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katrin2010 [14]

3 years ago

11

At some airports there are speed ramps to help passengers get from one place to another. A speed ramp is a moving conveyor belt

on which you can either stand or walk. Suppose a speed ramp has a length of 104 m and is moving at a speed of 2.1 m/s relative to the ground. In addition, suppose you can cover this distance in 84 s when walking on the ground. If you walk at the same rate with respect to the speed ramp that you walk on the ground, how long does it take for you to travel the 104 m using the speed ramp

2 answers:

mezya [45]3 years ago

8 0

Answer:

Time t = 31.16s

Therefore, it takes 31.16 seconds to travel 104 m using the speed ramp

Explanation:

Length of speed ramp = 104 m

Time taken to cover the length walking = 84s

Walking speed = length/time = 104/84 = 1.238 m/s

Speed ramp speed relative to the ground = 2.1m/s

Total speed of combined speed ramp and walking on speed ramp is;

Vt = 2.1 + 1.238 = 3.338 m/s

Time taken to cover 104 m with combined speed ramp and walking on speed ramp;

Time t = distance/speed = 104/3.338

Time t = 31.16s

Therefore, it takes 31.16 seconds to travel 104 m using the speed ramp

Amanda [17]3 years ago

4 0

Answer:

The time taken is $t_s = 31.16 s$

Explanation:

From the question we are told that

The length of the speed ramp is $L = 104m$

The speed of the speed ramp is $v_r = 2.1 m/s$

The time taken to cover the distance is $t = 84 s$

The walking speed( when not on the speed ramp) required to cover the 104m is mathematically represented as

$v_w = \frac{L}{t}$

Substituting values

$v_w = \frac{104}{84}$

$= 1.238 m/s$

The speed required to cover the 104m distance when on a speed speed ramp is mathematically represented as

$v = The\ walking\ speed + speed \ of \ the \ speed \ ramp$

substituting values

$v = 2.1 + 1.238$

$v = 3.338 m/s$

Now the time taken to travel the 104m distance when walking on the speed ramp is mathematically evaluated as

$t_s = \frac{L}{v}$

$= \frac{104}{3.338}$

$t_s = 31.16 s$

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Answer:

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Explanation:

a) This situation can be modelled by the Principle of Energy Conservation and the Work-Energy Theorem, where friction represents the only non-conservative force exerting on the crate in motion. Let consider the bottom of the straight ramp as the zero point. The energy equation for the crate is:

$U_{g,1}+K_{1} = U_{g,2}+K_{2}+ W_{fr}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final translational kinetic energy, measured in joules.

$W_{fr}$ - Work losses due to friction, measured in joules.

By applying the defintions of translational kinetic and gravitational potential energies and work, this expression is now expanded:

$m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta$

Where:

$m$ - Mass of the crate, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$y_{1}$ , $y_{2}$ - Initial and final height of the crate, measured in meters.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the crate, measured in meters per second.

$\mu_{k}$ - Kinetic coefficient of friction, dimensionless.

$\theta$ - Ramp inclination, measured in sexagesimal degrees.

The equation is now simplified and the coefficient of friction is consequently cleared:

$y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot \cos \theta$

$\mu_{k} = \frac{1}{\cos \theta} \cdot \left[y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) \right]$

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$y_{2} = (1.6\,m)\cdot \sin 30^{\circ}$

$y_{2} = 0.8\,m$

If $\theta = 30^{\circ}$ , $y_{1} = 0\,m$ , $y_{2} = 0.8\,m$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{1} = 5\,\frac{m}{s}$ and $v_{2} = 0\,\frac{m}{s}$ , the coefficient of friction is:

$\mu_{k} = \frac{1}{\cos 30^{\circ}}\cdot \left\{0\,m-0.8\,m+\frac{1}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}\cdot \left[\left(5\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] \right\}$

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Then, the magnitude of the friction force is:

$f =\mu_{k}\cdot m\cdot g \cdot \cos \theta$

If $\mu_{k} \approx 0.548$ , $m = 12\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\theta = 30^{\circ}$ , the magnitude of the force of friction is:

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The magnitude of the force of friction is 55.851 newtons.

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The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.

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Answer:

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Explanation:

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Now during the east trip,

Ve = Vp + Va

And Te = X/(Vp + Va)

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Thus, 4.9 hours = 4.9 x 60 x 60 = 17640 seconds

So, this time will be equal to the sum of that in the west and east directions.

Thus; T = Tw + Te

From above we know Tw and Te.

Let's substitute them into the equation;

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T = X[(Vp + Va + Vp - Va)/((Vp)² — (Va)²)

T = X[(2Vp)/((Vp)² — (Va)²)

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X = [T((Vp)² — (Va)²)]/(2Vp)

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