%3Cp%3Esand%20falls%26nbsp%3Bfrom%26nbsp%3Ba%26nbsp%3Bconveyor%26nbsp%3Bbelt%26n

Question

%3Cp%3Esand%20falls%26nbsp%3Bfrom%26nbsp%3Ba%26nbsp%3Bconveyor%26nbsp%3Bbelt%26nbsp%3Bat%26nbsp%3Ba%26nbsp%3Brate%26nbsp%3Bof%26nbsp%3B15%26nbsp%3Bm%5E3%2Fmin%26nbsp%3Bonto%26nbsp%3Bthe%26nbsp%3Btop%26nbsp%3Bof%26nbsp%3Ba%26nbsp%3Bconical%26nbsp%3Bpile.%26nbsp%3Bthe%26nbsp%3Bheight%26nbsp%3Bof%26nbsp%3Bthe%26nbsp%3Bile%26nbsp%3Bis%26nbsp%3Balways%26nbsp%3Bthree%26nbsp%3Beighths%26nbsp%3Bof%26nbsp%3Bthe%26nbsp%3Bdiameter%26nbsp%3Bof%26nbsp%3Bthe%26nbsp%3Bbase.%26nbsp%3Bhow%26nbsp%3Bfast%26nbsp%3Bis%26nbsp%3Bthe%26nbsp%3Bheight%26nbsp%3Bchaning%26nbsp%3Bwhen%26nbsp%3Bthe%26nbsp%3Bpile%26nbsp%3Bis%26nbsp%3B4m%26nbsp%3Bhigh%3F%3C%2Fp%3E

Explanation / Answer

V=piR^2h/3

Given h=3/8(2R)

h=3R/4

R=4h/3

So V=pi(16h^2/9)h/3

=16pih^3/27

Dedicating with respect to time we get

dV/dt=16pi/27(3h^2)(dh/dt)

At h=4

15=256pi/9(dH/dt)

dH/dt=225/256pi

=0.2799 m/min

So the rate of increase of height is 0.2799 metres/ min

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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