Fluorescent troffers are a type of

n200080 [17]

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.

5 0

2 years ago

Me ayudas plis noce

alekssr [168]

Answer:

Explanation:

7 0

1 year ago

The unit weight of a soil is 14.9kN/m3. The moisture content of the soil is17% when the degree of saturation is 60%. Determine:

Serggg [28]

Answer:

a) 2622.903 N/m^3

b) 1.38233

c)4.878811765

Explanation:

Find the void ratio using the formula:

$y = \frac{G_{s}*y_{w} + w*G_{s}*y_{w} }{1+e} ....... Eq1$

Here;

$G_{s}$ is specific gravity of soil solids

$y_{w}$ is unit weight of water = 998 kg/m^3

$w$ is the moisture content = 0.17

$e$ is the void ratio

$y$ is the unit weight of soil = 14.9KN/m^3

Saturation Ratio Formula:

$w*G_{s} = S*e ..... Eq2$

S is saturation rate

Substitute Eq 2 into Eq 1

$y = \frac{(\frac{S*e}{w}) * y_{w} + S*e*y_{w} }{1+e}$

$14900 = \frac{3522.352941*e + 598.8*e }{1+e} = \frac{4121.152941*e}{1+e}\\\\ e= 1.38233$

Specific gravity of soil solids

$G_{s} = \frac{S*e}{w} = \frac{0.6*1.38233}{0.17} = 4.878811765$

Saturated Unit Weight

$y_{s} = \frac{(G_{s} + e)*y_{w} }{1+e} \\=\frac{(4.878811765 + 1.38233)*998 }{1+1.38233}\\\\= 2622.902571 N/m^3$

7 0

2 years ago

A 35kg block of mass is subjected to forces F1=100N and F2=75N at agive angle thetha= 20° and 35° respectively.find the distance

Talja [164]

Answer:

21 m

Explanation:

Since F₁ = 100 N and acts at an angle of 20° to the horizontal, it has horizontal component F₁' = 100cos20° = 93.97 N and vertical component F₁" = 100sin20° = 34.2 N.

Also, F₂ = 75 N and acts at an angle of -35° to the horizontal, it has horizontal component F₂' = 75cos(-35°) = 75cos35° = 61.44 N and vertical component F₂" = 75sin(-35°) = -75sin35° = -43.02 N

The resultant horizontal force F₃' = F₁' + F₂' = 93.97 N + 61.44 N = 155.41 N

The resultant vertical force F₃" = F₁" + F₂" = 34.2 N - 43.02 N = -8.82 N

If f is the frictional force on the block, the net horizontal force on the block is F = F₃' - f.

Since f = μN where μ = coefficient of kinetic friction = 0.4 and N = normal force on the block.

For the block to be in contact with the surface, the vertical forces on the block must balance.

Since the normal force, N must equal the resultant vertical force F₃" and the weight, W = mg of the object for a zero net vertical force,

N = mg + F₃" (since both the weight and the resultant vertical force act downwards)

N = mg + F₃"

Since m = mass of block = 35 kg and g = acceleration due to gravity = 9.8 m/s² and F₃" = 8.82 N

So,

N = mg + F₃"

N = 35 kg × 9.8 m/s² + 8.82 N

N = 343 N + 8.82 N

N = 351.82 N

So, the net horizontal force F = F₃' - f.

F = 155.41 N - 0.4 × 351.82 N

F = 155.41 N - 140.728 N

F = 14.682 N

Since F = ma, where a = acceleration of block,

a = F/m = 14.682 N/35 kg = 0.42 m/s²

To find the distance the block moved, x we use the equation

x = ut + 1/2at² where u = initial speed of block = 0 m/s, t = time = 10 s and a = acceleration of block = 0.42 m/s²

Substituting the values of the variables into the equation, we have

x = ut + 1/2at²

x = 0 m/s × 10 s + 1/2 × 0.42 m/s² × (10 s)²

x = 0 m + 1/2 × 0.42 m/s² × 100 s²

x = 0.21 m/s² × 100 s²

x = 21 m

So, the distance moved by the block is 21 m.

4 0

2 years ago

- Scrap tire management is primarily regulated at the

kompoz [17]

Scrap tire management is primarily regulated at the state level.

3 0

2 years ago

Read 2 more answers

Fluorescent troffers are a type of _ lighting fixture