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11Alexandr11 [23.1K]

2 years ago

9

Which of the following explains the purpose of a convertible screwdriver?

1 answer:

aliya0001 [1]2 years ago

7 0

Answer:

The correct option is;

To be able to switch between different heads quickly and easily

Explanation:

A screwdriver is a tool or device that has a grip handle which surrounds an extension shaft that has a forged tip that fits into the complementary groove on a screw head such that by turning the handle while the forged head of the screwdriver sits in a screw head, the screw is turned

A convertible screw driver is built with the capability to easily change the tip or other attributes of the screwdriver quickly.

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PLEASE HELP!!! ILL GIVE BRANLIEST *EXTRA POINTS* dont skip :((

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The first one is the answer

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3 years ago

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The fracture toughness of a stainless steel is 137 MPa*m12. What is the tensile impact load sustainable before fracture that a r

Charra [1.4K]

Answer:

7.7 kN

Explanation:

The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.

It can be expressed by using the formula:

$K = \sigma Y \sqrt{\pi a}$

where;

fracture toughness K = 137 MPa $m^{1/2}$

geometry factor Y = 1

applied stress $\sigma$ = ???

crack length a = 2mm = 0.002

∴

$137 =\sigma \times 1 \sqrt{ \pi \times 0.002 }$

$137 =\sigma \times 0.07926$

$\dfrac{137}{0.07926} =\sigma$

$\sigma = 1728.489 MPa$

Now, the tensile impact obtained is:

$\sigma = \dfrac{P}{A}$

P = A × σ

P = 1728.289 × 4.5

P = 7777.30 N

P = 7.7 kN

7 0

3 years ago

State three active materials of a lead acid cell

igomit [66]

Answer:

lead dioxide,sulfate and lead acid

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3 years ago

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A spring-mass-damper instrument is employed for acceleration measurements. The spring constant is 12000 N/m. The mass is 5 g. Th

shepuryov [24]

Answer:

a) 246.56 Hz

b) 203.313 Hz

c) Add more springs

Explanation:

Spring constant = 12000 N/m

mass = 5g = 5 * 10^-3 kg

damping ratio = 0.4

<u>a) Calculate Natural frequency </u>

Wn = √k/m = $\sqrt{12000 / 5*10^{-3} }$

= 1549.19 rad/s ≈ 246.56 Hz

<u>b) Bandwidth of instrument </u>

W / Wn = $\sqrt{1-2(0.4)^2}$

W / Wn = 0.8246

therefore Bandwidth ( W ) = Wn * 0.8246 = 246.56 * 0.8246 = 203.313 Hz

C ) To increase the bandwidth we have to add more springs

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Technician A says it will save you money if you get a high interest rate from a bank on the loan. Technician B says it is always

zubka84 [21]

Is there a other option

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3 years ago