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2 years ago

13

As you push a toggle bolt into a wall, the

1 answer:

MakcuM [25]2 years ago

5 0

Answer:

Explanation:
One the object is attached the toggle it's placed onto the bolt. The wings should be open in the direction of the bolt head. The toggle is than collapsed and the entire toggle bolt is inserted into the hole in the wall. Once the wings are fully through the opening they will spring open on the other side.
Hopt it's help thanks..

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1. A thin plate of a ceramic material with E = 225 GPa is loaded in tension, developing a stress of 450 MPa. Is the specimen lik

mina [271]

Answer:

fracture will occur as the value is less than E/10 (= 22.5)

Explanation:

If the maximum strength at tip Is greater than theoretical fracture strength value then fracture will occur and if the maximum strength is lower than theoretical fracture strength then no fracture will occur.

$\sigma_m = 2\sigma_o [\frac{a}{\rho_t}]^{1/2}$

$= 2\times 750 (\frac{\frac{0.2mm}{2}}{0.001 mm}})^{1/2}$

= 15 GPa

fracture will occur as the value is less than E/10 = 22.5

7 0

3 years ago

Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held

77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

$\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} }$

${T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} } = 280.15 \times \left (9 \right )^{\frac{1.333-1}{1.333} } = 485.03\ K$

The heat supplied = $\dot {m}_f$ × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = $\dot m$ · $c_p(T_3 - T_2)$

$\dot m$ = 20 kg/s

The heat supplied = 20* $c_p(T_3 - T_2)$ = 21,350 kJ/s

$c_p$ = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

$\dfrac{T_5}{T_4} = \dfrac{2}{1.333 + 1}$

T₅ = 1208.42*(2/2.333) = 1035.94 K

$C_j = \sqrt{\gamma \times R \times T_5}$ = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = $\dot m$ × $C_j$ = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

5 0

2 years ago

A punch must cut a hole 30mm diameter in a sheet of steel 2mm thick. The ultimate shear

Anettt [7]

2+2=3

4+5=7

$tex]\purple{\rule{45pt}{7pt}}\blue{\rule{45pt}{999999pt}}tex]$

7 0

2 years ago

When brazing, the lowest effective brazing temperatures possible should be used to minimize the effects of heat on the base meta

nordsb [41]

Answer:

False

Explanation:

5 0

2 years ago

Three communications channels in parallel have independent failure modes of 0.1 failure per hour. These components must share a

ozzi

Answer:

the MTTF of the transceiver is 50.17

Explanation:

Given the data in the question;

failure modes = 0.1 failure per hour

system reliability = 0.85

mission time = 5 hours

Now, we know that the reliability equation for this situation is;

R(t) = [ 1 - ( 1 - $e^{-0.1t$ )³] $e^{-t/MTTF$

so we substitute

R(5) = [ 1 - ( 1 - $e^{-0.1(5)$ )³] $e^{-5/MTTF$ = 0.85

[ 1 - ( 1 - $e^{-0.5$ )³] $e^{-5/MTTF$ = 0.85

[ 1 - ( 0.393469 )³] $e^{-5/MTTF$ = 0.85

[ 1 - 0.06091 ] $e^{-5/MTTF$ = 0.85

0.9391 $e^{-5/MTTF$ = 0.85

$e^{-5/MTTF$ = 0.85 / 0.9391

$e^{-5/MTTF$ = 0.90512

MTTF = 5 / -ln( 0.90512 )

MTTF = 50.17

Therefore, the MTTF of the transceiver is 50.17

7 0

3 years ago