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2 years ago

What are the nominal dimensions for a 1x2 stick of lumber, a 2x4 stick of lumber and a standard sheet of plywood?

1 answer:

6 0

3/4 x 1 1/2 inches (19 x 38 mm) is the actual size for 1x2 stick of lumber,

1 1/2 x 3 1/2 inches (38 x 89 mm) is the actually size for a 2x4 stick of lumber,

Plywood is usually sold in 4 x 8-foot sheets. The most common nominal thicknesses of plywood are 1/2 inch and 3/4 inch, but once again the actual sizes are slightly different. A sheet of 1/2-inch plywood is really 15/32 inch thick, while a 3/4-inch sheet is 23/32 inch thick.

Hopefully this answers your question, I apologize if it doesn’t :)

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Come and look on my attachment

CaHeK987 [17]

Crazy Guy what do uh mean ?

4 0

2 years ago

Deidre has just moved from the sales department into the finance department. On her first day in her new role, she receives an e

melisa1 [442]

Answer:

d. The company uses role-based access control and her user account hasn't been migrated into the correct group(s) yet

Explanation:

Since Deidre is accessing her e-mail there appears to be nothing wrong with her account or password. Since her role is new, most likely the problem is associated with her new role.

3 0

2 years ago

A single fixed pulley is used to lift a load of 400N by the application of an effort of 480N in 10s through a vertical height of

Allushta [10]

Answer:

(a) the velocity ratio of the machine (V.R) = 1

(b) The mechanical advantage of the machine (M.A) = 0.833

Explanation:

Given;

load lifted by the pulley, L = 400 N

effort applied in lifting the, E = 480 N

distance moved by the effort, d = 5 m

(a) the velocity ratio of the machine (V.R);

since the effort applied moved downwards through a distance of d, the load will also move upwards through an equal distance 'd'.

V.R = distance moved by effort / distance moved by the load

V.R = 5/5 = 1

(b) The mechanical advantage of the machine (M.A);

M.A = L/E

M.A = 400 / 480

M.A = 0.833

$E = \frac{M.A}{V.R} \times 100\%\\\\E = 0.833 \ \times \ 100\%\\\\ E = 83.3 \ \%$

4 0

2 years ago

In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle tempe

KIM [24]

Answer:

$\eta$ =0.60

Explanation:

Given :Take $\gamma$ =1.4 for air

$P_1=100 KPa ,T_1=300K$

$\frac{V_1}{V_2}$ =r ⇒ r=16

As we know that

$T_2=T_1(r^{\gamma-1})$

So $T_2=300\times 16^{\gamma-1}$

$T_2$ =909.42K

Now find the cut off ration $\rho$

$\rho=\frac{V_3}{V_2}$

$\frac{V_3}{V_2}=\frac{T_3}{T_2}$

$\rho=\frac{2031}{909.42}$

$\rho=2.23$

So efficiency of diesel engine

$\eta =1-\dfrac{\rho^\gamma-1}{\gamma\times r^{\gamma-1}(\rho-1)}$

Now by putting the all values

$\eta =1-\dfrac{2.23^{1.4}-1}{1.4\times 16^{1.4-1}(2.23-1)}$

So $\eta$ =0.60

So the efficiency of diesel engine=0.60

7 0

3 years ago

What is the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 515 turns, its

kow [346]

<h2>Answer:</h2>

7532V

<h2>Explanation:</h2>

For a given transformer, the ratio of the number of turns in its primary coil ( $N_{p}$ ) to the number of turns in its secondary coil ( $N_{s}$ ) is equal to the ratio of the input voltage ( $V_{p}$ ) to the output voltage ( $V_{s}$ ) of the transformer. i.e