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tankabanditka [31]
2 years ago
13

Two basic types of mechanical fuel injector systems?​

Engineering
2 answers:
dem82 [27]2 years ago
7 0

Answer:

excretory system oxgen

REY [17]2 years ago
4 0

Answer:

Single-point or throttle body injection. Port or multipoint fuel injection.

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If a pendulum takes 2 sec to swing in each direction, find the period and the frequency of the swing
DaniilM [7]
The period ( measured in s) is the amount of time is takes for the pendulum to make a single cycle, i.e. how long it takes to swing in both direction. The frequency (measured in Hz) is the inverse of the period, the number of cycles it completes in one second.

So long as you know one you can find the other by diving 1 by it. Period = 1 / frequency, and frequency = 1 / period

If the pendulum takes 2s in each direction, then it has a period of 4s. So the frequency = 1 / 4s = 0.25Hz
8 0
3 years ago
What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 2.5×10-4
krok68 [10]

Answer:

1788.9 MPa

Explanation:

The magnitude of the maximum stress (σ) can be calculated usign the following equation:

\sigma = 2\sigma_{0} \sqrt{\frac{a}{\rho}}

<u>Where:</u>

<em>ρ: is the radius of curvature = 2.5x10⁻⁴ mm (0.9843x10⁻⁵ in)</em>

<em>σ₀: is the tensile stress = 100x10⁶ Pa (14500 psi) </em>

<em>2a: is the crack length = 4x10⁻² mm (1.575x10⁻³ in) </em>

Hence, the  maximum stress (σ) is:

\sigma = 2*100\cdot 10^{6} Pa \sqrt{\frac{(4 \cdot 10^{-2} mm)/2}{2.5 \cdot 10^{-4} mm}} = 1.79 \cdot 10^{6} Pa = 1788.9 MPa    

Therefore, the magnitude of the maximum stress is 1788.9 MPa.

I hope it helps you!

5 0
3 years ago
List everything wrong with 2020
natita [175]
Everything wrong with 2020 is WW3 that dump trump decided to start , Australia fires , Kobe passed away than Pop smoke :( corona virus got really big , quarantine started , riots & protesting started because of that dumb who’re racist cop ! Hope this helps
6 0
3 years ago
Read 2 more answers
The boost converter of Fig. 6-8 has parameter Vs 20 V, D 0.6, R 12.5 , L 10 H, C 40 F, and the switching frequency is 200 kHz. (
mr Goodwill [35]

Answer:

a) the output voltage is 50 V

b)

- the average inductor current is 10 A

- the maximum inductor current is 13 A

- the maximum inductor current is 7 A

c) the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components is 4 A

Explanation:

Given the data in the question;

a) the output voltage

V₀ = V_s/( 1 - D )

given that; V_s = 20 V, D = 0.6

we substitute

V₀ = 20 / ( 1 - 0.6 )

V₀ = 20 / 0.4

V₀ = 50 V

Therefore, the output voltage is 50 V

b)

- the average inductor current

I_L = V_s / ( 1 - D )²R

given that R = 12.5 Ω, V_s = 20 V, D = 0.6

we substitute

I_L = 20 / (( 1 - 0.6 )² × 12.5)

I_L = 20 / (( 0.4)² × 12.5)

I_L = 20 / ( 0.16 × 12.5 )

I_L = 20 / 2

I_L = 10 A

Therefore, the average inductor current is 10 A

- the maximum inductor current

I_{Lmax = [V_s / ( 1 - D )²R] + [ V

given that, R = 12.5 Ω, V_s = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

I_{Lmax = [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]

I_{Lmax = [20 / 2 ] + [ 60 / 20 ]    

I_{Lmax = 10 + 3

I_{Lmax = 13 A

Therefore, the maximum inductor current is 13 A

- The minimum inductor current

I_{Lmax = [V_s / ( 1 - D )²R] - [ V

given that, R = 12.5 Ω, V_s = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

I_{Lmin = [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]

I_{Lmin = [20 / 2 ] -[ 60 / 20 ]    

I_{Lmin = 10 - 3

I_{Lmin  = 7 A

Therefore, the maximum inductor current is 7 A

 

c)  the output voltage ripple

ΔV₀/V₀ = D/RCf

given that; R = 12.5 Ω, C = 40 μF = 40 × 10⁻⁶ F, D = 0.6, f = 200 Khz = 2 × 10⁵ Hz

we substitute

ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )

ΔV₀/V₀ = 0.6 / 100

ΔV₀/V₀ = 0.006 or 0.6%V₀

Therefore, the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components;

under ideal components; diode current = output current

hence the diode current will be;

I_D = V₀/R

as V₀ = 50 V and R = 12.5 Ω

we substitute

I_D = 50 / 12.5

I_D = 4 A

Therefore, the average current in the diode under ideal components is 4 A

7 0
3 years ago
An astronomer of 65 kg of mass hikes from the beach to the observatory atop the mountain in Mauna Kea, Hawaii (altitude of 4205
lara [203]

Answer:

0.845\ \text{N}

Explanation:

g = Acceleration due to gravity at sea level = 9.81\ \text{m/s}^2

R = Radius of Earth = 6371000 m

h = Altitude of observatory = 4205 m

Change in acceleration due to gravity due to change in altitude is given by

g_h=g(1+\dfrac{h}{R})^{-2}\\\Rightarrow g_h=9.81\times(1+\dfrac{4205}{6371000})^{-2}\\\Rightarrow g_h=9.797\ \text{m/s}^2

Weight at sea level

W=mg\\\Rightarrow W=65\times 9.81\\\Rightarrow W=637.65\ \text{N}

Weight at the given height

W_h=mg_h\\\Rightarrow W_h=65\times 9.797\\\Rightarrow W_h=636.805\ \text{N}

Change in weight W_h-W=636.805-637.65=-0.845\ \text{N}

Her weight reduces by 0.845\ \text{N}.

8 0
3 years ago
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