[sorry im doing this again] When does an electrically charged object attract another object?

MamaMia's Pizza purchases its pizza delivery boxes from a printing supplier. MamaMia's delivers on-average 200 pizzas each month

Ira Lisetskai [31]

Answer:

a) 138 units

b) 17 units

c) 17 units

d) Total Cost = $353.35

Explanation:

Given:

Average pizzas delivered = 200

Charge of inventory holding = 30% of cost

Lead time = 7 days

Now,

a) Economic Order Quantity = $\sqrt\frac{2\times\textup{Annual Demand}\times\textup{Cost per Order}}{\textup{Carrying cost}}$

also,

Annual Demand = 200 × 12 = 2400

Cost per Order = Cost of Box + Processing Costs

= 30 cents + $10

= $10.30

and, Carrying Cost = $\frac{\textup{Total Inventory Cost}}{\textup{total annual demand}}$

= $\frac{\textup{Total Cost per order}\times\textup{Annual demand}\times\frac{25}{100}}{\textup{Annual demand}}$

= $\frac{\$10.30\times2400}\times\frac{25}{100}}{2400}$

= $2.575

Therefore,

Economic Order Quantity = $\sqrt\frac{2\times\textup{2400}\times\textup{10.30}}{\textup{2.575}}$

= 138.56 ≈ 138 units

b) Reorder Point

= (average daily unit sales × the lead time in days) + safety stock

= ( $\frac{200}{30}\times7$

= 46.67 ≈ 47 units

c) Number of orders per year = $\frac{\textup{Annual Demand}}{\textup{Economic order quantity}}$

= $\frac{\textup{2400}}{\textup{138}}$

= 17.39 ≈ 17 units

d) Total Annual Cost (Total Inventory Cost)

= Ordering Cost + Holding Cost

Now,

The ordering Cost = Cost per Order × Total Number of orders per year

= $10.30 × 17

= $175.1

and,

Holding Cost = Average Inventory Held × Carrying Cost per unit

Average Inventory Held = $\frac{\textup{0+138}}{\textup{2}}$ = 69

Carrying Cost per unit = $2.575

Holding Cost = 69 × $2.575 =

$177.675

Therefore,

Total Cost = Ordering Cost + carrying cost

= $175.1 + $177.675 = $353.35

5 0

3 years ago

Read 2 more answers

How will the amount of power change if less work is done in more time?

yKpoI14uk [10]

The amount of power change if less work is done in more time"then the amount of power will decrease".

<u>Option: B</u>

<u>Explanation:</u>

The rate of performing any work or activity by transferring amount of energy per unit time is understood as power. The unit of power is watt

$Power = \frac{Work}{Time}$

Here this equation showcase that power is directly proportional to the work but dependent upon time as time is inversely proportional to the power i.e as time increases power decreases and vice versa.

This can be understood from an instance, on moving a load up a flight of stairs, the similar amount of work is done, no matter how heavy but when the work is done in a shorter period of time more power is required.

7 0

3 years ago

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8m/s2 . The accel

Zanzabum

Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.

Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m

Now you need the final speed to use it as initial speed of the next part.

Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s

Part B) Free fall

Maximum height, y max ==> Vf = 0

Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s

ymax = yo + Vo*t - g[t^2] / 2

ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m

Answer: ymax = 10084.2m

8 0

3 years ago

Which of the following statements is accurate? A. Sound waves passing through the air will do so as transverse waves, which vibr

egoroff_w [7]

The answer is, C. the wavelength is measured in parallel to the direction of the wave, at any point, under the same repetition for any type of wave.

7 0

3 years ago

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Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

$ma=\dfrac{kq^2}{r^2}$

$a=\dfrac{kq^2}{mr^2}$

$a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}$

$a=9.91\times 10^{15}\ m/s^2$

So, the acceleration of two isolated protons is $9.91\times 10^{15}\ m/s^2$ . Hence, this is the required solution.

3 0

3 years ago