Can someone please help me with this? A steel ball bearing is 5 cm in diameter a

Question

Can someone please help me with this?

A steel ball bearing is 5 cm in diameter at 29 degrees Celsius.  A bronze plate has a hole that is 4.995 cm. in diameter at 29 degrees Celsius. What common temperature must they have in order for the ball to just squeeze through the hole? Answer in units of degrees Celsius. Can someone please help me with this?

A steel ball bearing is 5 cm in diameter at 29 degrees Celsius.  A bronze plate has a hole that is 4.995 cm. in diameter at 29 degrees Celsius. What common temperature must they have in order for the ball to just squeeze through the hole? Answer in units of degrees Celsius.

Explanation / Answer

Alpha(a) for ac cu =17 *e-6 and ab brass 19*e-6

volume expansion r^3=r(cu) ^3 (1+3 ac dT)   gamma=3 alpha

are expansion r^2=r(brass)^2 (1+2 ab dT)      beta =2 alpha

the final r should be same

substuiting (5/4.995)^6=(1+2 ab dT)^3/(1+3 ac dT)^2                (1+a)^t=1+at for small a

1.006=1+6*ab *dT/1+6*ac*dT

0.006=6(19-17)*dT*E-6

dT=0.006/12 *1000000

dT=500

now 529 degrees celcius

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.