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3 years ago

What is the average speed in miles per hour of the car that traveled a total of 200 miles in 5.5 hours

Physics

2 answers:

Amanda [17]3 years ago

4 0

Answer:

Average speed of the car is 36.36 miles per hour

Explanation:

It is given that,

Total distance traveled by the car, d = 200 miles

Total time taken, t = 5.5 hours

In this case, we have to find the average speed of the car. Average speed of the car is defined as the total distance divided by total time taken. Mathematically, it can be written as :

$s=\dfrac{d}{t}$

$s=\dfrac{200\ miles}{5.5\ hours}$

s = 36.36 miles per hour

Hence, the average speed of the car is 36.36 miles per hour.

slega [8]3 years ago

3 0

36 miles per hour becuase if you do 200 divied by the hours your get 36 which is the speed

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Answer:

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a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

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$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

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The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$