy

ΔU = internal energy = 25000 kJ/kg/ 5 kg = 5000 KJ

P = 150 kPa = 150,000 Pa

V =  1 m3

ΔH=  5000 + (150,000 * 1)

ΔH=  155 KJ