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3 years ago

The unit of electric potential or electromotive force is the

Physics

1 answer:

goldenfox [79]3 years ago

3 0

Answer:

Volt

Explanation:

Electric Potential is the Voltage and it's measured in volt

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As a result of photosynthesis is the production of sugar molecules known as what?

Bezzdna [24]

Answer:

Glucose

Explanation:

4 0

3 years ago

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The acceleration of automobiles is often given in terms of the time it takes to go from 0 mi/h to 60 mi/h. One of the fastest st

Rudiy27

Answer:

9.934 m/s²

Explanation:

Given:

Initial speed of the Bugatti Veyron Super Sport = 0 mi/h

Final speed of the Bugatti Veyron Super Sport = 60 mi/h

Now,

1 mi/h = 0.44704 m / s

thus,

60 mi/h = 0.44704 × 60 = 26.8224 m/s

Time = 2.70 m/s

Now,

The acceleration (a) is given as:

$a=\frac{\textup{Change in speed}}{\textup{Time}}$

thus,

$a=\frac{26.8224 - 0 }{2.70}$

a = 9.934 m/s²

6 0

2 years ago

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

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Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1

hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

$\sum F=ma$

We know that the force between two bodies due to gravity is given by the following equation:

$F_{g} = G\frac{m_{1}m_{2}}{r^{2}}$

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

$F_{g} =G\frac{Mm}{r^{2}}$

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

$\omega = \frac{2\pi}{T}$

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

so:

$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

so now we can do the sum of forces:

$\sum F=ma$

$F_{g}=ma_{c}$

$G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})$

in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

$T_{1}^{2}>T_{2}^{2}$

So let's see what's going on there, we'll call:

$M_{1}$ = mass of Star 1

$M_{2}$ = mass of Star 2

So:

$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

$\frac{1}{M_{1}}>\frac{1}{M_{2}}$

and let's flip the inequality, so we get:

$M_{2}>M_{1}$

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0

2 years ago

Find the force between 2C and -1C separated by a distance 1m. Indicate if it is an attractive force (negative) or a repulsive fo

telo118 [61]

Answer:

Explanation:

Its definitely an Attractive force since the two charges are Unlike.

From Coulombs Law

F=kq1q2/R²

Given

K=9x10^9

R=1m

q1=2C

q2=-1C

F=(9x10^9 x 2 x -1)/1²

F= - 1.8x10^10N. (Attractive).

6 0

2 years ago