Which option identifies the tool best to use in the following scenario?

Engineering

1 answer:

ikadub [295]3 years ago

4 0

Answer: an Allen wrench

Explanation:

Based on the information given in the question, the option that identifies the tool that will be the best to use in the following scenario is the Allen key.

The Allen key also refered to as the hex key is a simple tool that is used in driving both screws, bolts and also used in driving the screws that have hexagonal sockets that is, six sided hole in their heads.

You might be interested in

For ceramic-matrix composites, high interfacial strength is desirable. ( True , False )

NemiM [27]

This is true as the answer

4 0

3 years ago

If a 110-volt appliance requires 20 amps, what is the total power consumed?

juin [17]

Answer:

2200 W

Explanation:

Use the given relation between current, voltage, and power to find the power requirement:

P = IV

P = (20 A)(110 V) = 2200 W

4 0

2 years ago

Read 2 more answers

Very thin films are usually deposited under vacuum conditions to prevent contamination and ensure that atoms can fly directly fr

katrin [286]

Answer:

a. 9947 m

b. 99476 times

c. 2*10^11 molecules

Explanation:

a) To find the mean free path of the air molecules you use the following formula:

$\lambda=\frac{RT}{\sqrt{2}\pi d^2N_AP}$

R: ideal gas constant = 8.3144 Pam^3/mol K

P: pressure = 1.5*10^{-6} Pa

T: temperature = 300K

N_A: Avogadros' constant = 2.022*10^{23}molecules/mol

d: diameter of the particle = 0.25nm=0.25*10^-9m

By replacing all these values you obtain:

$\lambda=\frac{(8.3144 Pa m^3/mol K)(300K)}{\sqrt{2}\pi (0.25*10^{-9}m)^2(6.02*10^{23})(1.5*10^{-6}Pa)}=9947.62m$

b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:

$n_{collision}=\frac{9947.62m}{0.05m}\approx198952\ times$

c) By using the equation of the ideal gases you obtain:

$PV=NRT\\\\N=\frac{PV}{RT}=\frac{(1.5*10^{-6}Pa)(\frac{4}{3}\pi(0.05m)^3)}{(8.3144Pa\ m^3/mol\ K)(300K)}=3.14*10^{-13}mol\\\\n=(3.14*10^{-13})(6.02*10^{23})\ molecules\approx2*10^{11}\ molecules$