What disadvantages does traditional high-speed internet have compared with space-based internet?( choose all that apply)
A. Traditional high-speed internet is often not available in rural areas.
B.Traditional high-speed internet can be delayed if signals have to travel long distances.
C.Traditional high-speed internet causes light pollution.
D.Traditional high-speed internet increases space debris.

The capacitive reactance of a DC series circuit increases when its capacitance decreases and vice-versa.

A DC series circuit can be defined as a type of circuit in which all of its resistive components are connected end to end, so as to form a single path for the flow of current.

This ultimately implies that, the same amount of current flows through a direct current (DC) series circuit.

Mathematically, the capacitive reactance of a DC series circuit is given by this formula:

[tex]X_C = \frac{1}{2\pi fC}[/tex]

Where:

From the above formula, we can deduce that the capacitive reactance of a DC series circuit is inversely proportional to both frequency and capacitance. Thus, the capacitive reactance of a DC series circuit increases when its capacitance decreases and vice-versa.

In conclusion, a capacitor would influence a simple DC series circuit by blocking the flow of direct current (DC) through it.

Read more on capacitance here: https://brainly.com/question/22989451

Answer:

The timing light is connected to the ignition circuit and used to illuminate the timing marks on the engine's crankshaft pulley or flywheel, with the engine running. The apparent position of the marks, frozen by the stroboscopic effect, indicates the current timing of the spark in relation to piston position.

Explanation:

:)

Answer: See explanation

Explanation:

Emerging technologies are simply referred to as the technical innovations which are done within a particular sector and typically brings about growth to such field or sector.

Some of the new and emerging technologies that are used in the electronics and telecommunications industry include:

1. Digital scent technology - This is a technology that is used for the sensing and the transmission of digital media such as video games, music etc that are event enabled.

2. Electronic nose - This refers to an electronic sensing device which is used in the detection of flavors.

3. Ambient intelligence - It is an electronic environments which shows sensitivity when there are people.

While changing a soil's basic texture is very difficult, you can improve its structure–making clay more porous, sand more water retentive–by adding amendments. The best amendment for soil of any texture is organic matter, the decaying remains of plants and animals.

Solution :

The correct size of the arc of a welding process depends upon the application and the electrode. As a rule, the arc length should not be more than a diameter of the core of the electrode.

As for the electrode of diameter size of 5/32" or 4 mm, the arc length should be more than its core diameter. Also for 5/32 " diameter electrode, the welding time for the one electrode must be one minute as well as the length of the weld be the same as the length of the electrode consumed.  

Answer:

Explanation:

, integrated circuit packaging is the final stage of semiconductor device fabrication, in which the block of semiconductor material is encapsulated in a supporting case that prevents physical damage and corrosion.

This question is incomplete, the complete question is;

Given the complex numbers A₁ = 6∠30 and A₂ = 4 + j5;

(a) convert A₁ to rectangular form

(b) convert A₂ to polar and exponential form

(c) calculate A₃ = (A₁ + A₂), giving your answer in polar form

(d) calculate A₄ = A₁A₂, giving your answer in rectangular form

(e) calculate A₅ = A₁/([tex]A^{*}[/tex]₂), giving your answer in exponential form.

Answer:

a) A₁ in rectangular form is 5.196 + j3

b) value of A₃  in polar form is 12.19∠41.02°

The polar form of A₂ is 6.403 ∠51.34°, exponential form of A₂ = 6.403[tex]e^{j51.34 }[/tex]

c) value of A₃  in polar form is 12.19∠41.02°

d) A₄ in rectangular form is 5.784 + j37.98

e) A₅ in exponential form is 0.937[tex]e^{j81.34 }[/tex]

Explanation:

Given data in the question;

a) A₁ = 6∠30

we convert A₁ to rectangular form

so

A₁ = 6(cos30° + jsin30°)

= 6cos30° + j6cos30°

= (6 × 0.866) + ( j × 6 × 0.5)

A₁  =  5.196 + j3

Therefore, A₁ in rectangular form is 5.196 + j3

b) A₂ = 4 + j5

we convert to polar and exponential form;

first we convert to polar form

A₂ = √((4)² + (5)²) ∠tan⁻¹( [tex]\frac{5}{4}[/tex] )

= √(16 + 25) ∠tan⁻¹( 1.25 )

= √41 ∠ 51.34°

A₂ = 6.403 ∠51.34°

The polar form of A₂ is 6.403 ∠51.34°

next we convert to exponential form;

A∠β can be written as A[tex]e^{j\beta }[/tex]

so, A₂  in exponential form will be;

A₂ = 6.403[tex]e^{j51.34 }[/tex]

exponential form of A₂ = 6.403[tex]e^{j51.34 }[/tex]

c) A₃ = (A₁ + A₂)

giving your answer in polar form

so, A₁ = 6∠30 = 5.196 + j3 and A₂ = 4 + j5

we substitute

A₃ = (5.196 + j3) + ( 4 + j5)

= 9.196 + J8

next we convert to polar

A₃ = √((9.196)² + (8)²) ∠tan⁻¹( [tex]\frac{8 }{9.196}[/tex] )

A₃ = √(84.566416 + 64) ∠tan⁻¹( 0.8699)

A₃ = √148.566416 ∠41.02°    

A₃ = 12.19∠41.02°

Therefore, value of A₃  in polar form is 12.19∠41.02°

d) A₄ = A₁A₂

giving your answer in rectangular form

we substitute

A₄ = (5.196 + j3) ( 4 + j5)

= 5.196( 4 + j5) + j3( 4 + j5)

= 20.784 + j25.98 + j12 - 15

A₄ = 5.784 + j37.98

Therefore, A₄ in rectangular form is 5.784 + j37.98

e) A₅ = A₁/([tex]A^{*}[/tex]₂)

giving your answer in exponential form

we know that [tex]A^{*}[/tex]₂ is the complex  conjugate of A₂

so

[tex]A^{*}[/tex]₂ = (6.403 ∠51.34° )*

= 6.403 ∠-51.34°

we convert to exponential form

A∠β can be written as A[tex]e^{j\beta }[/tex]

[tex]A^{*}[/tex]₂  = 6.403[tex]e^{-j51.34 }[/tex]

also

A₁ = 6∠30

we convert to polar form

A₁ = 6[tex]e^{j30 }[/tex]

so A₅ = A₁/([tex]A^{*}[/tex]₂)

A₅ = 6[tex]e^{j30 }[/tex] / 6.403[tex]e^{-j51.34 }[/tex]

A₅  = (6/6.403) [tex]e^{j(30+51.34) }[/tex]

A₅  = 0.937[tex]e^{j81.34 }[/tex]

Therefore A₅ in exponential form is 0.937[tex]e^{j81.34 }[/tex]

Given:

Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following for air at 300 K and engine oil at 380 K. Assume the free stream velocity is 3 m/s.

To Find:

a. The distance from the leading edge at which the transition will occur.

b. Expressions for the momentum and thermal boundary layer thicknesses as a function of x for a laminar boundary layer

c. Which fluid has a higher heat transfer

Calculation:

The transition from the lamina to turbulent begins when the critical Reynolds

number reaches [tex]5\times 10^5[/tex]

[tex](a). \;\text{Rex}_{cr}=5 \times 10^5\\\\\frac{\rho\;vx}{\mu}=5 \times 10^5\\\text{density of of air at}\;300K=1.16 \frac{kg}{m\cdot s}\\\text{viscosity of of air at}\;300K=1.846 \times 10^{-5} \frac{kg}{m\cdot s} \\v=3m/s\\\Rightarrow x=\frac{5\times 10^5 \times 1.846 \times 10^{-5} }{1.16 \times 3} =2.652 \;m \;\text{for air}\\(\text{similarly for engine oil at 380 K for given}\; \rho \;\text{and} \;\mu)\\[/tex]

[tex](b).\; \text{For the lamina boundary layer momentum boundary layer thickness is given by}:\\\frac{\delta}{x} =\frac{5}{\sqrt{R_e}}\;\;\;\;\quad\text{for}\; R_e <5 \times 10^5\\\\\text{for thermal boundary layer}\\\delta _t=\frac{\delta}{{P_r}^{\frac{1}{3}}}\quad\quad \text{where} \;P_r=\frac{C_p\mu}{K}\\\Rightarrow \delta_t=\frac{5x}{\sqrt{R_e}{P_r}^{\frac{1}{3}}}[/tex][tex](c). \frac{\delta}{\delta_t}={P_r}^{\frac{r}{3}}\\\text{For air} \;P_r \;\text{equivalent 1 hence both momentum and heat dissipate with the same rate for oil}\; \\P_r >>1 \text{heat diffuse very slowly}\\\text{So heat transfer rate will be high for air.}\\\text{Convective heat transfer coefficient will be high for engine oil.}[/tex]

Answer:

10.5

Explanation:

Measuring a Brake Drum, the Brake Micrometer is set to a Base Drum Diameter of 10 Inches plus four notches, and the dial reads 22.  Diameter of the drum is 10.5.

Any straight line segment that cuts through the center of a circle and has ends that are on the circle is considered a circle's diameter in geometry. It is also known as the circle's longest chord.

The diameter of a sphere can be defined using either of the two methods. In more recent usage, the diameter can also refer to a diameter's length, or displaystyle d.

As opposed to a diameter, which refers to the line segment itself, one uses the term "diameter" in this context since all diameters of a circle or sphere have the same length, which is equal to twice the radius "displaystyle r.

Therefore, Measuring a Brake Drum, the Brake Micrometer is set to a Base Drum Diameter of 10 Inches plus four notches, and the dial reads 22.  Diameter of the drum is 10.5.

To learn more about diameter, refer to the link:

https://brainly.com/question/5501950

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This question is incomplete, the complete question is;

Air entrainment is a process of entrapping tiny air bubbles in concrete mix in order to increase the durability of the hardened concrete in freeze-thaw climates. After 35 days, the breaking stress [in psi] was measured for concrete samples with and without air entrainment. Based on the data, does the air entrainment process increase the breaking stress of the concrete. { ∝ = 0.05 }, assuming population variances are equal.

Air Entrainment       No Air Entrainment

4479                                  4118

4436                                  4531

4358                                  4315

4724                                   4237

4414                                   3888

4358                                  4279

4487                                   4311

3984

4197

4327

Answer:

Since p-value ( 0.088) > significance level ( 0.05)

hence, Failed to reject Null hypothesis

It is then concluded that the null hypothesis H₀ is NOT REJECTED.

Therefore, there is no sufficient evidence to claim that population mean μ1 is greater than μ2 at 0.05 significance level.

We conclude that Air entrainment process can't increase the breaking stress of the concrete.

Explanation:

Given the data in the question;  

mean x" = (4479 + 4436 + 4358 + 4724 + 4414 + 4358 + 4487 + 3984 + 4197 + 4327) / 10

mean x"1 =  43764 / 10 = 4376.4

 x                 ( x - x" )             ( x - x" )²

4479              102.6              10526.76

4436              59.6               3552.16

4358             -18.4                 338.56

4724              347.6               120825.76

4414               37.6                 1413.76

4358             -18.4                 338.56

4487              110.6                 12232.36

3984            -382.4                153977.76

4197              -179.4                32184.36

4327             -49.4                  2440.36  

∑                                             337830.4

Standard deviation s1 = √( (∑( x - x" )²) / n -1  

Standard deviation s1 = √( 337830.4 / (10 - 1 ))

Standard deviation s1 = 193.74

 x2                 ( x2 - x"2 )           ( x2 - x"2 )²

4118                  -121.9               14859.61  

4531                   291.1               84739.21

4315                   75.1                5640.01

4237                  -2.9                 8.41    

3888                  -351.9            123833.61

4279                   39.1               1528.81

4311                     71.1                5055.21

∑                                              235664.87

mean x"2 = (4118 + 4531 + 4315 + 4237 + 3888 + 4279 + 4311) / 7

mean x"2 = 29679 / 7 = 4239.9  

Standard deviation s2 = √( (∑( x2 - x" )²) / n2 - 1  

Standard deviation s1 = √( 337830.4 / (7 - 1 ))

Standard deviation s1 = 198.19

so

Mean x"1 = 4376.4,   S.D1 = 193.74,  n1 = 10

Mean x"2 = 4239.9,   S.D2 = 198.19,   n2 = 7

so;

Null Hypothesis H₀ : μ1 = μ2

Alternative Hypothesis H₁ : μ1 > μ2

Lets determine our rejection region;

based on the data provided. the significance level ∝ = 0.005

with degree of freedom DF = n1 + n2 - 2 = 10 + 7 - 2 = 15

so, Critical Value = 1.753

The rejection region for this right -tailed is R = t:t > 1.753

Test statistics

since it is assumed that the population variances are equal, so we calculate pooled standard deviation;

Sp = √{ [ (n1 -1)S.D1² +  (n2 - 1)S.D2²] / [ n1 + n2 -2 ]

we substitute

Sp = √{ [ (10 -1)(193.74)² +  (7 - 1)(198.19)²] / [ 10 + 7 -2 ]

Sp = √ [ 573492.345 / 15 ]

Sp = 195.53

so the Test statistics will be;

t = (x"1 - x"2) / Sp√([tex]\frac{1}{n1}[/tex] + [tex]\frac{1}{n2}[/tex] )

t = (4376.4 - 4239.9) / 195.53√([tex]\frac{1}{10}[/tex] + [tex]\frac{1}{7}[/tex] )

t = 136.5 / 96.36

t = 1.42

so

P-value = 0.088

Since p-value ( 0.088) > significance level ( 0.05)

hence, Failed to reject Null hypothesis

It is then concluded that the null hypothesis H₀ is NOT REJECTED.

Therefore, there is no sufficient evidence to claim that population mean μ1 is greater than μ2 at 0.05 significance level.

We conclude that Air entrainment process can't increase the breaking stress of the concrete.