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3 years ago

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a. Replacing standard incandescent lightbulbs with energy-efficient compact fluorescent lightbulbs can save a lot of energy. Cal

culate the amount of energy saved over 10 h when one 60 W incandescent lightbulb is replaced with an equivalent 18 W compact fluorescent lightbulb. (3 points) b. How long could you run a 300 W plasma television set on the saved energy

1 answer:

monitta3 years ago

4 0

Answer:

a) 1512000 Joules

b) 5040 seconds = 84 minutes = 1.4 hours

Explanation:

a) Power saved by replacing bulbs = 60-18 = 42 W = 42 J/s

Time the bulb is used for = 10 hours

Energy saved during this time

42×10×60×60 = 1512000 Joules

Saved energy by replacing standard incandescent lightbulbs with energy-efficient compact fluorescent lightbulbs in 10 hours is 1512000 Joules

b) Power the plasma TV uses = 300 W = J/s

$\frac{1512000}{300}=5040\ s$

Time a plasma TV can be used for with the saved energy is 5040 seconds = 84 minutes = 1.4 hours.

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An ideal fluid flows through a pipe made of two sections with diameters of 1.0 and 3.0 inches, respectively. The speed of the fl

geniusboy [140]

Answer:

$(\frac{r_1}{r_2})^2=\frac{1}{9}$

Explanation:

From the question we are told that:

Diameter 1 $d_1=1.0$

Diameter 2 $d_2=3.0$

Generally the equation for Radius is mathematically given by

At Diameter 1

$r_{1}=\frac{1}{2} inch$

At Diameter 2

$r_{2}=\frac{3}{2} inch$

Generally the equation for continuity is mathematically given by

$A_1V_1=A_2V_2$

Therefore

$(\frac{r_1}{r_2})^2=(\frac{1/2}{3/2})^2$

$(\frac{r_1}{r_2})^2=\frac{1}{9}$

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3 years ago

A scale model is 4th the size of the pump. Determine the power ratio of the pump and its scale model if the ratio of the heads i

vredina [299]

Given:

size of scale model = 4(size of pump)

power ratio of pump and scale model = 5:1

Solution:

Let the diameter of scale model and pump be $d_{s}$ and $d_{p}$ respectively

and head be $H_{s}$ and $H_{p}$ respectively

Now, power, P is given as a function of head(H) and dischagre(Q)

P = $\rho gQH$ (1)

From eqn (1):

$P \propto QH$

and

$QH \propto \sqrt{H}D^{2}$

So,

$P \propto H^{\frac{3}{2}} D^{2}$

Therefore,

$\frac{P_{s}}{P_{p}}$ = $\frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}$

$\frac{P_{s}}{P_{p}}$ = $\frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}$

$\frac{P_{s}}{P_{p}}$ = $\frac{1^{2}\times 5^{\frac{3}{2}}}{4^{2}\times 1^{\frac{3}{2}}}$

$\frac{P_{s}}{P_{p}}$ = $\frac{5\sqrt{5}}{16}$

${P_{s}}:{P_{p}}$ = ${5\sqrt{5}}:{16}$

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3 years ago

) Chain drive ratio is a the ratio between the rotational speeds of the input and output _________ of a roller chain drive syste

Elis [28]

Answer:

Sprockets.

Explanation:

A chain drive is an efficient technique used for the transmission of mechanical power from one point to another. For example, it is used for transmitting power to the wheels of a bicycle, motorcycle, motor vehicle and other machineries such as chain saw etc.

Chain drive ratio is the ratio between the rotational speeds of the input and output sprockets of a roller chain drive system. This ultimately implies that, chain drive ratio is the ratio of the number of teeth on the driving sprocket (T1) divided to the number of teeth on the driven sprocket (T2).

Also, the chain drive ratio can be calculated by dividing the number of teeth on the large sprocket by the number of teeth on the small sprocket.

Additionally, the rotational speed of a sprocket is measured in revolutions per minute (RPM).

One of the issues with the roller chain is that, as the roller chain moves round the sprocket link by link, it affects its speed (surge) due to the change in acceleration and deceleration i.e the rise and fall of its pitch line.

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3 years ago

An organization sets its standards for quality according to the best product it can produce.

Marianna [84]

I believe it’s True, but please correct me if I’m wrong!

6 0

3 years ago

Read 2 more answers

A spur gearset has a module of 6 mm and a velocity ratio of 4. The pinion has 16 teeth. Find the number of teeth on the driven g

levacccp [35]

Answer:

NG=64 teeth

dG=384mm

dP=96mm

C=240mm

Explanation:

step one:

given data

module m=6mm

velocity ratio VR=4

number of teeth of pinion Np=16

<u>Step two:</u>

<u>Required</u>

1. Number of teeth on the driven gear

$N_G=N_P*V_R\\\\N_G=16*4\\\\N_G=64$

<em>The driven gear has 64 teeth</em>

2. The pitch diameters

The driven gear diameter

$d_G=N_G*m\\\\d_G=64*6\\\\d_G=384$

<em>The driven gear diameter is 384mm</em>

The pinion diameter

<em /> $d_P=N_P*m\\\\d_P=16*6\\\\d_P=96$ <em />

Pinion diameter is 96mm

3. Theoretical center-to-center distance

$C=\frac{d_G+d_P}{2} \\\\C=\frac{384+96}{2} \\\\C=\frac{480}{2}\\\\C=240$

The theoretical center-to-center distance is 240mm

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3 years ago