3. Which of the following is not a common impact

B1) 20 pts. The thickness of each of the two sheets to be resistance spot welded is 3.5 mm. It is desired to form a weld nugget

kap26 [50]

Answer:

minimum current level required = 8975.95 amperes

Explanation:

Given data:

diameter = 5.5 mm

length = 5.0 mm

T = 0.3

unit melting energy = 9.5 j/mm^3

electrical resistance = 140 micro ohms

thickness of each of the two sheets = 3.5mm

Determine the minimum current level required

first we calculate the volume of the weld nugget

v = $\frac{\pi }{4} * D^2 * l$ = $\frac{\pi }{4} * 5.5^2 * 5$ = 118.73 mm^3

next calculate the required melting energy

= volume of weld nugget * unit melting energy

= 118.73 * 9.5 = 1127.94 joules

next find the actual required electric energy

= required melting energy / efficiency

= 1127 .94 / ( 1/3 ) = 3383.84 J

TO DETERMINE THE CURRENT LEVEL REQUIRED use the relation below

electrical energy = I^2 * R * T

3383.84 / R*T = I^2

3383.84 / (( 140 * 10^-6 ) * 0.3 ) = I^2

therefore 8975.95 = I ( current )

4 0

3 years ago

Consider a system with two tasks, Task1 and Task2. Task1 has a period of 200 ms, and Task2 has a period of 300 ms. All tasks ini

Murrr4er [49]

<u>Explanation:</u>

Task 1 time period = 200ms, Task 2 time period = 300ms

Task ticked = $\frac{1000ms}{200ms}= 5$ → 5 times

Task 2 ticked = $\frac{1000ms}{300ms} = 3.33$ → 3 times

At 600 ms → 200ms 200ms 200ms

300ms → $\frac{30ms}{60ms}$

Largest time period = H.C.M of (200ms, 300ms)

= 600ms

4 0

2 years ago

A 60-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest acceleration anticipated i

dlinn [17]

Answer:

$h_{max} = 51.8 cm$

Explanation:

given data:

height of tank = 60cm

diameter of tank =40cm

accelration = 4 m/s2

suppose x- axis - direction of motion

z -axis - vertical direction

$\theta$ = water surface angle with horizontal surface

$a_x =$ accelration in x direction

$a_z =$ accelration in z direction

slope in xz plane is

$tan\theta = \frac{a_x}{g +a_z}$

$tan\theta = \frac{4}{9.81+0}$

$tan\theta =0.4077$

the maximum height of water surface at mid of inclination is

$\Delta h = \frac{d}{2} tan\theta$

$=\frac{0.4}{2}0.4077$

$\Delta h 0.082 cm$

the maximu height of wwater to avoid spilling is

$h_{max} = h_{tank} -\Delta h$

= 60 - 8.2

$h_{max} = 51.8 cm$

the height requird if no spill water is $h_{max} = 51.8 cm$

3 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

The regulated voltage of an alternator is stated as 13.6 to 14.6 volts at 3000 rpm with the

lions [1.4K]

Answer:

d) 1 volt

Explanation:

The allowable range is 1 volt. The allowed tolerance (deviation from nominal) depends on what the nominal voltage is.

5 0

3 years ago