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jolli1 [7]
3 years ago
10

What are the two reasons for a clear cut

Engineering
1 answer:
Inessa [10]3 years ago
3 0

Answer:

to clear land for agriculture and settlement and to use or sell timber for lumber, paper products, or fuel.

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In the following code, determine the values of the symbols here and there. Write the object code in hexadecimal. (Do not predict
allsm [11]

Answer:

Answer explained below

Explanation:

The value of here is 9

The value of there is hexadecimal value of DECO here, d = 0x39 aaaa (aaaa is the memory address of here )

We have the object code :-

let's take there address is 0x0007

0x0005 BR there :- 0x120020

0x0007 here: .WORD 9

310003 there: DECO here,d - 0x390007

310005 STOP

.END

4 0
3 years ago
What is another term for the notes that a reader can add to text in a word-processing document?
melamori03 [73]

Answer:

annotations maybe?

4 0
3 years ago
Read 2 more answers
(a) Determine the dose (in mg/kg-day) for a bioaccumulative chemical with BCF = 103 that is found in water at a concentration of
solmaris [256]

Answer:

0.064 mg/kg/day

6.25% from water, 93.75% from fish

Explanation:

Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.

The BCF = 10³, so the concentration of the chemical in the fish is:

10³ = x / (0.1 mg/kg)

x = 100 mg/kg

For 2 L of water and 30 g of fish:

2 kg × 0.1 mg/kg = 0.2 mg

0.030 kg × 100 mg/kg = 3 mg

The total daily intake is 3.2 mg.  Divided by the woman's mass of 50 kg, the dosage is:

(3.2 mg/day) / (50 kg) = 0.064 mg/kg/day

b) The percent from the water is:

0.2 mg / 3.2 mg = 6.25%

And the percent from the fish is:

3 mg / 3.2 mg = 93.75%

3 0
3 years ago
The parts of a feature control frame are the tolerance value, the datum references, and the
Elan Coil [88]

Answer:

d

Explanation:

4 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
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