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zysi [14]
3 years ago
12

A farmer wants to buy a 10 kg bag of fertilizer (organic soil). They have the choice between two merchants. Merchant A sells the

10 kg bag for $10 and the bag indicates that the fertilizer is completely dry. Merchant B sells the 10 kg bag for 8$, and the bag indicates that the fertilizer has a water content of 20%. If the farmer can only use the solid constituents of the bag, which merchant has the better deal?
Engineering
1 answer:
sergejj [24]3 years ago
5 0

Answer:

Since the farmer wants to buy a 10 kg bag of fertilizer, he should buy it from merchant A. However, Merchant A and B are selling at the same price for a unit value. In other words, Both Merchant A and B are selling 1kg of dry fertilizer for $1.

Explanation:

Which merchant has the better deal means which merchant offers the farmer a better deal.

For Merchant A,  10 kg bag = $10

meaning it contains a real 10 kg bag of dry fertilizer which the farmer can use without losing any Kg to drying.

While for Merchant B, 10 kg bag = $8

where the 10kg = 80% dry fertilizer + 20% water content

But the farmer can only use the solid constituents of the bag which means,

Merchant B is giving 80/100 x 10Kg of dry fertilizer for $8

That is, 8kg for $8

Since the farmer wants to buy a 10 kg bag of fertilizer, he should buy it from merchant A. However, Merchant A and B are selling at the same price for a unit value. In other words, Both Merchant A and B are selling 1kg of dry fertilizer for $1.

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A freshly annealed glass containing flaws of maximum length of 0.1 microns breaks under a tensile stress of 120 MPa. If a sample
almond37 [142]

Answer:

0.16 micron per day

Explanation:

Given:

The initial crack length, a₁ = 0.1 micron = 0.1 × 10⁻⁶ m

Initial tensile stress, σ₁ = 120 MPa

Final stress = 30 MPa

now from Griffith's equation, we have

\sigma=[\frac{G_cE}{\pi\ a}]^\frac{1}{2}

where,

Gc and E are the material constants

now,

for the initial stage

120=[\frac{G_cE}{\pi\ (0.1\times10^{-6}}]^\frac{1}{2}  ........{1}

and for the final case

30=[\frac{G_cE}{\pi\ a_2}]^\frac{1}{2}   ............{2}

on dividing 1 by 2, we get

\frac{120}{30}=[\frac{a_2}{0.1\times10^{-6}}]^\frac{1}{2}

or

a₂ = 4² × 0.1 × 10⁻⁶ m

or

a₂ = 1.6 micron

Now,

the change from 0.1 micron to 1.6 micron took place in 10 days

therefore, the rate at which the crack is growing = \frac{1.6-0.1}{10}

or

average rate of change of crack = 0.16 micron per day

6 0
3 years ago
The creation of designer drugs is outpacing the ability of society to enact laws to prohibit them. Many of these substances have
Andreas93 [3]

Answer:

The creation of "designer drugs" is outpacing the ability of society to enact laws to prohibit them.  Many of these substances have negative side effects, ranging from violent behavior to death.

 

40.  Which of the following responses to the problem would best fit the "crime control" philosophy?

a.  Government takes steps to limit the availability of ingredients used in the manufacture of designer drugs.

b.  Design a public awareness campaign to warn potential users of the dangers presented by use of these drugs.

c.  Partner with community leaders to identify underlying social issues promoting the drug subculture.

d.  Pass legislation and increase enforcement efforts to send a message of "zero tolerance" to those who manufacture, sell, and use designer drugs

The answer to the above question is

d.  Pass legislation and increase enforcement efforts to send a message of "zero tolerance" to those who manufacture, sell, and use designer drugs

Explanation:

State governments, naturally provide early response to the domestic issue of designer drug abuse than the federal government through the early banning of synthetic drug use.

State legislation on restricting designer drug use can be defined into three groups including

1. Specific synthetic substance ban

2. Ban on generic language

3. Ban based on the effect a substance has on the body.

It can be clearly demonstrated that legislation which combines this three categories is highly effective in reducing the supply and use of designer drugs.

3 0
3 years ago
28. What is the value of a resistor in a series circuit if you measure 0.5 amps flowing through it and 15 volts
Anna35 [415]
The girls name on my phone I am in my name is
7 0
3 years ago
a. Determine R for a series RC high-pass filter with a cutoff frequency (fc) of 8 kHz. Use a 100 nF capacitor. b. Draw the schem
Readme [11.4K]

Answer:

a) 199.04 ohms

b) attached in image

c) -0.696dB

Explanation:

We are given:

Fc = 8Khz = 8000hz

C = 100nF = 100*10^-^9F

a)Using the formula:

F_c = \frac{1}{2pie*Rc}

8000= \frac{1}{2*3.14*R*100*10^-^9}

R =\frac{1}{2*3.14*100*10^-^9*8000}

R = 199.04 ohms

b) diagram is attached

c) H(w) = \frac{V_out(w)}{Vin(w)} = \frac{1}{1-j\frac{wc}{w}}

H(F) = \frac{1}{1-j\frac{fc}{f}}

At F = 20KHz and Fc= 8KHz we have:

H(F)= \frac{1}{1-j\frac{8}{20}} = \frac{1}{1-j(0.4)}

|H(F)|= \frac{1}{\sqrt{1^2+(0.4)^2}}

=0.923

|H(F)| in dB = 20log |H(F)|

=20log0.923

= -0.696dB

5 0
3 years ago
Technician A says that the most commonly used combustion chamber types include hemispherical, and wedge. Technician B says that
Inessa05 [86]

Answer:

Technician A and Technician B both are correct.

Explanation:

Technician A accurately notes that perhaps the forms of combustion process most widely used are hemispherical and cross.

Technician B also correctly notes that in several cylinder heads, cooling system and greases gaps and pathways are found.

6 0
3 years ago
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