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Archy [21]
3 years ago
15

A car in an amusement park ride rolls without friction around a track (Fig. P7.42). The hB car starts from rest at point A at a

CR height h above the bottom of the loop. Treat the car as a particle. (a) What is the minimum value of h (in terms of R) such that the car moves around the loop without falling off at the top (point B)

Physics
1 answer:
MrRissso [65]3 years ago
6 0

Answer:

h>\dfrac{5}{2}R

Explanation:

Given that

Height = h

Radius = R

From energy conservation

KE_A+U_A=KE_B+U_B

At point B

The minimum speed to complete the   the circle

V_B=\sqrt{gR}\ m/s

So the kinetic energy at point B

KE_B=\dfrac{1}{2}mV^2

KE_B=\dfrac{1}{2}mgR

KE_A+U_A=KE_B+U_B

0+mgh=\dfrac{1}{2}mgR+2mgR

Without falling off at the top (point B)

0+mgh>\dfrac{1}{2}mgR+2mgR

mg(h-2R)>\dfrac{1}{2}mgR

g(h-2R)>\dfrac{1}{2}gR

h>\dfrac{5}{2}R

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Need an answer, please need an answer, please​
Karolina [17]

9514 1404 393

Answer:

  • moderate low: 82
  • moderate high: 112.75
  • vigorous low: 123
  • vigorous high: 174.25

Explanation:

When calculations are repetitive, I find it convenient to use a calculator that can work with tables.

The PMHR is (220 -15) = 205.

Each of the other heart rates is computed as the formula shows. For example, the low moderate heart rate is 205×0.40 = <u> 82 </u> bpm

The other rates are shown in the attached table. They are computed the same way.

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true

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