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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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When the pendulum and roller coaster move to the top, its has more potential energy whereas when comes to the bottom has more kinetic energy.

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1 year ago
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3 years ago
How much current must be applied across a 60 Ω light bulb filament in order for it to consume 55 W of power? Unserious answers w
sergij07 [2.7K]

Answer: current I = 0.96 Ampere

Explanation:

Given that the

Resistance R = 60 Ω 

Power = 55 W

Power is the product of current and voltage. That is

P = IV ...... (1)

But voltage V = IR. From ohms law.

Substitutes V in equation (1) power is now

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Substitute the above parameters into the formula to get current I

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Make I^2 the subject of formula

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I^2 = 0.92

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