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chubhunter [2.5K]
3 years ago
13

At what minimum speed must a roller coaster be traveling when upside down at the top of a circle (fig. 5-34) so that the passeng

ers will not fall out? assume a radius of curvature of 9.6 m.
Physics
1 answer:
almond37 [142]3 years ago
4 0
At top, 2 forces act on passengers: normal force, force of gravity. The two act downward in the direction of the center of the circle. 
To answer the problem we would be using this formula which is apply N2L: 
Fnet = ma 
mg + N = mv^2/r 
minimum speed occurs when N = 0 
mg = mv^2/r 
v = sqrt (rg) 
= √94.09
v = 9.7 m/s
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gravity is 9.8kg

so 3 times 3 times 9.8 = 88.2J

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Why does a projectile fired along a horizontal not follow a straight path?​
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Explanation:

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What is the difference between heat exhaustion and heat stroke?
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6 0
4 years ago
Read 2 more answers
Calculate the de Broglie wavelength of (a) a mass of 1.0 g traveling at 1.0 m s−1 , (b) the same, traveling at 1.00 × 105 km s−1
lesantik [10]

Answer:

a)\lambda=6.63\times10^{-31}m

b)\lambda=6.63\times10^{-39}m

c)\lambda=9.97\times10^{-11}m

d)\lambda=4.03\times10^{-36}m

e)λ=∞

Explanation:

De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

\lambda=\frac{h}{P}=\frac{h}{mv}

with h the Plank's constant (6.63\times10^{-34}\frac{m^{2}kg}{s}) and P the momentum of the object that is mass (m) times velocity (v).

a)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}

\lambda=6.63\times10^{-31}m

b)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}

\lambda=6.63\times10^{-39}m

c)\lambda=\frac{6.63\times10^{-34}}{(6.65\times10^{-27}*1000)}

\lambda=9.97\times10^{-11}m

d)\lambda=\frac{6.63\times10^{-34}}{(74*2.22)}

\lambda=4.03\times10^{-36}m

e) \lambda=\frac{6.63\times10^{-34}}{(74*0)}

λ=∞

6 0
4 years ago
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