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nikklg [1K]
2 years ago
5

Reverse masking forms a soft edge on the panel.

Engineering
1 answer:
valina [46]2 years ago
7 0
True I think I’m not sure?
You might be interested in
If you are setting up a race car. What is the cross weight? Does it matter?
lara31 [8.8K]

Answer:

cross-weight is used to tighten it up.

Explanation:

and yes this is important because Cross-weight percentage compares the diagonal weight totals to the car's total weight.

hope  this help

(mark this answer as an brainliest answer)

7 0
3 years ago
Chemical manufacturers must present which information on the product's label?
Wewaii [24]

Answer:

hazardous chemicals leaving the workplace is labeled, tagged or marked with the following information: product identifier; signal word; hazard statement

Explanation:

this is so you know what chemicals are in it

6 0
3 years ago
Drivers education - Unit 3
melamori03 [73]

The following scenarios are pertinent to driving conditions that one may encounter. See the following rules of driving.

<h3>What do you do when the car is forced into the guardrail?</h3>

Best response:

  • I'll keep my hands on the wheel and slow down gradually.
  • The reason I keep my hands on the steering wheel is to avoid losing control.
  • This will allow me to slowly back away from the guard rail.
  • The next phase is to gradually return to the fast lane.
  • Slamming on the brakes at this moment would result in a collision with the car behind.

Scenario 2: When driving on a wet road and the car begins to slide

Best response:

  • It is not advised to accelerate.
  • Pumping the brakes is not recommended.
  • Even lightly depressing and holding down the brake pedal is not recommended.
  • The best thing to do is take one foot off the gas pedal.
  • There should be no severe twists at this time.

Scenario 3: When you are in slow traffic and you hear the siren of an ambulance behind

Best response:

  • The best thing to do at this moment is to go to the right side of the lane and come to a complete stop.
  • This helps to keep the patient in the ambulance alive.
  • It also provide a clear path for the ambulance.
  • Moving to the left is NOT recommended.
  • This will exacerbate the situation. If there is no place to park on the right shoulder of the road, it is preferable to stay in the lane.

Learn more about rules of driving. at;

brainly.com/question/8384066

#SPJ1

4 0
2 years ago
Tahir travel twice as far as ahmed, but onley one third as fast. Ahmed starts travel on tuesday at noon at point x to point z 30
shepuryov [24]

Answer:

6:00 pm the next day

Explanation:

Given that

Tahir traveled twice as far as Ahmed. We say,

Ahmed traveled a distance, D

Tahir would travel a distan, 2D

Tahir traveled 1/3 as fast as Ahmed, so we say

Ahmed traveled at a speed, S

Tahir would travel at a speed, S/3

If Ahmed starts travel on tuesday at noon at point x to point z 300km, by 9:00pm,

Time taken by Ahmed to travel is

9:00 pm - 12:00 pm = 9 hours

Ahmed, traveled 300 km in 9 hours, meaning he traveled at 33.3 km in an hour.

Speed, S that Ahmed traveled with is 33.3 km/h

Remember, we stated that Tahir travels at a speed of S/3, that is, The speed of Tahir is

33.3/3 = 11.1 km/h.

300 km would then be traveled in 300 km/11.1 km/h = 27 hours.

Tahir started traveling, 3 hours after Ahmed, that is 12:00 pm + 3:00 hrs = 3:00 pm, and if he's to spend 27 hours on the journey he would reach destination z at 6:00 pm the next day

7 0
2 years ago
What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 3 × 10-4
Vladimir [108]

Answer:

maximum stress is 2872.28 MPa

Explanation:

given data

radius of curvature = 3 × 10^{-4} mm

crack length = 5.5 × 10^{-2} mm

tensile stress = 150 MPa

to find out

maximum stress

solution

we know that  maximum stress formula that is express as

\sigma m = 2 ( \sigma o ) \sqrt{\frac{a}{\delta t}}     ......................1

here σo is applied stress and a is half of internal crack and t is radius of curvature of tip of internal crack

so put here all value in equation 1 we get

\sigma m = 2 ( \sigma o) \sqrt{\frac{a}{\delta t}}  

\sigma m = 2(150) \sqrt{ \frac{\frac{5.5*10^{-2}}{2}}{3*10^{-4}}}  

σm = 2872.28 MPa

so maximum stress is 2872.28 MPa

8 0
3 years ago
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