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rjkz [21]
3 years ago
11

Two common message delivery metrics that measure how much of the target market is exposed to the advertisement and the number of

times they are exposed are:
Engineering
1 answer:
Nostrana [21]3 years ago
3 0

Answer:

Effective reach and Frequency

Explanation:

Effective Reach is percentage of target audience that is exposed to a particular ad and receives given message to affect sales and purchase who are reached at or above effective frequency level. Here effective frequency level is the number of exposures necessary to make an impact and attain communication goal.

Effective reach is used in application of statistics to advertising and media analysis to calculate the effectiveness of ad and means used for ad. Effective reach is a time-dependent summary of aggregate audience behaviour.

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A cable in a motor hoist must lift a 700-lb engine. The steel cable is 0.375in. in diameter. What is the stress in the cable?
UkoKoshka [18]

Answer:43.70 MPa

Explanation:

Given

mass of engine 700 lb \approx 317.515 kg

diameter of cable 0.375 in.\approx 9.525 mm

A=\frac{\pi d^2}{4}=71.26 mm^2

we know stress(\sigma)=\frac{load\ applied}{area\ of\ cross-section}

\sigma =\frac{317.515\times 9.81}{71.26\times 10^{-6}}=43.70 MPa

7 0
4 years ago
In a tensile test on a steel specimen, true strain = 0.12 at a stress of 250 MPa. When true stress = 350 MPa, true strain = 0.26
scZoUnD [109]

Answer:

The strength coefficient is 625 and the strain-hardening exponent is 0.435

Explanation:

Given the true strain is 0.12 at 250 MPa stress.

Also, at 350 MPa the strain is 0.26.

We need to find  (K) and the (n).

\sigma =K\epsilon^n

We will plug the values in the formula.

250=K\times (0.12)^n\\350=K\times (0.26)^n

We will solve these equation.

K=\frac{250}{(0.12)^n} plug this value in 350=K\times (0.26)^n

350=\frac{250}{(0.12)^n}\times (0.26)^n\\ \\\frac{350}{250}=\frac{(0.26)^n}{(0.12)^n}\\  \\1.4=(2.17)^n

Taking a natural log both sides we get.

ln(1.4)=ln(2.17)^n\\ln(1.4)=n\times ln(2.17)\\n=\frac{ln(1.4)}{ln(2.17)}\\ n=0.435

Now, we will find value of K

K=\frac{250}{(0.12)^n}

K=\frac{250}{(0.12)^{0.435}}\\ \\K=\frac{250}{0.40}\\\\K=625

So, the strength coefficient is 625 and the strain-hardening exponent is 0.435.

5 0
3 years ago
An assembly line in a modern business compared to one from Henry Ford's time is more likely to rely on which of the following?
Pavel [41]

Answer:

d. fixed layouts

Explanation:

7 0
3 years ago
In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resis
Rufina [12.5K]

Question:

In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resistance R.

For a power line that supplies power to 10 000 households, we can conclude that

a) IV < I²R

b) I²R = 0

c) IV = I²R

d) IV > I²R

e) I = V/R

Answer:

d) IV > I²R

Explanation:

In a typical transmission line, the current I is very small and the voltage V is very high as to minimize the I²R losses in the transmission line.

The power delivered to households is given by

P = IV

The losses in the transmission line are given by

Ploss = I²R

Therefore, the relation IV > I²R  holds true, the power delivered to the consumers is always greater than the power lost in the transmission line.

Moreover, losses cannot be more than the power delivered. Losses cannot be zero since the transmission line has some resistance. The power delivered to the consumers is always greater than the power lost in the transmission.

6 0
3 years ago
Tech A says that bench bleeding a master cylinder will prevent having to bleed air from the brake lines during replacement. Tech
cupoosta [38]

Answer:

  Tech B

Explanation:

Bleeding the master cylinder on the bench does nothing for the air in the brake lines on the vehicle. The bench bleeding is a preferred first step, but bleeding the rest of the brake system is also required. A final check of proper operation on the vehicle should also be accomplished.

Tech B is correct.

6 0
3 years ago
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