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3 years ago

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How much will it cost to train the entire company to use a recycling program if the training includes paper handouts? (Remember,

the training will take one hour and a team of three people will take three hours each to create the training program.) With the current budget, is it possible to train the company and buy recycling carts if recycling pickup in your city is free? Assume the best choice for recycle carts in your area was $80 per cart.

1 answer:

bezimeni [28]3 years ago

8 0

Answer:

With the current budget of $1,000, it is possible to train the company and buy 3 recycling carts given that recycling pickup within the area is free

Explanation:

The office waste management budget that is not being spent = $1,000

The given expense parameters are;

The training cost per hour per employee = $12.00

The number of people in the entire company = 45

The cost of preparing the training material per hour = $20.00

The number of people to create the training material = 3 people

The time it will take each person in creating the training material = 3 hours

The cost of recycling cart = $80.00 per cart

The cost of paper handouts per employee = $0.05

The cost of materials include;

The total cost for the training = $12.00 × 45 = $540

The cost for the handout = $0.05 × 45 = $2.25

The cost of preparing the materials = $20.00 × 3 × 3 = $180.00

The total costs of the training = $540 + $4 + $180.00 = $724

The amount available to buy cart = $1,000 - $724 = $276

Therefore;

The amount available to buy cart = $276

The number of carts that can be bought = 276/80 = 3.45 carts

Therefore, we round down to get;

The number of carts that can be bought = 3 carts

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Realiza las siguientes conversiones.

amid [387]

Answer:

a) 4 hectómetros cuadrados equivalen a 400 decámetros cuadrados.

b) 21345 centímetros cuadrados equivalen a 2,135 metros cuadrados.

c) 0,592 kilómetros cuadrados equivalen a 592000 metros cuadrados.

d) 0,102 metros cuadrados equivalen a 1020 centímetros cuadrados.

e) 23911 kilómetros cuadrados equivalen 2391100 hectómetros cuadrados.

Explanation:

a) <em>4 hectómetros cuadrados a decámetros cuadrados:</em>

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un hectómetro cuadrado equivale a 100 decámetros cuadradps. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

$x = 4\,Hm^{2}\times\frac{100\,Dm^{2}}{1\,Hm^{2}}$

$x = 400\,Dm^{2}$

4 hectómetros cuadrados equivalen a 400 decámetros cuadrados.

b) <em>21345 centímetros cuadrados a metros cuadrados:</em>

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un metro cuadrado equivale a 10000 centímetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

$x = 21345\,cm^{2}\times \frac{1\,m^{2}}{10000\,cm^{2}}$

$x = 2,135\,m^{2}$

21345 centímetros cuadrados equivalen a 2,135 metros cuadrados.

c) <em>0,592 kilómetros cuadrados a metros cuadrados:</em>

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un kilómetro cuadrado equivale a 1000000 metros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

$x = 0,592\,km^{2}\times \frac{1000000\,m^{2}}{1\,km^{2}}$

$x = 592000\,m^{2}$

0,592 kilómetros cuadrados equivalen a 592000 metros cuadrados.

d) <em>0,102 metros cuadrados a centímetros cuadrados:</em>

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un metro cuadrado equivale a 10000 centímetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

$x = 0,102\,m^{2}\times \frac{10000\,cm^{2}}{1\,m^{2}}$

$x = 1020\,cm^{2}$

0,102 metros cuadrados equivalen a 1020 centímetros cuadrados.

e) <em>23911 kilómetros cuadrados a hectómetros cuadrados:</em>

Según las unidades de área y sus escalas utilizadas por el Sistema Internacional de Pesos y Medidas, un kilómetro cuadrado equivale a 100 hectómetros cuadrados. Entonces, obtenemos el dato equivalente por la siguiente regla de tres simple:

$x = 23911\,km^{2}\times \frac{100\,Hm^{2}}{1\,km^{2}}$

$x = 2391100\,Hm^{2}$

23911 kilómetros cuadrados equivalen 2391100 hectómetros cuadrados.

7 0

4 years ago

thermodynamics A nuclear power plant based on the Rankine cycle operates with a boiling-water reactor to develop net cycle power

IrinaK [193]

Answer:

(a) the percent thermal efficiency is 27.94%

(b) the temperature of the cooling water exiting the condenser is 31.118°C

Explanation:

3 0

3 years ago

The closed tank of a fire engine is partly filled with water, the air space above being under pressure. A 6 cm bore connected to

skelet666 [1.2K]

Answer:

The air pressure in the tank is 53.9 $kN/m^{2}$

Solution:

As per the question:

Discharge rate, Q = 20 litres/ sec = $0.02\ m^{3}/s$

(Since, 1 litre = $10^{-3} m^{3}$ )

Diameter of the bore, d = 6 cm = 0.06 m

Head loss due to friction, $H_{loss} = 45 cm = 0.45\ m$

Height, $h_{roof} = 2.5\ m$

Now,

The velocity in the bore is given by:

$v = \frac{Q}{\pi (\frac{d}{2})^{2}}$

$v = \frac{0.02}{\pi (\frac{0.06}{2})^{2}} = 7.07\ m/s$

Now, using Bernoulli's eqn:

$\frac{P}{\rho g} + \frac{v^{2}}{2g} + h = k$ (1)

The velocity head is given by:

$\frac{v_{roof}^{2}}{2g} = \frac{7.07^{2}}{2\times 9.8} = 2.553$

Now, by using energy conservation on the surface of water on the roof and that in the tank :

$\frac{P_{tank}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{tank} = \frac{P_{roof}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{roof} + H_{loss}$

$\frac{P_{tank}}{\rho g} + 0 + 0 = \0 + 2.553 + 2.5 + 0.45$

$P_{tank} = 5.5\times \rho \times g$

$P_{tank} = 5.5\times 1\times 9.8 = 53.9\ kN/m^{2}$

4 0

3 years ago

Please read and an<br><br> 3. Many Jacks use hydraullc power.<br> A) O True<br> B) O False

drek231 [11]

Answer:

A) True. I hope this helps

5 0

3 years ago

The force of T = 20 N is applied to the cord of negligible mass. Determine the angular velocity of the 20-kg wheel when it has r

horrorfan [7]

Image of wheel is missing, so i attached it.

Answer:

ω = 14.95 rad/s

Explanation:

We are given;

Mass of wheel; m = 20kg

T = 20 N

k_o = 0.3 m

Since the wheel starts from rest, T1 = 0.

The mass moment of inertia of the wheel about point O is;

I_o = m(k_o)²

I_o = 20 * (0.3)²

I_o = 1.8 kg.m²

So, T2 = ½•I_o•ω²

T2 = ½ × 1.8 × ω²

T2 = 0.9ω²

Looking at the image of the wheel, it's clear that only T does the work.

Thus, distance is;

s_t = θr

Since 4 revolutions,

s_t = 4(2π) × 0.4

s_t = 3.2π

So, Energy expended = Force x Distance

Wt = T x s_t = 20 × 3.2π = 64π J

Using principle of work-energy, we have;

T1 + W = T2

Plugging in the relevant values, we have;

0 + 64π = 0.9ω²

0.9ω² = 64π

ω² = 64π/0.9

ω = √64π/0.9

ω = 14.95 rad/s

4 0

4 years ago