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Wewaii [24]
2 years ago
8

A person is trying to ride a bike all the way round the inside of a pipe for a stunt in a film. The filmmaker wants to know what

speeds are involved. The pipe has a diameter of 8 m. The mass of the bike and rider is 400 kg. The rider goes at a constant speed of 5 m/s.
a) What is its acceleration at the bottom?
b) What is the force on the bike at an angle of 30° up from the bottom?
c) What is the minimum velocity at the top for the bike and rider to stay moving in a circle?
d) Do the bike and rider have sufficient velocity to stay movins a on a circle at the top?
​
Physics
1 answer:
nirvana33 [79]2 years ago
4 0

Answer:

1) The acceleration will be g = 9.8 m/s²

2) F = Wsinα = mgsinα = 400kg·9.8m/s²·1/2 = 2000N

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andriy [413]

Answer:

\rm KE\pm \Delta KE = 17.6\pm 6.8\ J.

Explanation:

<u>Given:</u>

  • Mass, \rm m\pm\Delta m = 1.3\pm 0.4\ kg.
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where,

\rm \Delta m,\ \Delta v are the uncertainties in mass and velocity respectively.

The kinetic energy is given by

\rm KE = \dfrac 12 mv^2 = \dfrac 12 \times 1.3\times 5.2^2=17.576\approx 17.6\ J.

The uncertainty in kinetic energy is given as:

\rm \dfrac{\Delta KE}{KE}=\dfrac{\Delta m}{m}+\dfrac{2\Delta v}{v}\\\dfrac{\Delta KE}{17.6}=\dfrac{0.4}{1.3}+\dfrac{2\times 0.2}{5.2}\\\dfrac{\Delta KE}{17.6}=0.384\\\Rightarrow \Delta KE = 17.6\times 0.384 = 6.7854\ J\approx6.8\ J\\\\Thus,\\\\KE\pm \Delta KE = 17.6\pm 6.8\ J.

7 0
3 years ago
What are the unit for acceleration
Flura [38]
<h3>Answer</h3>

m/s^2 (meter per sec square)

Explanation:

acc = change in velocity/time

= distance/time

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time

= m/s

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s

=m/s^2

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A track begins at 0 meters and has a total distance of 100 meters. Juliet starts at the 10-meter mark while practicing for a rac
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Hot air rises = it's less dense than cold air = falls

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A particle cannot generally be localized to distances much smaller than its de Broglie wavelength. This fact can be taken to mea
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The De broglie wavelength of a thermal neutron at room temperature 300K = 1.5 × A°

<h3>How is the De broglie wavelength of a thermal neutron at room temperature calculated?</h3>

Temperature, T = 300K

Momentum, p = mv

Therefore v = p/m

Energy, E= 1/2 m( p/m) ²

Boltzman Energy= 3/2 KT

3/2KT = 1/2 m(p/m)²

Therefore p =  \sqrt{3RTm}

According to De broglie hypothesis, P = h ÷ λ

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                     = 0.15 × 10^{-9}

Therefore the De broglie wavelength of a thermal neutron at room temperature 300K = 1.5 × A°

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