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katrin2010 [14]
2 years ago
15

a metal rod 24mm diameter and 2m long is subjected to an axial pull of 40 kN. If the rod is 0.5mm, then find the stress-induced

and value of the youngs modulus
Engineering
1 answer:
ozzi2 years ago
8 0

Answer:

i dont know but i will take the points tho hahah

Explanation:

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What gadgets are charge coupled devices used in?
Alinara [238K]

Answer:

They are used in imaging application gadgets such as video cameras,TV, surveillance cameras and document scanners

Explanation:

A charge couple device (CCDs) are highly capable in imagery detector.Its common application is in video and digital imaging.The quality of a charge couple device is determined by factors such as the dynamic range, dark charge level and the quantum efficiency.These devices serve the purpose of detecting optical images though some are installed with applications for data storage.

5 0
2 years ago
An LED camping headlamp can run for 18 hours, powered by three AAA batteries. The batteries each have a capacity of 1000 mAh, an
KIM [24]

Answer:

a) the power consumption of the LEDs is 0.25 watt

b) the LEDs drew 0.0555 Amp current

Explanation:

Given the data in the question;

Three AAA Batteries;

<---- 1000mAh [ + -] 1.5 v ------1000mAh [ + -] 1.5 v --------1000mAh [ + -] 1.5 v------

so V_total = 3 × 1.5 = 4.5V

a) the power consumption of the LEDs

I_battery = 1000 mAh / 18hrs    { for 18 hrs}

I_battery = 1/18 Amp    { delivery by battery}

so consumption by led = I × V_total

we substitute

⇒ 1/18 × 4.5

P = 0.25 watt

Therefore the power consumption of the LEDs is 0.25 watt

b) How much current do the LEDs draw

I_Draw = I_battery = 1/18 Amp = 0.0555 Amp

Therefore the LEDs drew 0.0555 Amp current

5 0
2 years ago
A fluid of density 900 kg/m3 passes through a converging section of an upstream diameter of 50 mm and a downstream diameter of 2
NISA [10]

Answer:

Q= 4.6 × 10⁻³ m³/s

actual velocity will be equal to 8.39 m/s

Explanation:

density of fluid = 900 kg/m³

d₁ = 0.025 m

d₂ = 0.05 m

Δ P = -40 k N/m²

C v = 0.89

using energy equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2

under ideal condition v₁² = 0

v₂² = 88.88

v₂ = 9.43 m/s

hence discharge at downstream will be

Q = Av

Q = \dfrac{\pi}{4}d_1^2 \times v

Q = \dfrac{\pi}{4}0.025^2 \times 9.43

Q= 4.6 × 10⁻³ m³/s

we know that

C_v =\dfrac{actual\ velocity}{theoretical\ velocity }\\0.89 =\dfrac{actual\ velocity}{9.43}\\actual\ velocity = 8.39m/s

hence , actual velocity will be equal to 8.39 m/s

6 0
3 years ago
The production process of rods from machine "A" yields specimen with the following specs. Mean: µ(LA)=20.00mm, STD: s(LA)=0.50mm
Oxana [17]

Answer: the standard deviation STD of machine B is s (Lb) = 0.4557

Explanation:

from the given data, machine A and machine B produce half of the rods

Lt = 0.5La + 0.5Lb

so

s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)

but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)

so we substitute

s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)

0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)

0.64 = 0.25 + s²(Lb) + 0.4s(Lb)

s²(Lb) + 0.4s(Lb) - 0.39 = 0

s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2

s (Lb) = 0.4557

therefore the standard deviation STD of machine B is s (Lb) = 0.4557

8 0
2 years ago
If he wants to keep the height the same, what could the other dimensions be for him to get the volume he wants?
Fiesta28 [93]

tbm queria saber essa pergunta

8 0
3 years ago
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