It has been suggested that rotating cylinders several miles in length and several miles in diameter be placed in space and used as colonies. Inhabitants of the space colonies would live on the inside surface of the cylinder. Inertial effects would resemble gravity's influence and keep them 'plastered to the surface.' Suppose that you are an inhabitant of a space colony which is 1070 miles in length and 4.86 miles in diameter. How many revolutions per hour must the cylinder have in order for the occupants to experience a centripetal acceleration equal to the acceleration of gravity

the required revolution per hour is 28.6849

Explanation:

Given the data in the question;

we know that the expression for the linear acceleration in terms of angular velocity is;

= rω²

ω² = / r

ω = √( / r )

where r is the radius of the cylinder

ω is the angular velocity

given that; the centripetal acceleration equal to the acceleration of gravity a  = g = 9.8 m/s²

so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m

Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m

so we substitute

ω = √( 9.8 m/s² / 3909.87 m )

ω = √0.002506477 s²  

ω = 0.0500647 ≈ 0.05 rad/s  

we know that; 1 rad/s = 9.5493 revolution per minute

ω = 0.05 × 9.5493 RPM

ω = 0.478082 RPM  

1 rpm = 60 rph  

so  

ω = 0.478082 × 60

ω = 28.6849  revolutions per hour  

Therefore, the required revolution per hour is 28.6849

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

its a 5.5 n

glad that