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3 years ago

Which type of central heating system involves fans and ducts to circulate warm air?

Physics

2 answers:

attashe74 [19]3 years ago

8 0

Answer: forced-air heating

Explanation:

satela [25.4K]3 years ago

3 0

Forced-Air would be the type of central heating system

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When asked how to create an electromagnet, the best answer would be:

fredd [130]

Answer:

You can create an electromagnet by wrapping an insulated wire around a metal with ferromagnetic properties and applying an electric current."

Explanation:

Electromagnets are made by wrapping an insulated wire around a metal with ferromagnetic properties (example is iron), to form a loop, and then applying a current through the wire. Electromagnets can generate magnetism with a strong force field, and unlike normal magnets, their strength can be varied by varying the amount of current flowing through the coil. Their main disadvantage, which is also their most utilized property is that their magnetism is lost once the current flowing through the wire is cut-off.

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3 years ago

On the decibel scale, an increase of 10 decibels means an intensity increase of __________times?

Sonbull [250]

Answer:

Explanation:

The decibel or decibel, is a unit that relates two values of sound pressure, and electrical power. the base unit is the bel, but given the amplitude but for practicality, a submultiple, the decibel, is used. It is a scalar expression .

Zero belios is the reference value used as a base, hence a bel is equivalent to 10 decibels and means that there is a power increase of 10 times over the reference value.

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3 years ago

Read 2 more answers

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )