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2 years ago

11

Is it okay to keep an albino gecko

1 answer:

arlik [135]2 years ago

5 0

Answer:

Yes

Explanation:

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28J because mechanical energy is the sum of the potential energy and kinetic energy

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How dose storing waste underground can be effected by ground water in aquifers<br><br>

DiKsa [7]

Answer:

it can contaminate the water

Explanation:

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2 years ago

"When Baby Ian is held horizontally on his back, he sometimes makes an embracing motion by arching his back, extending his legs,

Burka [1]

Answer:

Moro reflex

Explanation:

The baby has a series of reflexes in its first months of life and they are very important for the good development of the little ones. One of them is the Moro Reflex. In this reflection, the baby spreads his arms wide, stretches his legs and extends his neck. Moro's reflex begins from birth and continues until about four months of the baby. It occurs when the baby abruptly shifts position or falls backwards or feels in an insecure position, at which time the baby makes a hug movement by arching his back, extending his legs, throwing his arms out and then bringing his arms. towards the body.

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3 years ago

Which image illustrates refraction?

maks197457 [2]

Answer:

B illustrates refraction

3 0

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:

a

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

b

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

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

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

=> $R_1 = 0.0189 \ \Omega$

3 0

3 years ago