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siniylev [52]
3 years ago
13

1. advantages of 2 pulley system

Engineering
1 answer:
n200080 [17]3 years ago
5 0

Answer:

Advantages

The main advantage in the use of pulleys is that the effort becomes less as compared to the normal lifting of the weights. In other words, it reduces the amount of actual force required to lift heavy objects. It also changes the direction of the force applied. These two advantages in the use pulleys make them an important tool for heavy lifting. It also provides a mechanical advantage.

The other advantage in the use of pulleys is that the distance between the operator and weight. There is a safe distance between them which avoids any disaster. Pulleys are easy to assemble and cost-effective. The combination of different directional pulleys can change the position of the load with little effort. Though there are moving parts in the pulley system they require less or no lubrication after installation.

Disadvantages

Apart from the above-said advantages while using pulley systems, there are several disadvantages in their use. The main disadvantage in the use of the pulley system is that it requires large space to install and operate. The mechanical advantage of pulleys can go to higher values but need more space to install them.

In some cases, the ropes/belts move over the wheel with no grooves, the chances of the slip of ropes/belts from the wheel are inevitable. If the system is installed to use for a long time, they require maintenance and regular check-up of ropes/cables as the friction between the wheels and cables/ropes occur causing wear and tear to them. Continuous use of the system makes the ropes weak. The rope may break while using the system causing damages to the operator, surrounding place and the load which is being lifted.

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I gave 15 min to finish this java program
lisov135 [29]

Answer:

class TriangleNumbers

{

public static void main (String[] args)

{

 for (int number = 1; number <= 10; ++number) {

  int sum = 1;

  System.out.print("1");

  for (int summed = 2; summed <= number; ++summed) {

   sum += summed;

   System.out.print(" + " + Integer.toString(summed));

  }

  System.out.print(" = " + Integer.toString(sum) + '\n');

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}

}

Explanation:

We need to run the code for each of the 10 lines. Each time we sum  numbers from 1 to n. We start with 1, then add numbers from 2 to n (and print the operation). At the end, we always print the equals sign, the sum and a newline character.

4 0
3 years ago
When measuring a Brake Drum, the Brake Micrometer is set to a Base Drum Diameter of 10 Inches plus four notches, and the dial re
kozerog [31]

Answer:

10.5

Explanation:

7 0
2 years ago
A hot brass plate is having its upper surface cooled by impinging jet of air at temperature of 15°C and convection heat transfer
gulaghasi [49]

Answer:

809.98°C

Explanation:

STEP ONE: The first step to take in order to solve this particular Question or problem is to find or determine the Biot value.

Biot value = (heat transfer coefficient × length) ÷ thermal conductivity.

Biot value = (220 × 0.1)÷ 110 = 0.2.

Biot value = 0.2.

STEP TWO: Determine the Fourier number. Since the Biot value is greater than 0.1. Tis can be done by making use of the formula below;

Fourier number = thermal diffusivity × time ÷ (length)^2.

Fourier number = (3 × 60 × 33.9 × 10^-6)/( 0.1)^2 = 0.6102.

STEP THREE: This is the last step for the question, here we will be calculating the temperature of the center plane of the brass plate after 3 minutes.

Thus, the temperature of the center plane of the brass plane after 3 minutes = (1.00705) (0.89199) (900- 15) + 15.

= > the temperature of the center plane of the brass plane after 3 minutes = 809.98°C.

5 0
2 years ago
A sheet of steel 1.5 mm thick has nitrogen (N2) atmospheres on both sides at 1200°C and is permitted to achieve steady-state dif
Gelneren [198K]

Answer:

do the wam wam

Explanation:

3 0
2 years ago
Consider the diffusion of water vapor through a polypropylene (PP) sheet 1 mm thick. The pressures of H2O at the two faces are 3
Neko [114]

Answer:

\boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}

Explanation:

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The pressure difference will be 10,000 Pa- 3000 Pa= 7000 Pa

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Since the thickness of sheet is given as 1mm= 0.1 cm then

J=38\times 10^{-13}(cm^{3}. STP)(cm)/(cm^{2}.s.Pa)\times \frac {7000 pa}{0.1cm}=0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}

Therefore, the diffusion flux is \boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}

7 0
2 years ago
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