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3 years ago

Why are conductors and insulators both required to construct the electrical wiring in our home

1 answer:

8 0

Conductors are materials with many free electrons, so they allow electrical current to flow through them. Therefore, conductors are required in order to bring electricity to every room of the house.

Insulators, instead, are materials with few or no free electrons, so electrical currents do not flow through them. In the electrical wiring of the houses, they are used in order to isolate the conductive elements of the wire from other conductive materials (in fact, if the conductive elements touch other conductive elements of the house, part of the current would be dissipated)

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When making fitness purchases what should you not do?

astraxan [27]

Of these options the one which best answers the question is 'Use whatever professionals claim to use.'
It is not that professional recommendations should not be consired or valued. The reason you should not just use whatever professionals claim to use is that their needs, as professionals, may be very different from your own. In order to know whether your own needs will be met you need to know what those needs are and whether the product meets those needs. All of the other options given would help you to figure this out, but the testimony of professionals would not.

7 0

3 years ago

Read 2 more answers

Which statement about convection in the Earth's mantle is true?

marissa [1.9K]

Answer:

All of these

Explanation:

Mantle convection causes tectonic plates to move around and passes heat on the earth surface in forms of lava and magma.

Hope it helps!

4 0

3 years ago

A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan

Nesterboy [21]

Answer:

Compression distance: $d \approx 0.102\,m$

Explanation:

According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy ( $K$ ), in joules, at the expense of elastic potential energy ( $U$ ), in joules, and overcoming work losses due to friction ( $W_{l}$ ), in joules:

$K + W_{l} = U$ (1)

By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$ (2)

Where:

$m$ - Mass of the block, in kilograms.

$v$ - Final velocity of the block, in meters per second.

$\mu$ - KInetic coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$s$ - Width of the rough patch, in meters.

$k$ - Spring constant, in newtons per meter.

$d$ - Compression distance, in meters.

If we know that $m = 1.2\,kg$ , $v = 2.3\,\frac{m}{s}$ , $\mu = 0.44$ , $g = 9.807\,\frac{m}{s^{2}}$ , $s = 0.05\,m$ and $k = 730\,\frac{N}{m}$ , then the compression distance of the spring is:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$

$m\cdot v^{2} + 2\cdot m\cdot g \cdot s = k\cdot d^{2}$

$d = \sqrt{\frac{m\cdot (v^{2}+2\cdot g\cdot s)}{k} }$

$d \approx 0.102\,m$

4 0

3 years ago

Explain the causes of yaa asantewaa war of 1900

Anna11 [10]

Answer: the conflict began when a british representative - Sr. Frederick Mitchell Hodgson sat on the golden stool.

Explanation:

since the stool wasn't a throne, when Yaa Asantewaa found out, he led a rebellion which killed 1000 British and allied African soldiers and 2,000 Ashanti.

good luck with your assignment :)

8 0

3 years ago

The hydraulic oil in a car lift has a density of 8.53 x 102 kg/m3. The weight of the input piston is negligible. The radii of th

navik [9.2K]

Answer:

(a) the input force is 36.56 N

(b) the input force is 37.49 N

Explanation:

Given;

density of hydraulic oil, ρ = 8.53 x 10² kg/m³

radius of plunger, r₁ = 0.135 m

radius of piston, r₂ = 5.43 x 10⁻³ m

Part (a) The input force needed to support 22600-N weight, when the bottom surfaces of the piston and plunger are at the same level;

$P =\frac{F}{A}$

Where;

P is pressure

F is force

A is circular area = πr²

$\frac{F_1}{A_1} =\frac{F_2}{A_2} \\\\F_2 = \frac{F_1*A_2}{A_1} =\frac{F_1* \pi r_2^2}{\pi r_1^2} = \frac{F_1* r_2^2}{ r_1^2} \\\\F_2 = \frac{22600*(5.43*10^{-3})^2 }{(0.135)^2}\\\\F_2 = 36.56 \ N$

Part (b) The input force needed to support 22600-N weight, when the bottom surface of the output plunger is 1.20 m above that of the input plunger

$P_2 = P_1 + \rho gh$

But, F = PA and A = πr²

$F_2 = F_1(\frac{A_2}{A_1} ) + \rho gh*A_2\\\\F_2 = F_1(\frac{r_2^2}{r_1^2} )+\rho gh(\pi r_2^2)\\\\F_2 = 22600(\frac{5.43*10^{-3}}{0.135})^2 \ + 853*9.8*1.2*\pi (5.43*10^{-3})^2\\\\F_2=36.56 + 0.93\\\\F_2 = 37.49 \ N$

4 0

3 years ago