Ask question

Ask question

All categories

3 years ago

6

What is the safest way to vent building with backdraft potential

1 answer:

Alja [10]3 years ago

3 0

vertical ventilation

Hope This Helped

You might be interested in

Calculate the speed of light if the wavelength is 3cm and the period is 1*10^-10

ioda

First you find the frequency which is. 1/T(period) so u get 1*10^10
Now that you know frequency u can solve for speed: s= wavelength* frequency, but don't forget to convert 3cm into meters: 3/100= .03m so now you do (1*10^10)(.3)= 300,000,000 m/s or 3*10^8 m/s

Hope this helped :))

6 0

3 years ago

An object is at x = 0 at t = 0 and moves along the x axis according to the velocity–time graph in Figure P2.50.(a) What is the o

bezimeni [28]

Picture? I may be able to answer if you have a chart or some kind of graph as a referral to the question

8 0

3 years ago

a vertical solid steel post 29cm in diameter and 2.0m long is required to support a load of 8200kg, ignore the weight of the pos

erik [133]

Answer:

The stress is $\sigma = 1.218*10^{6} \ N/m^2$

Explanation:

From the question we are told that

The diameter of the post is $d = 29 \ cm = 0.29 \ m$

The length is $L = 2.0 \ m$

The weight of the loading mass

Generally the radius of the post is mathematically represented as

$r = \frac{0.29}{2}$

=> $r = 0.145 \ m$

Generally the area of the post is

$A = \pi r^2$

=> $A = 3.14 * 0.145 ^2$

=> $A = 0.066 \ m^2$

Generally the weight exerted by the load is mathematically represented as

$F = m * g$

=> $F = 8200 * 9.8$

=> $F = 80360 \ N$

Generally the stress is mathematically represented as

$\sigma = \frac{F}{A}$

=> $\sigma = \frac{80360 }{0.066}$

=> $\sigma = 1.218*10^{6} \ N/m^2$

7 0

2 years ago

at room temperature, iron is a solid and mercury is a liquid. based on this information we can infer thata. iron has a higher bo

olchik [2.2K]

A. iron has a higher boiling point thanmercury.

8 0

3 years ago

Read 2 more answers

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

MatroZZZ [7]

Answer:

The wavelength is $\lambda_2 = 534 *10^{-9} \ m$

Explanation:

From the question we are told that

The wavelength of the first light is $\lambda _ 1 = 587 \ nm$

The order of the first light that is being considered is $m_1 = 10$

The order of the second light that is being considered is $m_2 = 11$

Generally the distance between the fringes for the first light is mathematically represented as

$y_1 = \frac{ m_1 * \lambda_1 * D}{d}$

Here D is the distance from the screen

and d is the distance of separation of the slit.

For the second light the distance between the fringes is mathematically represented as

$y_2 = \frac{ m_2 * \lambda_2 * D}{d}$

Now given that both of the light are passed through the same double slit

$\frac{y_1}{y_2} = \frac{\frac{m_1 * \lambda_1 * D}{d} }{\frac{m_2 * \lambda_2 * D}{d} } = 1$

=> $\frac{ m_1 * \lambda _1 }{ m_2 * \lambda_2} = 1$

=> $\lambda_2 = \frac{m_1 * \lambda_1}{m_2}$

=> $\lambda_2 = \frac{10 * 587 *10^{-9}}{11}$

=> $\lambda_2 = 534 *10^{-9} \ m$

4 0

2 years ago