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Shalnov [3]
4 years ago
6

1. I drop a penny from the top of the tower at the front of Fort Collins High School and it takes 1.85 seconds to hit the ground

. Calculate the velocity in m/s after 1.10 seconds of freefall and calculate the velocity at impact in mi/hr.
Physics
2 answers:
kakasveta [241]4 years ago
7 0


what is the height of Fort Collins High School?

If found, use the formula, V=x/t-----------------(x=distance; t=time)

Svetradugi [14.3K]4 years ago
5 0
You have three known variables:

Acceleration - 9.8m/s^2
Time - 1.85s
Initial Velocity - 0m/s

For the first part of your question:

v = u + at

v = (0)+(9.8)(1.1)

v = 10.78 m/s  v=11 m/s (2 significant figures)

For the second part of your question:

v=u+at

v=(0)+(9.8)(1.85)
v=18.13 m/s



This still needs to be converted to m/h:

18.13m/s = 18.13\times 3600 metres/h=65628  metres/hour

65628 metres/hour = 65628\div1600 mi/h = 40.7925 mi/h

= 41 mi/hr (2 significant figures)





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What is the ideal banking angle (in degrees) for a gentle turn of 2.00 km radius on a highway with a 125 km/h speed limit (about
nikitadnepr [17]

An "ideal" banking angle assumes no friction is required to keep a car on the road as it turns. Let <em>θ</em> denote the banking angle, and consult the attached free-body diagram for a car making the turn. There are only 2 relevant forces acting on the car,

• the normal force with magnitude <em>n</em>

• the car's weight with magnitude <em>w</em>

and the net force points toward the center of the circle made by the turn, with centripetal acceleration

<em>a</em> = (125 km/h)² / (2.00 km) = 7812.5 km/h² ≈ 0.603 m/s²

Split up the forces into components acting perpendicular (⟂) and parallel (//) to the banked curve, so that by Newton's second law,

∑ <em>F</em> (⟂) = <em>N</em> + <em>W</em> (⟂) = <em>m</em> <em>a</em> (⟂)

and

∑ <em>F</em> (//) = <em>W</em> (//) = <em>m a</em> (//)

Let the direction of <em>N</em> be the positive perpendicular axis, and down the incline and toward the center of the circle the positive parallel axis. The net force vector and acceleration both make an angle <em>θ</em> with the banked curve, and <em>W</em> makes the same angle with the negative perpendicular axis, so that the equations above reduce to

<em>N</em> - <em>m g</em> cos(<em>θ</em>) = <em>m</em> <em>a</em> sin(<em>θ</em>)

and

<em>m g</em> sin(<em>θ</em>) = <em>m a</em> cos(<em>θ</em>)

The second equation is all we need at this point to find the ideal <em>θ</em>. The mass <em>m</em> cancels out, and we can solve for <em>θ</em> to get

tan(<em>θ</em>) = <em>a</em>/<em>g</em> ≈ (0.603 m/s²) / (9.80 m/s²) ≈ 0.0615

→   <em>θ</em> ≈ 3.52°

8 0
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