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olga nikolaevna [1]

3 years ago

11

The radius of the planent venus is nearly the same as that of the earth,but its mass is only eighty percent that of the earth. I

f an object weighs w on the earth,what does it wiegh on venus

2 answers:

oee [108]3 years ago

7 0

Weight=mg
g=GM/r^2
g on venus is 0.80w as radius is kept constant
m of object is kept constant
w α g
w(venus( is 0.8w

TiliK225 [7]3 years ago

7 0

Size, Mass and Orbit:

Whereas Earth has a mean radius of 6,371 km and a mass 5,972,370,000 quadrillion kg, Venus has a mean radius of about 6,052 km and a mass of 4,867,500,000 quadrillion kg. This means that Venus is roughly 0.9499 the size of Earth and 0.815 as massive.Apr 7, 2016

Volume: 10.8321×10 km 9.2843×10 km

Mean Radius: 6,371.0 km 6,051.8 ± 1.0 km

Gravity: 9.8 m/s2 8.87 m/s

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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

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

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

=> $R_1 = 0.0189 \ \Omega$

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