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pshichka [43]
3 years ago
12

A market research survey has 15 questions and will be sent to 500 people. What is the total cost to conduct survey if it has a $

1,000 base fee plus $5 per question and $2 per person survey
Engineering
1 answer:
Dimas [21]3 years ago
8 0

Answer:

$39,500

Explanation:

In order to find the total cost, we proceed as follows:

1)

First, the base fee is:

b=\$1000

2)

Then we calculate the total number of questions: we have 15 questions per survey, and a total of 500 surveys (for 500 people), therefore the total number of questions is

Q=15\cdot 500 =7500

Since the cost is $5 per question, the total cost of the questions is

q=\$5\cdot 7500 =\$37,500

3)

Finally, we have 500 persons, and there is a cost of $2 per each person, so the cost due to this is

p=\$2\cdot 500 = \$1000

Therefore, the total cost is:

T=b+q+p=1000+37,500+1000=\$39,500

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The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

<h3>Thickness of the aluminum</h3>

The thickness of the aluminum can be determined using from distance of closest approach of the particle.

K.E = \frac{2KZe^2}{r}

where;

  • Z is the atomic number of aluminium  = 13
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<h3>For 2.5 MeV electrons</h3>

r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (1.6\times 10^{-19})^2}{2.5 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m

<h3>For 2.5 MeV protons</h3>

Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.

<h3>For 10 MeV alpha-particles</h3>

Charge of alpah particle = 2e

r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (2 \times 1.6\times 10^{-19})^2}{10 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m

Thus, the thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

Learn more about closest distance of approach here: brainly.com/question/6426420

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