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3 years ago

Define light yearend explain how and why light years are used to measure distances in the universe

1 answer:

3 0

Light years are the distance light can travel in a full year. They are used because there are really big distances in the universe that are easier to measure with a really large form of measurement (light years).

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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

What happens to its surface temperature as a star changes color from red to blue

MAVERICK [17]

The surface temperature of the star increases when it changes its colour changes from red to blue.
Blue>white>yellow>orange>red
The given series is the temperature decreasing order of stars. Now our sun is considered as a yellow star.

5 0

3 years ago

Two loudspeakers are 1.60 m apart. A person stands 3.00 m from one speaker and 3.50 m from the other. (a) What is the lowest fre

VMariaS [17]

Answer:

Explanation:

Given

Distance between two loud speakers $d=1.6\ m$

Distance of person from one speaker $x_1=3\ m$

Distance of person from second speaker $x_2=3.5\ m$

Path difference between the waves is given by

$x_2-x_1=(2m+1)\cdot \frac{\lambda }{2}$

for destructive interference m=0 I.e.

$x_2-x_1=\frac{\lambda }{2}$

$3.5-3=\frac{\lambda }{2}$

$\lambda =0.5\times 2$

$\lambda =1\ m$

frequency is given by

$f=\frac{v}{\lambda }$

where $v=velocity\ of\ sound\ (v=343\ m/s)$

$f=\frac{343}{1}=343\ Hz$

For next frequency which will cause destructive interference is

i.e. $m=1$ and $m=2$

$3.5-3=\frac{2\cdot 1+1}{2}\cdot \lambda$

$\lambda =\frac{1}{3}\ m$

frequency corresponding to this is

$f_2=\frac{343}{\frac{1}{3}}=1029\ Hz$

for $m=2$

$3.5-3=\frac{5}{2}\cdot \lambda$

$\lambda =\frac{1}{5}\ m$

Frequency corresponding to this wavelength

$f_3=\frac{343}{\frac{1}{5}}$

$f_3=1715\ Hz$

8 0

3 years ago

X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

a) $\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

3 0

3 years ago

Read 2 more answers

How does the structure of mitochondrial on affect its structure

anyanavicka [17]

It is itself. This question does not make sense.

7 0

3 years ago