Particle A has half the mass and eight times the kinetic energy of particle B. What is the speed ratio vA / vB?

Use the equation "KE=½mv²", and use some algebra. > "Particle A has two times the mass...of particle B" mA = 2mB > "Particle A has...8 situations the kinetic power of particle B" KE_A = 8(KE_B) or: ½(mA)(vA)² = 8(½(mB)(vB)²) the rest is uncomplicated algebra: in basic terms sparkling up the above equation for "vA/vB". (hint: start up by utilising dividing the two factors by utilising "(mB)(vB)²". Then make the substitution: mA/mB = 2 (from the 1st eq0.5 ma *  VA^2 = 8 * 0.5 * mb * VB^2

0.5 * 0.5*mb *  VA^2 = 8 * 0.5 * mb * VB^2

0.5 * 0.5*  VA^2 = 8 * 0.5 * VB^2

VA/VB = 4uation))

Explanation:

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y