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Ymorist [56]
3 years ago
7

The formula v = √2.5r models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radiu

s of curvature r r, in feet. A highway crew measures the radius of curvature at an exit ramp on a highway as 380 feet. What is the maximum safe speed?
Physics
1 answer:
mel-nik [20]3 years ago
4 0

Answer:

The maximum safe speed of the car is 30.82 m/s.

Explanation:

It is given that,

The formula that models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radius of curvature r r, is in feet is given by :

v=\sqrt{2.5\ r}.........(1)

A highway crew measures the radius of curvature at an exit ramp on a highway as 380 feet, r = 380 feet

Put the value of r in equation (1) as :

v=\sqrt{2.5\ \times 380}

v = 30.82 m/s

So, the maximum safe speed of the car is 30.82 m/s. Hence, this is the required solution.

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I shared a picture of the problem. It’s a basic Physics question and an Algebra question.
julia-pushkina [17]

Hence the expression of ω in terms of m and k is

\omega = \sqrt{\frac{k}{m}

Given the expressions;

T_s = 2 \pi \sqrt{\frac{m}{k} } \ and \ T_s = \frac{2 \pi}{\omega}

Equating both expressions we will have;

2 \pi \sqrt{\frac{m}{k} }  = \frac{2 \pi}{\omega}

Divide both equations by 2π

\frac{2 \pi\sqrt{\frac{m}{2 \pi} } }{2 \pi}=\frac{\frac{2 \pi}{\omega} }{2\pi}\\\sqrt{\frac{m}{2 \pi} } = \frac{1}{\omega}\\

Square both sides

(\sqrt{\frac{m}{k} } )^2 = (\frac{1}{\omega} )^2\\\frac{m}{k} = \frac{1}{\omega ^2} \\\omega ^2 = \frac{k}{m}

Take the square root of both sides

\sqrt{\omega ^2} =\sqrt{\frac{k}{m} } \\\omega = \sqrt{\frac{k}{m}

Hence the expression of ω in terms of m and k is

\omega = \sqrt{\frac{k}{m}

3 0
3 years ago
Identify some sports or activities you can do to improve your overall health and explain how physical activity can improve your
Alexxx [7]

Answer:

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Explanation:

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Hope this helps

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dybincka [34]

Answer:

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Explanation:

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3 years ago
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Ksju [112]
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