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Describe 2 ways you could increase the efficiency of a household central heating system

chubhunter [2.5K]

Answer:

Arrange an annual service. Treat your boiler like your car. ...

Keep your boiler clean. ...

Bleed your radiators. ...

Top up the pressure. ...

Use a Powerflush. ...

Insulate your pipes. ...

Turn the heating on. ...

If all else fails…

Explanation:

6 0

3 years ago

What is the magnitude of δv12 if the bulb is removed from the socket (i.e. the circuit is not closed)?

andreev551 [17]

Since the circuit is incomplete or not closed, no current flows in the circuit. as per ohm's law , Voltage is directly proportional to current and is given as

V = Voltage = i R where i = current , R = resistance

as no current flows in the circuit, i = 0

the resistance R can not be zero. hence

V = 0 (R)

V = 0 Volts

so the magnitude of the Voltage is zero Volts

4 0

3 years ago

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Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass

Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

$\Delta W = \Delta PE + \Delta KE$

Where,

PE = Potential Energy

KE = Kinetic Energy

$\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2$

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

$P = \frac{\Delta W}{\Delta t}$

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

The rate of mass flow is,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

Where,

$\rho_w$ = Density of water

A = Area of the hose $\rightarrow A=\pi r^2$

The given radius is 0.83cm or $0.83 * 10^{-2}$ m, so the Area would be

$A = \pi (0.83*10^{-2})^2$

$A = 0.0002164m^2$

We have then that,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

$\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)$

$\frac{\Delta m}{\Delta t} = 1.16856kg/s$

Final the power of the pump would be,

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

$P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)$

$P = 57.1192W$

Therefore the power of the pump is 57.11W

6 0

3 years ago

What term refers to the part of a spacecraft that is occupied by the crew for takeoff and landing?

Semenov [28]

Command module ✅

service module

lunar module

annum module

7 0

2 years ago

We know that every object exerts an attraction on every other object and the heavier the object ___ the attraction.

hichkok12 [17]

I don't know what the exact word is, but I do know that the bigger an objects mass is the more it will attract other objects toward it, mainly smaller objects with less mass. it might be gravity or something around those lines....is it a multiple choice question?

6 0

3 years ago