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2 years ago

Which is a valid inertial frame of reference?

Physics

1 answer:

kow [346]2 years ago

6 0

The correct answer to the question will be C. a hot air balloon moving at 30° east of north with no net force

EXPLANATION:

Before coming into any conclusion, first we have to understand an inertial frame of reference.

An inertial frame of reference is a frame of reference which is an unaccelerated frame. In this frame of reference, Newton's laws are valid.

A falling rock can not be considered as an inertial frame as it is accelerating. When a body falls, the forces that act on the body are gravity and air resistance. The gravity acts in vertical downward direction while air resistance acts in vertical upward direction. There is a net force in the downward direction which means it is experiencing an acceleration.

The frictionless spinning merry-go-round can not be considered as an inertial frame of reference. It is so because a rotating frame can not be considered as an inertial frame. It is non inertial in nature.

A hot air balloon moving 30 degree east of north with no net force can be considered as an inertial frame as it is not accelerating.

A space shuttle whose boosters just ignited for take off can not be also inertial frame as it is an accelerated frame.

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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

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kirill115 [55]

Answer:

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3 years ago

Which soil horizon is located closest to the earth's crust?

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Answer: O horizon

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Explanation:

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