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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:

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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1. (15) A truck scale is made of a platform and four compression force sensors, one at each corner of the platform. The sensor i

Elanso [62]

Answer:

a). 139498.24 kg

b). 281.85 ohm

c). 10.2 ohm

Explanation:

Given :

Diameter, d = 22 m

Linear strain, $\epsilon$ = 3%

= 0.03

Young's modulus, E = 30 GPa

Gauge factor, k = 6.9

Gauge resistance, R = 340 Ω

a). Maximum truck weight

σ = Eε

σ = $0.03 \times 30 \times 10^9$

$\frac{P}{A} =0.03 \times 30 \times 10^9$

$P = 0.03 \times 30 \times 10^9\times \frac{\pi}{4}\times (0.022)^2$

= 342119.44 N

For the four sensors,

Maximum weight = 4 x P

= 4 x 342119.44

= 1368477.76 N

Therefore, weight in kg is $m=\frac{W}{g}=\frac{1368477.76}{9.81}$

m = 139498.24 kg

b). Change in resistance

$k=\frac{\Delta R/R}{\Delta L/L}$

$\Delta R = k. \epsilon R$ , since $\epsilon= \Delta L/ L$

$\Delta R = 6.9 \times 0.03 \times 340$

$\Delta R = 70.38$ Ω

For 4 resistance of the sensors,

$\Delta R = 70.38 \times 4 = 281.52$ Ω

c). $k=\frac{\Delta R/R}{\epsilon}$

If linear strain,

$\frac{\Delta R}{R} \approx \frac{\Delta L}{L}$ , where k = 1

$\Delta R = \frac{\Delta L}{L} \times R$

$\Delta R = 0.03 \times 340$

$\Delta R = 10.2$ Ω

4 0

3 years ago

An automotive fuel cell consumes fuel at a rate of 28m3/h and delivers 80kW of power to the wheels. If the hydrogen fuel has a h

EastWind [94]

Answer:

The efficiency of this fuel cell is 80.69 percent.

Explanation:

From Physics we define the efficiency of the automotive fuel cell ( $\eta$ ), dimensionless, as:

$\eta = \frac{\dot W_{out}}{\dot W_{in}}$ (Eq. 1)

Where:

$\dot W_{in}$ - Maximum power possible from hydrogen flow, measured in kilowatts.

$\dot W_{out}$ - Output power of the automotive fuel cell, measured in kilowatts.

The maximum power possible from hydrogen flow is:

$\dot W_{in} = \dot V\cdot \rho \cdot L_{c}$ (Eq. 2)

Where:

$\dot V$ - Volume flow rate, measured in cubic meters per second.

$\rho$ - Density of hydrogen, measured in kilograms per cubic meter.

$L_{c}$ - Heating value of hydrogen, measured in kilojoules per kilogram.

If we know that $\dot V = \frac{28}{3600}\,\frac{m^{3}}{s}$ , $\rho = 0.0899\,\frac{kg}{m^{3}}$ , $L_{c} = 141790\,\frac{kJ}{kg}$ and $\dot W_{out} = 80\,kW$ , then the efficiency of this fuel cell is:

(Eq. 1)

$\dot W_{in} = \left(\frac{28}{3600}\,\frac{m^{3}}{s}\right)\cdot \left(0.0899\,\frac{kg}{m^{3}} \right)\cdot \left(141790\,\frac{kJ}{kg} \right)$

$\dot W_{in} = 99.143\,kW$

(Eq. 2)

$\eta = \frac{80\,kW}{99.143\,kW}$

$\eta = 0.807$

The efficiency of this fuel cell is 80.69 percent.

3 0

3 years ago

What is the first thing to do when you make a three-point turn?.

Jlenok [28]

Answer:

1. Move as far right as possible, check traffic, and signal a left turn.

2. Turn the steering wheel sharply to the left and move forward slowly.

3. Shift to reverse, turn your wheels sharply to the right, check traffic, and back your vehicle to the right curb, or edge of roadway.

5 0

3 years ago

Determine whether w is in the span of the given vectors v1; v2; : : : vn

Brrunno [24]

Determine whether w is in the span of the given vectors v1; v2; : : : vn

. If your answer is yes, write w as a linear combination of the vectors v1; v2; : : : vn and enter the coefficients as entries of the matrix as instructed is given below

Explanation:

1.Vector to be in the span means means that it contain every element of said vector space it spans. So if a set of vectors A spans the vector space B, you can use linear combinations of the vectors in A to generate any vector in B because every vector in B is within the span of the vectors in A.

2.And thus v3 is in Span{v1, v2}. On the other hand, IF all solutions have c3 = 0, then for the same reason we may never write v3 as a sum of v1, v2 with weights. Thus, v3 is NOT in Span{v1, v2}.

3.In the theory of vector spaces, a set of vectors is said to be linearly dependent if at least one of the vectors in the set can be defined as a linear combination of the others; if no vector in the set can be written in this way, then the vectors are said to be linearly independent.

4.Given a set of vectors, you can determine if they are linearly independent by writing the vectors as the columns of the matrix A, and solving Ax = 0. If there are any non-zero solutions, then the vectors are linearly dependent. If the only solution is x = 0, then they are linearly independent.

7 0

3 years ago

A train starts from rest at station A and accelerates at 0.4 m/s^2 for 60 s. Afterwards it travels with a constant velocity for

mash [69]

<h3><u>The distance between the two stations is</u><u> </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>

$\\$

Explanation:

<h2>Given:</h2>

$a_1 \:=\:0.4\:m/s²$

$t_1 \:=\:60\:s$

$v_{i1} \:=\:0\:m/s$

$a_2 \:=\:0\:m/s²$

$t_2 \:=\:25\:min\:=\:1500\:s$

$a_3 \:=\:-0.8\:m/s²$

$v_{f3} \:=\:0\:m/s$

$\\$

<h2>Required:</h2>

Distance from Station A to Station B

$\\$

<h2>Equation:</h2>

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v\:=\:\frac{d}{t}$

$\\$

<h2>Solution:</h2><h3>Distance when a = 0.4 m/s²</h3>

Solve for $v_{f1}$

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$0.4\:m/s²\:=\:\frac{v_f\:-\:0\:m/s}{60\:s}$

$24\:m/s\:=\:v_f\:-\:0\:m/s$

$v_f\:=\:24\:m/s$

$\\$

Solve for $v_{ave1}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v_{ave}\:=\:\frac{0\:m/s\:+\:24\:m/s}{2}$

$v_{ave}\:=\:12\:m/s$

$\\$

Solve for $d_1$

$v\:=\:\frac{d}{t}$

$12\:m/s\:=\:\frac{d}{60\:s}$

$720\:m\:=\:d$

$d_1\:=\:720\:m$

$\\$

<h3>Distance when a = 0 m/s²</h3>

$v_{f1}\:=\:v_{i2}$

$v_{i2}\:=\:24\:m/s$

$\\$

Solve for $v_{f2}$

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$0\:m/s²\:=\:\frac{v_f\:-\:24\:m/s}{1500\:s}$

$0\:=\:v_f\:-\:24\:m/s$

$v_f\:=\:24\:m/s$

$\\$

Solve for $v_{ave2}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v_{ave}\:=\:\frac{24\:m/s\:+\:24\:m/s}{2}$

$v_{ave}\:=\:24\:m/s$

$\\$

Solve for $d_2$

$v\:=\:\frac{d}{t}$

$24\:m/s\:=\:\frac{d}{1500\:s}$

$36,000\:m\:=\:d$

$d_2\:=\:36,000\:m$

$\\$

<h3>Distance when a = -0.8 m/s²</h3>

$v_{f2}\:=\:v_{i3}$

$v_{i3}\:=\:24\:m/s$

$\\$

Solve for $v_{f3}$

$a\:=\:\frac{v_f\:-\:v_i}{t}$

$-0.8\:m/s²\:=\:\frac{0\:-\:24\:m/s}{t}$

$(t)(-0.8\:m/s²)\:=\:-24\:m/s$

$t\:=\:\frac{-24\:m/s}{-0.8\:m/s²}$

$t\:=\:30\:s$

$\\$

Solve for $v_{ave3}$

$v_{ave}\:=\:\frac{v_i\:+\:v_f}{2}$

$v_{ave}\:=\:\frac{24\:m/s\:+\:0\:m/s}{2}$

$v_{ave}\:=\:12\:m/s$

$\\$

Solve for $d_3$

$v\:=\:\frac{d}{t}$

$12\:m/s\:=\:\frac{d}{30\:s}$

$360\:m\:=\:d$

$d_3\:=\:360\:m$

$\\$

<h3>Total Distance from Station A to Station B</h3>

$d\:= \:d_1\:+\:d_2\:+\:d_3$

$d\:= \:720\:m\:+\:36,000\:m\:+\:360\:m$

$d\:= \:37,080\:m$

$d\:= \:37.08\:km$

$\\$

<h2>Final Answer:</h2><h3><u>The distance between the two stations is </u><u>3</u><u>7</u><u>.</u><u>0</u><u>8</u><u> km</u></h3>

7 0

3 years ago