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Sophie [7]
3 years ago
7

Two ways of sawing logs are?​

Engineering
1 answer:
zmey [24]3 years ago
7 0

Answer:

Plain Sawn. At least 95 percent of all hardwood lumber commercially produced in the U.S. is flat or plain sawn. ...

Quarter Sawn. In this method, the log actually is cut into quarters, then sawn quarter by quarter. ...

Explanation:

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How much work, in Newtons, is required to lift a 20.4-kg (45lb) plate from the ground to a stand that is 1.50 meters up?
nataly862011 [7]

Answer:

Explanation:

Work, U, is equal to the force times the distance:

U = F · r

Force needed to lift the weight, is equal to the weight: F = W = m · g

so:

U = m · g · r

   = 20.4kg · 9.81 \frac{N}{kg} · 1.50m

   = 35.316 \frac{N}{m}

   = 35.316 W

4 0
2 years ago
The yellow rectangle area is 25% (or 1/4) the area of the blue rhombus. The height (H) of the yellow rectangle is twice as long
Kitty [74]

Answer:

I don't know sry

Explanation:

6 0
3 years ago
Use the results of Prob. 5–82 for plane strain near the tip with u 5 0 and n 5 13. If the yieldstrength of the plate is Sy, what
rosijanka [135]

Answer:

Kindly follow the steps as shown below.

Explanation:

8 0
3 years ago
Urgent please help me...
Pepsi [2]
Scuze , am incercat , dat chiar nu stiu. Prefer sa nu raspund si sa scriu asta decat sa raspund gresit.❤️
7 0
3 years ago
An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, p
Veseljchak [2.6K]

Answer:

Ts =Ta E)- 300(

569.5 K

5

Q12-W12 = -4014.26

Mol

AU2s = Q23= 5601.55

Mol

AUs¡ = Ws¡ = -5601.55

Explanation:

A clear details for the question is also attached.

(b) The P,V and T for state 1,2 and 3

P =1 bar Ti = 300 K and Vi from ideal gas Vi=

10

24.9x10 m

=

P-5 bar

Due to step 12 is isothermal: T1 = T2= 300 K and

VVi24.9 x 10x-4.9 x 10-3 *

The values at 3 calclated by Uing step 3l Adiabatic process

B-P ()

Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=

14

Ps-1x(4.99 x 103

P-1x(29x 10)

9.49 barr

And Ts =Ta E)- 300(

569.5 K

5

(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and

Work done for Isotermal process define as

8.314 x 300 In =4014.26

Wi2= RTi ln

mol

And fromn first law of thermodynamic

AU12= W12 +Q12

Q12-W12 = -4014.26

Mol

F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and

Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-

AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53

Inol

Now from first law of thermodynamic the Q23

AU2s = Q23= 5601.55

Mol

For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0

and

AH

C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18

mol

AU=C, (T¡-T)= x 8.314 (300

-5601.55

569.5)

mol

Now from first law of thermodynamie the Ws1

J

mol

AUs¡ = Ws¡ = -5601.55

3 0
3 years ago
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