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3 years ago

Which of the following allows natural selection and evolution to occur?

Physics

2 answers:

leonid [27]3 years ago

6 0

Answer:

D. differences in the characteristics of individual trees

Explanation:

Natural selection and evolution can only occur if there are differences in the characteristics of individual organisms. Organisms with advantageous adaptations, such as deep roots, are more likely to survive and reproduce than organisms without the advantageous adaptations. Continuous natural selection of one trait over another causes organisms to evolve so the advantageous trait eventually becomes more common in a population.

Black_prince [1.1K]3 years ago

4 0

Answer:

i think its B

Explanation:

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Conduction is usually faster/slower in liquids and certain solids than in gases.

KIM [24]

Conduction is slower in liquids and gases.

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Materials which are good conductors of thermal energy are called thermal conductors.

metals are good thermal conductors.

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3 years ago

Reduce<br> Reflects<br> Transfers<br> Stops

Bond [772]

I think transfers is the answer

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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

What is the force in N of an object that has a mass of 7 kilograms and an acceleration of 4 m/s/s

VMariaS [17]

Newton's 2nd law of motion: Force = (mass) x (acceleration)

If you want to move a 7-kg object with an acceleration of 4 m/s²,
then you will need to push it with (7 x 4) = 28 newtons of force.

4 0

3 years ago

The diagram below shows a model of the Earth's orbit around the Sun.

Sladkaya [172]

I would say that I and IV because summer is I and fall is IV

7 0

3 years ago

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