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IgorLugansk [536]
4 years ago
5

A runner runs 4875 ft in 6.85 minutes. what is the runners average speed in miles per hour?

Physics
2 answers:
yanalaym [24]4 years ago
8 0

The average speed can be easily calculated by taking the ratio of distance and time. That is:

average speed = distance / time

 

so calculating:

average speed = 4875 ft / 6.85 minutes

<span>average speed = 711.68 ft / min</span>

zaharov [31]4 years ago
7 0

<u>Answer</u>:

Speed of the runner is 8.08 miles per hour approximately.

<u>Explanation</u>:

Given:

A runner runs =4875 ft

Time Taken = 6.85 minutes.

To Find:  

Average speed of the runner in miles per hour=?

Solution:

Distance = Speed x time

Conveting feet into miles:

=>Distance= 4875\times\frac{1}{5280} miles

=>Distance = 0.9232955 miles

Converting minutes into hours:

time = 6.85 minutes

=> 6.85\times \frac{1}{60}  hours

=> \frac{6.85}{60}= 0.1141667 hours.

Now, we know that, distance = speed x time  

Then, 0.9232955 miles = speed x 0.1141667 hours  

=>speed = \frac{0.9232955}{0.1141667}

=>speed = 8.08725749

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A mass of 0.75 kilograms is attached to a spring/mass oscillator. A force of 5 newtons is required to stretch the spring 0.5 met
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b > 66.41 kg/s

Explanation:

The spring force F = -kx, where k = spring constant, the damping force f = -bv. The net force F' = F + f

F + f = ma

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-kx -bdx/dt = md²x/dt².

Re-arranging the equation, we have

So, md²x/dt² + bdx/dt + kx = 0

Dividing through by m, we have

d²x/dt² + (b/m)dx/dt + (k/m)x = 0

This is a second-order differential equation. The characteristic equation is thus,

D² + (b/m)D + (k/m) = 0

Using the quadratic formula, we find D.

D = \frac{-(b/m) +/- \sqrt{(b/m)^{2} - 4k/m} }{2}

For an overdamped system,

(b/m)^{2} - 4k/m} >   0

(b/m)^{2} >   4k/m}\\(b/m) >   \sqrt{4k/m}} \\(b/m) >   2\sqrt{k/m}} \\b >   2\sqrt{km}}

Now, k = F/x. Since the weight of the object causes the spring to stretch a distance of 0.5 m, k = mg/x where m = mass of object = 0.75 kg, g = 9.8 m/s² and x = x₀ =0.5 m.

Substituting k = mg/x into the inequality for b, we have

b > 2√{(mg/x₀)m}

b > 2√{(m²g/x₀)}

b > 2m√{g/x₀)}

b > 2 × 0.75 kg√{9.8 m/s²/0.5 m)}

b > 1.5 kg√{19.6/s²)}

b > 1.5 kg × 4.427/s

b > 66.41 kg/s

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