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MAXImum [283]
3 years ago
13

A 20 kg bike accelerates at 10 m/s2. With what force was the person pedaling if friction

Physics
2 answers:
denpristay [2]3 years ago
8 0

Answer:200N

Explanation:

Molodets [167]3 years ago
4 0
Look it up plz cuz like I’m slow so
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An engineer wishes to design a roller coaster so that
nataly862011 [7]

To prevent cars from falling, the radius at the top of the circle should be

small such cars inverted at the top remain attached during motion.

Correct response;

The radius of the coaster can be <u>C) 20 m</u>

<h3>Method by which the above option is selected</h3>

Mass of roller coaster car, m = 500 kg

Speed at the top of the circle, v = 20 m/s

Required:

The maximum radius of the circular path the roller coaster car.

Solution:

\displaystyle Centrifugal \ force, \, F_{c} = \mathbf{\frac{m \cdot v^2}{r}}

Where;

r = The radius of the circular path.

Weight of the roller coaster car = m·g = The centripetal force

Where;

g = Acceleration due to gravity = 9.81 m/s²

At equilibrium, we have;

Centrifugal force = Centripetal force

\displaystyle \frac{m \cdot v^2}{r}  = \mathbf{ m \cdot g}

Therefore;

\displaystyle r = \mathbf{ \frac{v^2}{g}}

Which gives;

\displaystyle r = \frac{20^2}{9.81} \approx 40.77

The maximum radius for safety of a roller coaster, r ≈ 40.77 meters

\displaystyle Range \ of \ radius \ of \ the \ circle = \frac{40.77}{4}  \leq Radius \ of \ circle \leq 40.77

Which gives;

Range of the radius of the circle = 10.2 ≤ Radius of circle ≤ 40.77

The correct option for safety considerations is therefore;

  • <u>C) 20 m</u>

<em>The possible question options are;</em>

<em>A) 5 m  B) 10 m  C) 20 m  D) 40 m  E) 80 m</em>

Learn more about centripetal force here:

brainly.com/question/12674230

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3 years ago
Objects falling through the air experience a type of friction called air resistance. true or false
oee [108]

Answer:

True

Explanation:

Objects falling through the air experience a force of opposition called air resistance

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2 years ago
How do you calculate the net force when there are multiple forces in different directions?​
Artyom0805 [142]

To find F_{net} we need to use vector addition and use the x and y components. First we subtract vector 2 from vector 5 which results in a vector with a  length of 3 pointing directly east, then we use the distance formula to find the length of the net force F_{net} = \sqrt{(3)^2+(4)^2} \\  which gives F_{net} = 5. We now have a magnitude but we also need a direction, since vector 4 and vector 5 are perpendicular. Using \theta = \tan^{-1} (\frac{4}{3})   where tan^-1(y/x) we get an angle of 53 degrees. The resultant force vector is 5 distance with an angle of 53 degrees north east.

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