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Alina [70]
3 years ago
11

A handful of professional skaters have taken a skateboard through an inverted loop in a full pipe. For a typical pipe with a dia

meter of 14 feet (4.27 meters), what is the minimum speed the skater must have at the very top of the loop?
Physics
1 answer:
Bingel [31]3 years ago
5 0

Answer

given,

diameter of the pipe is  =  (14 ft)4.27 m

minimum speed of the skater must have at very top = ?

At the topmost point of the pipe the  normal force will be equal to zero.

F = mg

centripetal force acting on the skateboard

F = \dfrac{mv^2}{r}

equating both the force equation

mg = \dfrac{mv^2}{r}

v = \sqrt{gr}

r = d/2 = 14/ 2 = 7 ft

or

r = 4.27/2 = 2.135 m

g = 32 ft/s²   or g = 9.8 m/s²

v = \sqrt{32 \times 7}

v = 14.96 ft/s

or

v = \sqrt{9.8 \times 2.135}

v = 4.57 m/s

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. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of
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Answer:

See Explanation

Explanation:

Given

s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}

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s_0=s

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Solving (b): Units and dimension of v_0

Remove the other terms of the equation, we have:

s=v_0t

Rewrite as:

v_0t = s

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Solving (c): Units and dimension of a_0

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s=\frac{j_0t^3}{6}

Rewrite as:

\frac{j_0t^3}{6} = s

Make j_0 the subject

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Replace s and t with their units [Ignore all constants]

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j_0 = LT^{-3}

Hence:

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Remove the other terms of the equation, we have:

s=\frac{S_0t^4}{24}

Rewrite as:

\frac{S_0t^4}{24} = s

Make S_0 the subject

S_0 = \frac{24s}{t^4}

Replace s and t with their units [ignore all constants]

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S_0 = LT^{-4

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S_0 = LT^{-4 --- dimension

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Ignore other terms of the equation, we have:

s=\frac{ct^5}{120}

Rewrite as:

\frac{ct^5}{120} = s

Make c the subject

c = \frac{120s}{t^5}

Replace s and t with their units [Ignore all constants]

c = \frac{L}{T^5}

c = LT^{-5}

Hence:

c \to LT^{-5} --- dimension

c \to m/s^5 --- units

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