All categories

3 years ago

Bruuh its sum werid people on brainly

Engineering

2 answers:

ale4655 [162]3 years ago

8 0

Answer:

what happend??

Explanation:

agasfer [191]3 years ago

4 0

Ikr, people are so darn odd these days it’s just an app

You might be interested in

In a wheatstone bridge three out of four resistors have of 1K ohm each ,and the fourth resistor equals 1010 ohm. If the battery

Dima020 [189]

Answer:

248.756 mV

49.7265 µA

Explanation:

The Thevenin equivalent source at one terminal of the bridge is ...

voltage: (100 V)(1000/(1000 +1000) = 50 V

impedance: 1000 || 1000 = (1000)(1000)/(1000 +1000) = 500 Ω

The Thevenin equivalent source at the other terminal of the bridge is ...

voltage = (100 V)(1010/(1000 +1010) = 100(101/201) ≈ 50 50/201 V

impedance: 1000 || 1010 = (1000)(1010)/(1000 +1010) = 502 98/201 Ω

The open-circuit voltage is the difference between these terminal voltages:

(50 50/201) -(50) = 50/201 V ≈ 0.248756 V . . . . open-circuit voltage

The current that would flow is given by the open-circuit voltage divided by the sum of the source resistance and the load resistance:

(50/201 V)/(500 +502 98/201 +4000) = 1/20110 A ≈ 49.7265 µA

8 0

3 years ago

While supporting load P, the maximum contraction permitted in a 4.75-m long pipe column is 7 mm. Given that E =180 GPa for the c

Alexeev081 [22]

Answer:

$\mathbf{stress \ \sigma = 264.6 \ Mpa}$

Explanation:

From the concept of Hooke's Law,

$E =\dfrac{ stress \ \sigma}{ strain \ \varepsilon}$

where;

$strain \ \varepsilon = \dfrac{change \ in \ dimension }{original \ dimension}$

$strain \ \varepsilon = \dfrac{7 \ mm }{4.75 \times 10^{3} \ mm}$

$strain \ \varepsilon =0.00147368$

Recall:

$E =\dfrac{ stress \ \sigma}{ strain \ \varepsilon}$

$stress \ \sigma = E \times { strain \ \varepsilon}$

$stress \ \sigma = 180 \times 10^{3} \ Mpa \times 0.00147$

$\mathbf{stress \ \sigma = 264.6 \ Mpa}$

Thus, the stress in the pipe at the maximum allowable contraction = 264.6 Mpa

7 0

3 years ago

A cylindrical rod of a metal alloy is stressed elastically in tension. The original diameter was 11 mm and a force of 55 kN was

andrey2020 [161]

Answer:

10.984mm

Explanation:

by elastic modulus

stress=modulus of elasticity*strain

stress=loading/area area" cross-section"

11mm=0.011m

area=π(d/2)^2=π(0.011/2)^2=9.503*10^-5 square meter

stress=55000/(9.503*10^-5)=578.745 MPa

convert MPa and GPa to pascal.

strain=stress/modulus=(578.745*10^6)/(125*10^9)=0.00463............axial strain

v=Poisson ratio

lateral strain=(-v)*axial strain= -0.31*0.00463

lateral strain= -1.4353*10^-3=change in diameter/ original diameter

change in diameter=(-1.4353*10^-3)*0.011= -1.57883*10^-5 m

negative indicates decrease in diameter.

decrease in dia.=0.01578mm

new diameter=11-0.01578= 10.984mm

3 0

3 years ago

If I were to build a fully functional robotic leg what type of circuit board should I use?

garik1379 [7]

Answer:

A good one

Explanation:

Rotatianal degrees and stuf for different types of movement would take some proccesing power and if it were linked to brain neurons you would need one not prone to frying

5 0

3 years ago

The average grain diameter of an aluminum alloy is 14 mu m with a strength of 185 MPa. The same alloy with an average grain diam

timama [110]

Answer:

See explanations for completed answers

Explanation:

Given that; The average grain diameter of an aluminum alloy is 14 mu m with a strength of 185 MPa. The same alloy with an average grain diameter of 50 mu m has a strength of 140 MPa. (a) Determine the constants for the Hall-Patch equation for this alloy, (b) How much more should you reduce the grain size if you desired a strength of 220 MPa

See attachment for completed solvings

6 0

3 years ago