All categories

3 years ago

When is a handrail required for stairs?

Engineering

1 answer:

Leviafan [203]3 years ago

5 0

Answer:

after 8 stepshddnffuddbnggkbdbkloyr

You might be interested in

How to calculate tension.

Evgen [1.6K]

Answer:

Tension can be easily explained in the case of bodies hung from chain, cable, string

Explanation

uniform speed, tension; T = W.

T=m(g±a)

3 0

2 years ago

A circular specimen of MgO is loaded in three-point bending. Calculate the minimum possible radius of the specimen without fract

Hitman42 [59]

Answer:

radius = 9.1 × $10^{-3}$ m

Explanation:

given data

applied load = 5560 N

flexural strength = 105 MPa

separation between the support = 45 mm

solution

we apply here minimum radius formula that is

radius = $\sqrt[3]{\frac{FL}{\sigma \pi}}$ .................1

here F is applied load and is length

put here value and we get

radius = $\sqrt[3]{\frac{5560\times 45\times 10^{-3}}{105 \times 10^6 \pi}}$

solve it we get

radius = 9.1 × $10^{-3}$ m

8 0

3 years ago

Create a set of questions a company designing a new motorcycle for a younger audience might want to ask external customers to ma

Brut [27]

Answer:

Who would pay for my product or service?

Who has already bought from me?

Am I overestimating my reach?.

What does my network think?.

Am I making assumptions based on my personal knowledge and experience?

What's my revenue model

How will I sell my product or service?

How did my competitors get started?

How will I find my customers?

Is there room to expand my target market?

3 0

2 years ago

Vehicles begin to arrive at a parking lot at 8:10 am at a constant rate of 6 veh/min until 8:25 am. There is no arrival from 8:2

tresset_1 [31]

1.i am superman

2.175 175 175

3.em dilisues

4.bye classmate magbabalik pa ako

8 0

2 years ago

A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft), is subjected to an acceleration a = {6ti + 12t^2k} ft/s^

Sindrei [870]

Answer:

The particle's position at t = 2 is

r = (11ft, 2ft, 21ft) = (11, 2, 21) ft

Explanation:

r₀ = (3, 2, 5) ft

a = (6t, 0, 12t²) ft/s²

a = dv/dt

dv/dt = (6tî + 0j + 12t²ķ)

dv = (6tî + 0j + 12t²ķ) st

Integrating the left hand side from 0 to v (the particle was originally at rest) and the right hand side from 0 to t,

We obtain,

v = (3t²î + 0j + 4t³ķ) ft/s

v = dr/dt

dr/dt = (3t²î + 0j + 4t³ķ)

dr = (3t²î + 0j + 4t³ķ) st

Integrating the left hand side from r₀ (the original position) to r and the right hand side from 0 to t,

r - r₀ = (t³î + 0j + t⁴ķ) ft

r = (t³î + 0j + t⁴ķ) + r₀

r = (t³î + 0j + t⁴ķ) + (3î + 2j + 5ķ)

At t=2s, t³ = 8 and t⁴ = 16

r = (8î + 0j + 16ķ) + (3î + 2j + 5ķ)

r = (11î + 2j + 21ķ) ft

r = (11ft, 2ft, 21ft)

Hope this helps!

4 0

3 years ago