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3 years ago

In the construction of a timber-framed house, select and justify three timber based

1 answer:

3 0

Timber frame construction uses timber studs and rails, together with a structural sheathing board, to form a structural frame that transmits all vertical and horizontal loads to the foundations.

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3–102 One of the common procedures in fitness programs is to determine the fat-to-muscle ratio of the body. This is based on the

gayaneshka [121]

Answer:

x_fat = [ 0.5*(Wsa + Wsw) - p_muscle*V ] / V*( p_fat - p_muscle )

Explanation:

Given:

- The total volume of body = V

- The average density of the body = p_avg

- The density of muscle = p_muscle

- The density of fat = p_fat

Find:

Obtain a relation for the volume fraction of body fat x_fat

Solution:

- The volume of the fat is given by:

V_fat = x_fat*V

- The volume of the muscle is given by:

V_muscle = V - V_fat

= V - x_fat*V

=V*( 1 - x_fat )

- We will use the conservation of mass for the body related as:

mass_fat + mass_muscle = Total average mass

p_fat*V_fat + p_muscle*V_muscle = p_avg*V

p_fat*x_fat*V + p_muscle*V*( 1 - x_fat ) = p_avg*V

p_fat*x_fat + p_muscle*( 1 - x_fat ) = p_avg

- To determine p_1 we weigh the body in air:

Weight reading (Wsa) = m = p_1*V

p_1 = Wsa / V*g

- To determine p_2 we weigh the body in water:

Weight reading (Wsw) = m - p_w*V= p_1*V - p_w*V

Weight reading (Wsw) = V*(p_1 - p_w) = V*(p_2)

Where, p_2 = p_1 - p_water

p_2 = Wsw / V

- The average density p_avg:

p_avg = 0.5*(p_1 + p_2)

p_avg = 0.5*(Wsa / V + Wsw / V)

p_avg = 0.5*(Wsa + Wsw) / V

- Plug in the mass equation:

p_fat*x_fat + p_muscle*( 1 - x_fat ) = 0.5*(Wsa + Wsw) / V

x_fat*( p_fat - p_muscle ) = 0.5*(Wsa + Wsw) / V - p_muscle

x_fat = [ 0.5*(Wsa + Wsw) - p_muscle*V ] / V*( p_fat - p_muscle )

6 0

3 years ago

Determine the combined moment about O due to the weight of the mailbox and the cross member AB. The mailbox weighs 3.2 lb and th

koban [17]

Answer:

Attached is the complete question but the weight of the mailbox and cross bar differs from the given values which are : weight of mail box = 3.2 Ib, weight of the uniform cross member = 10.3 Ib

Answer : moment of inertia = 186.7 Ib - in

Explanation:

Given data

weight of the mailbox = 3.2 Ib

weight of the uniform cross member = 10.3 Ib

The origin is of mailbox and cross member is 0

The perpendicular distance from Y axis of centroid of the mailbox

= 4 + (25/2) = 16.5"

The centroid of the bar =( ( 1 + 25 + 4 + 4 ) / 2 ) - 4 = 13"

therefore The moment of Inertia( Mo) = (3.2 * 16.5) + ( 10.3 * 13)

= 52.8 + 133.9 = 186.7 Ib-in

8 0

3 years ago

Please help me with this. Plzzz.

Drupady [299]

Answer:

450,000m = 450km = 4.5E5

32,600,000W = 32.6MW = 3.26E7

59,700,000,000cal = 59.7Gcal = 5.97E10

0.000000083s = 83ns = 8.3E-8

35,000Ω = 35kΩ = 3.5E4

Explanation:

Giga = 1,000,000,000

Mega = 1,000,000

kilo = 1,000

unit = 1

deci = .1

centi = .01

milli = .001

micro = .000001

nano = .0000000001

pico = .000000000001

You should be able to look at these and convert between them in seconds if you want to pursue anything in engineering.

7 0

3 years ago

The domain of discourse is the members of a chess club. The predicate B(x, y) means that person x has beaten person y at some po

satela [25.4K]

Answer:A. No one has ever beat Nancy.

Explanation:

The dormain of discourse in a simple language is the set of entities upon which our discussions are based when discussing about something.

The dormain of discourse is also known simply as universe, can also be said to be a set of entities o

upon which certain variables of interest in some formal treatment may range.

The dormain of discourse is generally attributed to Augustus De Morgan, it was also extensively used by George Boole in his Laws of Thought.

THE LOGICAL UNDERSTANDING OF THE THE QUESTION IS THAT NO ONE HAS EVER BEAT NANCY.

8 0

3 years ago

What engineers call moment, scientists call

Svet_ta [14]

Answer:

yes

Explanation:

7 0

3 years ago