One cubic meter (1.00 m3) of aluminum has a mass of 2.70 103 kg, and the same volume of iron has a mass of 7.86 103 kg. find the radius of a solid aluminum sphere that will balance a solid iron sphere of radius 1.82 cm on an equal-arm balance.

1.28 cm

Explanation:

Density (ρ) is an intensive property resulting from the quotient of mass (m) and volume (V).

The density of Al is ρAl = 2.70 × 10³ kg/m³

The density of Fe is ρFe = 7.86 × 10³ kg/m³

A Fe sphere of radius 1.82 cm has the following volume.

V = 4/3 × π × r³

V = 4/3 × π × (0.0182 cm)³

V = 2.53 × 10⁻⁵ m³

The mass corresponding to this sphere is:

2.53 × 10⁻⁵ m³ × 2.70 × 10³ kg/m³ = 0.0683 kg

The Al sphere has the same mass, so its volume is:

0.0683 kg × (1 m³/7.86 × 10³ kg) = 8.69 × 10⁻⁶ m³

The radius corresponding to an Al sphere of 8.69 × 10⁻⁶ m³ is:

V = 4/3 × π × r³

8.69 × 10⁻⁶ m³ = 4/3 × π × r³

r = 0.0128 m = 1.28 cm

answer: the direction of the wind is expressed as the direction from which the wind is blowing. for example, easterly winds blow from east to west, while westerly winds blow from west to east.

explanation:

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.