Two pulleys, one with radius r 1r1 and the other with radius r 2r2, are connected by a belt. The pulley with radius r 1r1 rotates at omega 1ω1 revolutions per minute, whereas the pulley with radius r 2r2 rotates at omega 2ω2 revolutions per minute. Show that StartFraction r 1 Over r 2 EndFraction equals StartFraction omega 2 Over omega 1 EndFraction r1 r2

r1/r2 = 2ω2/ω1

Step-by-step explanation:

The velocity of each pulley is expressed as v = ωr where;

v is the linear velocity of the pulley

ω is the angular velocity of the pulley

r is the radius of the pulley.

For the two pulleys, the velocity I'd both pulleys are the same.

v1 = v2

v1 is the linear velocity of first pulley

v2 is the linear velocity of the second pulley.

v1 = ω1r1

v2 = 2ω2(r2)

r1 and r2 is the radius of pulley 1 and pulley 2 respectively.

Since v1 = v2

ω1r1 = 2ω2(r2)

Divide both sides by r2

ω1r1/r2 = 2ω2(r2)/r2

ω1r1/r2 = 2ω2

Divide both sides by ω1

ω1r1/r2/ω1= 2ω2/ω1

r1/r2 = 2ω2/ω1

answer:

the position of equilibrium will shift to the left.

step-by-step explanation:

the general equation for the dissociation of a weak base is

b + h₂o   ⇌   bh⁺ + oh⁻

according to le châtelier's principle, when a stress is applied to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

if you apply a stress by adding more bh⁺ or oh⁻,   the system will respond by moving in a direction that gets rid of the added ions.

the position of equilibrium will shift to the left, forming more of the undissociated base.

a reflection across the x-axis , followed by the by a reflection across the y-axis , and then a scale dilation of 0.5 .