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2 years ago

a low velocity fastening system that is used to drive steel pins or threaded studs into a masonry and steel is a

1 answer:

5 0

A low velocity fastening system that is used to drive steel pins or threaded studs into a masonry and steel is a powder-actuated tool.

<h3>What is a fastening system?</h3>

It should be noted that a fastening system is important to fixing rails to railroad ties.

In this case, a low velocity fastening system that is used to drive steel pins or threaded studs into a masonry and steel is a powder-actuated tool.

Learn more about fasteners on:

brainly.com/question/16350719

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Air pressure is higher above an airfoil.<br> true or false

attashe74 [19]

Answer: true

Explanation:

it flows faster over the top of the wing because the top is more curved than the bottom of the wing. However

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3 years ago

Tech a says the higher the numarical gear ratio (4:1), the more torque that will be applied to the wheels. Tech b says that the

Zepler [3.9K]

Answer:

Tech A

Explanation:

The amount of energy required to apply the same force with a 1:1 ratio is divided into 4, so you can apply 4 times as much force than a 1:1 ratio. efficiency and speed come into play here, but assuming the machine powering the gear can run at a unlimited RPM, 4:1 will have more force and a slower output speed than a 2:1 ratio.

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3 years ago

Read 2 more answers

The drag force, Fd, imposed by the surrounding air on a

Ad libitum [116K]

Answer:

a) 23.551 hp

b) 516.89 hp

Explanation:

<u>given:</u>

$F_{d} =\frac{1}{2} C_{d} A_{p} V^{2} \\V_{a}=25 m/hr-->25*\frac{5280}{3600} =36.67ft/s\\V_{b}=70 m/hr-->70*\frac{5280}{3600} =102.67ft/s\\\\C_{d}=.28\\A=25 ft^2\\p=.075lb/ft^2$

<u>required:</u>

the power in hp

<u>solution:</u>

$(F_{d})_{a} =\frac{1}{2} C_{d} A_{p} V_{a} ^{2}$ .............(1)

by substituting in the equation (1)

=353.27 lbf

$(F_{d})_{b} =\frac{1}{2} C_{d} A_{p} V_{b} ^{2}$ ..........(2)

by substituting in the equation (2)

= 2769.29 lbf

power is defined by

P=F.V

$P_{a}=$ 353.27*36.67

=12954.411 lbf.ft/s

=12954.411*.001818

=23.551 hp

$P_{a}=$ 2769.29*102.67

= 284323 lbf.ft/s

= 284323*.001818

= 516.89 hp

3 0

3 years ago

Rubber bushings are used on suspensions to

Harlamova29_29 [7]

D. All of the above

4 0

3 years ago

Calculate the viscosity(dynamic) and kinematic viscosity of airwhen

nikitadnepr [17]

Answer:

(a) dynamic viscosity = $1.812\times 10^{-5}Pa-sec$

(b) kinematic viscosity = $1.4732\times 10^{-5}m^2/sec$

Explanation:

We have given temperature T = 288.15 K

Density $d=1.23kg/m^3$

According to Sutherland's Formula dynamic viscosity is given by

${\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}$ , here

μ = dynamic viscosity in (Pa·s) at input temperature T,

$\mu _0$ = reference viscosity in(Pa·s) at reference temperature T0,

T = input temperature in kelvin,

$T_0$ = reference temperature in kelvin,

C = Sutherland's constant for the gaseous material in question here C =120

$\mu _0=4\pi \times 10^{-7}$

$T_0$ = 291.15

$\mu =4\pi \times 10^{-7}\times \frac{291.15+120}{285.15+120}\times \left ( \frac{288.15}{291.15} \right )^{\frac{3}{2}}=1.812\times 10^{-5}Pa-s$ when T = 288.15 K

For kinematic viscosity :

$\nu = \frac {\mu} {\rho}$

$kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec$

3 0

3 years ago