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3 years ago

Technician A says that an A-pillar may be designed to transfer collision energy

1 answer:

7 0

C both A and b cause they are technician both technicians so they both measure out the floor pan reinforcement be designed to transfer collision energy so I say both A and B

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What word is typically written at the bottom of a cover letter to indicate an

Licemer1 [7]

Answer:

i my have been taught differntly but it may be enclosure

Explanation:

7 0

3 years ago

Read 2 more answers

Determine the minimum required wire radius assuming a factor of safety of 3 and a yield strength of 1500 MPa.

lakkis [162]

This question is incomplete, the complete question is;

A large tower is to be supported by a series of steel wires. It is estimated that the load on each wire will be 11,100 N.

Determine the minimum required wire radius assuming a factor of safety of 3 and a yield strength of 1500 MPa.

answer in mm please

Answer:

the minimum required wire radius is 5.3166 mm

Explanation:

Given that;

Load F = 11100N

N = 3

∝y = 1500 MPa

∝workmg = ∝y / N = 1500 / 3 = 500 MPa

now stress of Wire:

∝w = F/A

500 × 10⁶ = 11100 / A

A = 22.2 × 10⁻⁶ m²

(π/4)d² = A

(π/4)d² = 22.2 × 10⁻⁶

d² = 2.8265 × 10⁻⁵

d = 5.3165 7 × 10⁻³ m³

now we convert to mm(millimeters)

d = 5.3166 mm

Therefore the minimum required wire radius is 5.3166 mm

5 0

3 years ago

The current at resonance in a series L-C-R circuit is 0.2mA. If the applied voltage is 250mV at a frequency of 100 kHz and the c

iVinArrow [24]

Answer:

The resistance of the circuit is 1250 ohms

The inductance of the circuit is 0.063 mH.

Explanation:

Given;

current at resonance, I = 0.2 mA

applied voltage, V = 250 mV

resonance frequency, f₀ = 100 kHz

capacitance of the circuit, C = 0.04 μF

At resonance, capacitive reactance ( $X_c$ ) is equal to inductive reactance ( $X_l$ ),

$Z = \sqrt{R^2 + (X_ l - X_c)^2} \\\\But \ X_l= X_c\\\\Z = R$

Where;

R is the resistance of the circuit, calculated as;

$R = \frac{V}{I} \\\\R = \frac{250 \ \times \ 10^{-3}}{0.2 \ \times \ 10^{-3}} \\\\R = 1250 \ ohms$

The inductive reactance is calculated as;

$X_l = X_c = \frac{1}{\omega C} = \frac{1}{2\pi f_o C} = \frac{1}{2\pi (100\times 10^3)(0.04\times 10^{-6} ) } = 39.789 \ ohms\\$

The inductance is calculated as;

$X_l = \omega L = 2\pi f_o L\\\\L = \frac{X_l}{2\pi f_o}\\\\L = \frac{39.789}{2\pi (100 \times 10^3)} \\\\L= 6.3 \ \times \ 10^{-5} \ H\\\\L = 0.063 \times \ 10^{-3} \ H\\\\L = 0.063 \ mH$