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irina1246 [14]
3 years ago
12

Suppose a vacuum cleaner uses 120 J of electrical energy. If 45 J are used to pull air into the vacuum cleaner, how efficient is

the vacuum cleaner.
Physics
1 answer:
postnew [5]3 years ago
3 0
I believe that if you multiply 120 x 45 you would get 5,400 why? Because when a problem has "of" means to multiply.

So... the vacuum cleaner is 5,400 percent efficient :)

Sorry if it's wrong I'm only in fifth grade
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A child accidentally drops a toy from his apartment window. It falls for 1.05 seconds. What is the velocity of the toy just befo
AVprozaik [17]

Sorry!

This cannot be answered. We don't have weight, height, etc.

6 0
3 years ago
5. In a local bar, a customer slides an empty beer mug on the counter for a refill. The bartender does not see the mug, which sl
pishuonlain [190]

Answer:

sorry about this question

5 0
3 years ago
What kind of organisms are able to convert sunlight into usable energy?
Tamiku [17]

Answer:

Photosynthetic organisms

Explanation:

The electromagnetic energy of sunlight is converted to chemical energy in the chlorophyll-containing cells of photosynthetic organisms. In eukaryotic cells these reactions occur in the organelle known as the chloroplast

Hope this helps! :)

4 0
3 years ago
Read 2 more answers
What is the volume in liters of the basketball?
babymother [125]

he basket ball diameter used by NBA men players is around 9.55 inches

So diameter = 9.55 inch = 0.243 m

So radius of basket ball = 0.1215 m

Volume , V=\frac{4}{3} \pi r^3

V = \frac{4}{3}* \pi *0.1215^3=0.0075 m^3=7.5 L

So volume of basket ball used by men NBA players = 7.5 L

   

5 0
3 years ago
Read 2 more answers
A 2.4-m high 200-m2 house is maintained at 22°C by an air-conditioning system whose COP, is 3.2. It is estimated that the kitch
kotegsom [21]

Answer:

The amount that is "vented" out by "the fans" is <u>$0.50</u> for 10 hours.

Option: a

<u>Explanation</u>:

"Energy discharged by air in every hour" can be determined by,

\mathrm{Q}=\mathrm{m}_{\mathrm{air}} \mathrm{C}_{\mathrm{p}} \Delta \mathrm{T}

Q = heat energy (Joules, J)  

m = mass of a substance (kg)  

c = specific heat (units J/kg∙K)  

\mathrm{m}_{\mathrm{air}}=\rho \mathrm{v}

\text { Density of air } \rho=1.20 \mathrm{kg} / \mathrm{m}^{3}

\text { Density of air } \rho=1.20 \times 200 \times 2.4

\text { Density of air } \rho=576 \mathrm{kg}

∆T = 10 hours

\text { Specific Heat Capacities of Air. The nominal values used for air at } 300 \mathrm{K} \text { are } \mathrm{C_P}=1.00 \mathrm{kJ} / \mathrm{kg} . \mathrm{K}

Q = 576 × 1.00 × 10

Q = 5760 kJ/hours

W = 1.6 kwh

We know that, “Coefficient of performance” (COP)

\mathrm{Cop}=\frac{Q}{w}

\mathrm{W}=\frac{Q}{\mathrm{cop}}

Given that, COP = 3.2

\mathrm{W}=\frac{1.6}{3.2}

W = 0.5 kwh

The unit cost of electricity is $0.10/kWh

The unit cost of electricity is $0.10/kWh

Unit electricity cost for 10 hours = 0.5 × 10 × 0.1$

Unit electricity cost for 10 hours = $0.5

The amount that is "vented out" by "the fans" is $0.50 for 10 hours.

6 0
3 years ago
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