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3 years ago

13

Answer?...................

1 answer:

torisob [31]3 years ago

5 0

Answer:

The correct option is;

c. Leaving the chuck key in the drill chuck

Explanation:

A Common safety issues with a drill press leaving the chuck key in the drill chuck

It is required that, before turning the drill press power on, ensure that chuck key is removed from the chuck. A self ejecting chuck key reduces the likelihood of the chuck key being accidentally left in the chuck.

It is also required to ensure that the switch is in the OFF position before turning plugging in the power cable

Be sure that the chuck key is removed from the chuck before turning on the power. Using a self-ejecting chuck key is a good way of insuring that the key is not left in the chuck accidentally. Also to avoid accidental starting, make sure the switch is in the OFF position before plugging in the cord. Always disconnect the drill from the power source when making repairs.

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3 years ago

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3 years ago

A rigid canister with a radius of 5 in and a height of 10 in is filled with air. The initial pressure and temperature of air in

Elza [17]

Answer:1.458 Btu

Explanation:

Given

radius of canister $\left ( r\right )=5in$

Height of canister $\left ( h\right )=10 in.$

Initial pressure $\left ( P_i\right )=14.7 Psi$

Initia ltemprature $\left ( T_i\right )=70^{\circ}F$

Final pressure $\left ( P_f\right )=30Psi$

as canister is rigid therefore change in volume is zero

therefore

$\frac{P_i}{T_i}$ = $\frac{P_f}{T_f}$

$\frac{14.7}{70}$ = $\frac{30}{T_f}$

$T_f=142.85^{\circ}F$

volume of canister= $\pi \times r^{2}\times h$

= $\pi \times 5^{2}\times 10=250\pi$

volume of canister=12872,038.8 $mm^3$

now calculating mass of air

PV=mRT

substituting values

$\left ( 14.7Psi\right )\left ( 12872,038.8 mm^3\right )=m\left ( 0.287\right )\left ( 70^{\circ}F\right )$

m=21.0192 gm

Therefore heat transferred = $mc_p\left ( T_f-T_i\right )$

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3 0

4 years ago

A wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 27 . It has b

Mamont248 [21]

Answer:

The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa

Explanation:

Given data;

Let,

critical stress required for initiating crack propagation Cc = 112MPa

plain strain fracture toughness = 27.0MPa

surface length of the crack = a

dimensionless parameter = Y.

Half length of the internal crack, a = length of surface crack/2 = 8.8/2 = 4.4mm = 4.4*10-³m

Also for 6.2mm length of surface crack;

Half length of the internal crack = length of surface crack/2 = 6.2/2 = 3.1mm = 3.1*10-³m

The dimensionless parameter

Cc = Kic/(Y*√pia*a)

Y = Kic/(Cc*√pia*a)

Y = 27/(112*√pia*4.4*10-³)

Y = 2.05

Now,

Cc = Kic/(Y*√pia*a)

Cc = 27/(2.05*√pia*3.1*10-³)

Cc = 135.78MPa

The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa

For more understanding, I have provided an attachment to the solution.

4 0

4 years ago