All categories

1 year ago

3. This material is considered flammable.

Engineering

2 answers:

almond37 [142]1 year ago

4 0

Answer:

All of the above

Explanation:

Most of materials are flammable because of the chemicals theyre made up of. Most materials that are bought say to keep them at room temature for a reason.

Ray Of Light [21]1 year ago

4 0

I'm not sure try oil

hoped I helped!

Have a wonderful day-

Bye!

You might be interested in

Output all combinations of character variables a, b, and c. If a = 'x', b = 'y', and c = 'z', then the output is: xyz xzy yxz yz

gladu [14]

Answer & Explanation:

//written in java

public class Main {

public static void main(String[] args) {

//declare a char variable for a, b, c

char a;

char b;

char c;

//assign a b and c

//a b and c can be replaced for with

// '#', '$', '%', then with '1', '2', '3'

// for further testing

a = 'x';

b = 'y';

c = 'z';

//output for all possible combination for a, b, c.

System.out.println("" + a + b + c + " " + a + c + b + " " + b + a + c +

" " + b + c + a + " " + c + a + b + " " + c + b + a);

}

7 0

2 years ago

If a ball is dropped from a height (H) its velocity will increase until it hits the ground (assuming that aerodynamic drag due

enot [183]

Answer

Explanation:

so the velocity is 39 feet per sec so the impact is 720 cm from the ground

take 720 * 39 sq

4 0

2 years ago

A solid titanium alloy [G 114 GPa] shaft that is 720 mm long will be subjected to a pure torque of T 155 N m. Determine the mini

madam [21]

Answer:

Part 1: The diameter of the shaft so that the shear stress is not more than 150 MPa is 17.3 mm.

Part 2: The diameter of the shaft so that the twist angle is not more than 7° is 16.9 mm.

Explanation:

Part 1

The formula is given as

$\dfrac{T}{J}=\dfrac{\tau}{R}$

Here T is the torque which is given as 155 Nm

J is the rotational inertia which is given as $\dfrac{\pi d^4}{32}$

τ is the shear stress which is given as 150 MPa

R is the radius which is given as d/2 so the equation becomes

$\dfrac{T}{J}=\dfrac{\tau}{R}\\\dfrac{155}{\pi d^4/32}=\dfrac{150 \times 10^6}{d/2}\\\dfrac{155 \times 32}{\pi d^4}=\dfrac{300 \times 10^6}{d}\\\dfrac{1578.82}{d^4}=\dfrac{300 \times 10^6}{d}\\d^3=\dfrac{1578.82}{300 \times 10^6}\\d^3=5.26 \times 10^{-6}\\d=0.0173 m \approx 17.3 mm$

So the diameter of the shaft so that the shear stress is not more than 150 MPa is 17.3 mm.

Part 2

The formula is given as

$\dfrac{T}{J}=\dfrac{G\theta}{L}$

Here T is the torque which is given as 155 Nm

J is the rotational inertia which is given as $\dfrac{\pi d^4}{32}$

G is the torsional modulus which is given as 114 GPa

L is the length which is given as 720 mm=0.720m

θ is the twist angle which is given as 7° this is converted to radian as

$\theta=\dfrac{7*\pi}{180}\\\theta=0.122 rad\\$

so the equation becomes

$\dfrac{T}{J}=\dfrac{G\theta}{L}\\\dfrac{155}{\pi d^4/32}=\dfrac{114 \times 10^9\times 0.122}{0.720}\\\dfrac{1578.81}{d^4}=1.93\times 10^{10}\\d^4=\dfrac{1578.81}{1.93\times 10^{10}}\\d=(\dfrac{1578.81}{1.93\times 10^{10}})^{1/4}\\d=0.0169 m \approx 16.9mm$

So the diameter of the shaft so that the twist angle is not more than 7° is 16.9 mm.

6 0

2 years ago

Factors such as brake shoe orientation, pin location, and direction of rotation determine whether a particular brake shoe is con

Annette [7]

Answer:

Self energizing brake shoe produces more Torque

Explanation:

Attached below is the sketches of the various cases

<u>The cases are :</u>

case 1 ; deenergizing

case 2 ; self energizing

case 3 ; produces high braking torque

For a brake to be self energizing when the Frictional torque and braking torque are in the same direction

<em>summary from equation 3 </em>

when: b > uc the brake is controllable

b = uc It is self locking

b < uc The brake is uncontrollable

8 0

1 year ago

Which of the following statements do not correctly describe pull manufacturing? (1). Material flow is determined by the need of

weeeeeb [17]

Answer: option B.(1) and (3)

(1). Material flow is determined by the need of the downstream workstation

(3). A single piece flow is ideal for a push system and not for a pull system

Explanation:

A pull system is a Lean manufacturing principle created to reduce waste in the production process. A pull system is to build products based on actual demand and not on forecasts.

One of the most important principles in the Lean Philosophy is the Pull Principle. The Pull principle is based on the assumption that one should only produce what is asked for by the customer. The Pull principle can be applied in more than just production, it can be applied in the office and even in life.

5 0

2 years ago