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Readme [11.4K]
3 years ago
9

Materials to be used to build a watch tower​

Engineering
1 answer:
gizmo_the_mogwai [7]3 years ago
8 0

Answer:

Wood, brick or stone. I you want to make it for a school project then you can use sticks.

Hope it helps!!!

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a motor has torque of 200 nm if i motors attached to an arm with a length of 25 cm and a string is too attached to the end appro
kogti [31]

Answer:

noescompa

Explanation:

puntos

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3 years ago
A thick steel slab ( 7800 kg/m3, 480 J/kg·K, 50 W/m·K) is initially at 300°C and is cooled by water jets impinging on one of its
AleksandrR [38]

Answer: 67.392s

Explanation: detailed calculation is shown below

4 0
3 years ago
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Solve the inequality. Then graph your solution.<br> -9v – 10 &lt; 7y +6
Cerrena [4.2K]
16

if you add 9+10 you get 18 - 7+6
5 0
2 years ago
A wooden pallet carrying 540kg rests on a wooden floor. (a) a forklift driver decides to push it without lifting it.what force m
kicyunya [14]

Answer:

The appropriate solution is "1481.76 N".

Explanation:

According to the question,

Mass,

m = 540 kg

Coefficient of static friction,

\mu_s = 0.28

Now,

The applied force will be:

⇒ F=\mu_s mg

By substituting the values, we get

       =0.28\times 540\times 9.8

       =1481.76 \ N

8 0
2 years ago
g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress
zvonat [6]

Answer:

a) \mathbf{\sigma _ 1 = 4800 psi}

     \mathbf{ \sigma _2 = 0}

b)\mathbf{\sigma _ 1 = 6000 psi}

  \mathbf{ \sigma _2 = 3000 psi}

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the  thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = 0

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 4800 psi}

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = \frac{Pd}{4t}

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 6000 psi}

\sigma _2 = \frac{Pd}{4t} \\ \\  \sigma _2 = \frac{250(12)}{4(0.25)}

\mathbf{ \sigma _2 = 3000 psi}

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

8 0
3 years ago
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