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3 years ago

CO2 , NaCl , HCl can be classified as ?

Physics

1 answer:

ExtremeBDS [4]3 years ago

4 0

I suppose they can all be classified as compounds

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Electrical energy is used to turn the blades of a fan. The amount of energy transformed is seen here : 750 J electrical energy i

Charra [1.4K]

Electrical energy is used to run the fan

Here as per given condition 750 J of electrical energy is used to run the fan which is converted into Kinetic energy as 400 J

So here we can see that 350 J of energy is lost against many other type of frictional and resistive loses.

So here we can say that out of 750 J of energy only 400 J is used to run the fan and rest amount of energy is lost against friction.

also we can say that efficiency of this fan will be

$\eta = \frac{Useful}{Total}$

$\eta = \frac{400}{750}$

$\eta = \frac{8}{25}$

3 0

2 years ago

A retail gasoline pump is able to fill a 14.0 gallon automobile tank in 1.50 minutes. What is the mass flow rate deliverd by the

Ludmilka [50]

We have volume of gasoline = 14.0 gallon

Time taken to fill automobile tank = 1.50 minutes

So volume rate = 14.0 gallon/1.50 minutes = 9.33 gallon/ minute

We have density of gasoline = 0.77 kg/L = 6.073 lb/US gal

Mass rate = Density * Volume rate

= 9.33 gallon/ minute*6.073 lb/US gal = 56.68 lb/min

So mass flow rate delivered by the gasoline pump in lbm/min = 56.68

6 0

3 years ago

Read 2 more answers

What are the 3 tools used to collect weather data

sergey [27]

Anemometer, Psychrometer, <span>Barometer</span>

5 0

2 years ago

Take a close look at the energy transfers and transformations shown in the above diagram. Which type of energy is transformed in

Elanso [62]

Answer:

kinetic energy

Explanation:

a certain amount of energy is transferred by the kick. The ball gains an equal amount of energy, mostly in the form of kinetic energy.

4 0

2 years ago

Read 2 more answers

A charged object traveling 7 m in a uniform electric field of 5 N/C experiences a 4 J increase in Kinetic Energy.

Travka [436]

To solve this problem it is necessary to apply the principles of conservation of Energy in order to obtain the final work done.

The electric field in terms of the Force can be expressed as

$E = \frac{F}{q} \rightarrow F=Eq$

Where,

F = Force

E= Electric Field

q = Charge

Puesto que el trabajo realizado es equivalente al cambio en la energía cinetica entonces tenemos que

KE = W

KE = F*d

In the First Case,

$4 = (qE)d\\q = \frac{4}{Ed}\\q = \frac{4}{5*7}\\q = 0.1142C$

In Second Case,

$KE = q E'd$

$KE = (0.1142)(40)(7)$

$KE = 31.976J$

The total energy change would be subject to,

$\Delta KE = 31.976-4$

$\Delta KE = 27.976J$

Therefore the Kinetic Energy change of the charged object is 27.976J

3 0

2 years ago