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3 years ago

Which waves can u see

Physics

2 answers:

Vlad1618 [11]3 years ago

8 0

-- ocean waves

-- Slinky waves

-- visible light waves

-- guitar-string waves

-- waves on a rope

NemiM [27]3 years ago

3 0

light waves. the other waves like gamma rays or infrared waves or radio waves are all not visible to the eye. light is the only thing out of those you can see.

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What is the radius of a nucleus of an atom?

Advocard [28]

Answer:

i think its 1-10 ´ 10-15 m

Explanation:

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3 years ago

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A6 cm object is 8 cm from a convex lens that has a focal length of 2.7 cm. The image is 4 cm from the lens. The height of the im

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Answer:

7.8cm

Explanation:

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2 years ago

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An object of mass 8kg is attached to massless string of length 2m and swum with a tangential velocity of 3 what is the tension o

Paul [167]

Answer:

36 N

Explanation:

If the object of mass, m = 8 kg is swung in a horizontal circle of radius, r = 2m = length of string with tangential velocity v = 3 m/s, the tension in the string is the centripetal force which is T = mv²/r

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3 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

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3 years ago

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3 years ago