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3 years ago

What part of an electric circuit are the wires that carry the electric current

Physics

1 answer:

bezimeni [28]3 years ago

8 0

Answer:

okay so, An electric circuit is a closed loop through which current can actually flow. All electric circuits must have a voltage source, for example a battery, and a conductor, which is usually a wire.

how does it flow or how does .current flow in a circuit ? When a wire is connected to battery terminals which is electrons, electrons flow from negative to positive. Electrons have a negative charge in case you didnt know. they are repelled from the negative and attracted to the positive.

if this helped please mark this as brainliest!!! :)

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2 years ago

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2 years ago

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2 years ago

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Suppose your surface body temperature averaged 90 degrees F. How much radiant energy in W/m^2 would be emitted from your body?

Debora [2.8K]

$493 \; \text{W}\cdot \text{m}^{-2}$ .

<h3>Explanation</h3>

The Stefan-Boltzmann Law gives the energy radiation <em>per unit area</em> of a black body:

$\dfrac{P}{A} = \sigma \cdot T^{4}$

where,

$P$ the total power emitted,
$A$ the surface area of the body,
$\sigma$ the Stefan-Boltzmann Constant, and
$T$ the temperature of the body in degrees Kelvins.

$\sigma = 5.67 \times 10^{-8} \;\text{W}\cdot \text{m}^{-2} \cdot \text{K}^{-4}$ .

$T = 90 \; \textdegree{}\text{F} = (\dfrac{5}{9} \cdot (90-32) + 273.15) \; \text{K} = 305.372 \; \text{K}$ .

$\dfrac{P}{A} = \sigma \cdot T^{4} = 5.67 \times 10^{-8} \times 305.372^{4} = 493\; \text{W}\cdot \text{m}^{-2}$ .

Keep as many significant figures in $T$ as possible. The error will be large when $T$ is raised to the power of four. Also, the real value will be much smaller than $493\; \text{W}\cdot \text{m}^{-2}$ since the emittance of a human body is much smaller than assumed.

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2 years ago

A spring is hanging from the ceiling. When a 250 gram of mass is attached to the free end, the spring elongates by 5 cm. The spr

lord [1]

Answer:

k = 49 N/m

Explanation:

Given that,

Mass, m = 250 g = 0.25 kg

When the mass is attached to the end of the spring, it elongates 5 cm or 0.05 m. We need to find the spring constant. Let it is k.

The force due to mass is balanced by its weight as follows :

mg=kx

$k=\dfrac{mg}{x}\\\\k=\dfrac{0.25\times 9.8}{0.05}\\\\k=49\ N/m$

So, the spring constant of the spring is 49 N/m.

8 0

3 years ago