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vredina [299]
3 years ago
11

The map shows the distribution of dairy farms across the lower 48 of the United States. Each dot on the map represents approxima

tely ten dairy farms.
Engineering
1 answer:
yan [13]3 years ago
6 0

Answer:

1. Large swathes of land is being devoted to the cultivation of corn and cultivation of nutritious grasses.

2. Vegetation and Rainfall

Explanation:

1. One characteristic land-use feature associated with the dairy farms is the use of large swaths of land to grow corn and nutritious alfalfa grass. Earlier, these lands were devoted to the cultivation of temperate wheat. However, the success of commercial dairy farming in the region has led to the introduction of extensive farming methods where large farms are devoted to corn and alfalfa production. Application of manures, usage of machines is very high and the overall yield of the farms is very high but yield per unit of land is low.

2. Two geographic features not shown on the map that corroborates to the above concentration pattern are-

Vegetation- Temperate grassland (Prairie) of these regions provides large and extensive ground for the cultivation of fodder grass. Prairie supports the cultivation of these grasses.

Rainfall- These areas receive plenty of rainfall from humongous freshwater lakes in their vicinity. This rainfall proves crucial for cultivation for animal feeds.

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A pitfall cited in Section 1.10 is expecting to improve the overall performance of a computer by improving only one aspect of th
Oxana [17]

Answer:

a) For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

(1-0.2)*70 s =56s

The reduction on this case is 70-56 s=14s

And since the new total time would be given by 250-14=236 s

b) For this case the total time is reduced 20%  so that means that the new total time would be (1-0.2)=0.8 times the original total time (1-0.2) *250s =200 s

The original time for INT operations is calculated as:

250 = 70+85+40 +t_{INT}

t_{INT}=55s

For this part the only time that was changed is assumed the INT operations so then:

200 = 70+85+40 \Delta t_{INT}

And then: \Delta t_{INT}= 200-70-85-40=5 s

c) A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

Explanation:

From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.

Part 1

For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

(1-0.2)*70 s =56s

The reduction on this case is 70-56 s=14s

And since the new total time would be given by 250-14=236 s

Part 2

For this case the total time is reduced 20%  so that means that the new total time would be (1-0.2)=0.8 times the original total time (1-0.2) *250s =200 s

The original time for INT operations is calculated as:

250 = 70+85+40 +t_{INT}

t_{INT}=55s

For this part the only time that was changed is assumed the INT operations so then:

200 = 70+85+40 \Delta t_{INT}

And then: \Delta t_{INT}= 200-70-85-40=5 s

And we can quantify the decrease using the relative change:

\% Change = \frac{5s}{55 s} *100 = 9.09\% of reduction

Part 3

A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

8 0
3 years ago
When converting liquid level units to sensor output signal units, you should first convert the liquid level units to _____ units
dangina [55]
The answer is 2 because I just took the test!
7 0
2 years ago
(TCO 4) A system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without
steposvetlana [31]

Answer:

The frequency that the sampling system will generate in its output is 70 Hz

Explanation:

Given;

F = 190 Hz

Fs = 120 Hz

Output Frequency = F - nFs

When n = 1

Output Frequency = 190 - 120 = 70 Hz

Therefore, if a system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without further modification, the frequency that the sampling system will generate in its output is 70 Hz

5 0
3 years ago
ANSWER QUICK<br>Why did Winston Churchill take over for Neville Chamberlain shortly after ww2? ​
kenny6666 [7]

Answer:neville chamberlain died

Explanation:

4 0
3 years ago
An 800-kg drag racer accelerates from rest to 390 km/hr in 5.8 s. What is the net impulse applied to the racer in the first 5.8
marissa [1.9K]

Answer:

Impulse =14937.9 N

tangential force =14937.9 N

Explanation:

Given that

Mass of car m= 800 kg

initial velocity u=0

Final velocity v=390 km/hr

Final velocity v=108.3 m/s

So change in linear momentum P= m x v

           P= 800 x 108.3

 P=86640 kg.m/s

We know that impulse force F= P/t

So F= 86640/5.8 N

F=14937.9 N

Impulse force F= 14937.9 N

We know that

v=u + at

108.3 = 0 + a x 5.8

a=18.66\ m/s^2

So tangential force F= m x a

F=18.66 x 800

F=14937.9 N

6 0
3 years ago
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