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mestny [16]
2 years ago
14

Select three types of lines that engineers use to help represent the shape of a design in a sketch.

Engineering
1 answer:
Vikki [24]2 years ago
7 0

Hidden lines

  • Used to describe the in shown lines (like diagonals inside cubes)

Extension lines:-

  • Used to explain the expansion of structures like building

Object lines

  • Used to describe the structure of objects and the lining to show borders
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Conclusion. What process is responsible for the bubbling action of the organism? What is your evidence?
noname [10]

Answer:

Explanation:

Hands-on Activity Bubbling Plants Experiment to Quantify Photosynthesis ... After running the experiment, students pool their data to get a large sample ... Explain that photosynthesis is a process that plants use to convert light ... Describe a simple experiment that provides indirect evidence that photosynthesis is occurring.

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6 0
3 years ago
Technician A says that a voltage drop of 0.8 volts on the starter ground circuit is within specifications. Technician B says tha
Romashka-Z-Leto [24]

Answer:

Technician A is wrong

Technician B is right

Explanation:

voltage drop of 0.8 volts on the starter ground circuit is not within specifications. Voltage drop should be within the range of 0.2 V to 0.6 V but not more than that.

A spun bearing can seize itself around the crankshaft journal causing it not to move. As the car ignition system is turned on, the stater may draw high current in order to counter this seizure.

8 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
Rosita is planning an investigation to determine how a lifeboat's shape affects its
KiRa [710]
Yes! Is right, just did the work!
3 0
3 years ago
The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam
lisabon 2012 [21]

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy

T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG  > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

6 0
3 years ago
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