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Viktor [21]
2 years ago
11

A 68 kg hiker walks at 5.0 km/h up a 9% slope. The indicated incline is the ratio of the vertical distance and the horizontal di

stance expressed as percentage.
What is the necessary metabolic power?

Hint: You can model her power needs as the sum of the 380 W power to walk on level ground plus the power needed to raise her body by the appropriate amount. Assume that the efficiency of the body in using energy is 25%.
Physics
1 answer:
expeople1 [14]2 years ago
6 0

The power is calculated using the formula:

Power = Work / Time

where Work = Force * Distance , therefore:

Power = Force * Distance / Time

Power = Force * Velocity<span>


Converting the velocity in units of m/s:</span>

Velocity = (5km / h) (1000m / km) (1h / 3600s)<span>
Velocity = 1.39 m/s 

The force is equal to the y-component of the hikers weight:</span>

Force = Wy * g = W * sin θ * g

 

Let us first find the angle θ. By definition the slope is equivalent to:

slope = tan θ = ratio of vertical and horizontal distance

tan θ = 0.09

θ = 5.14˚

Therefore, Force = 68 kg * sin(5.14˚) * 9.8 m/s^2<span>
Force = 666.4 * sin (5.14) = 59.73 N 

</span>

<span>Calculating for power:
<span>Power = 59.73 N * 1.39 m/s = 82.96 W
Since the hikers efficiency is 25%, to determine the metabolic power: 
<span>Metabolic Power = 82.96 W / 0.25 = 331.83 watts </span></span></span>
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Answer:

10.791 m/s

5.93505 m

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From the momentum principle we have

\Delta P=F\Delta t

Force

F=mg

So,

m(v_f-v_i)=mg(t_f-t_i)\\\Rightarrow v_i=v_f-g(t_f-t_i)\\\Rightarrow v_i=0-(-9.81)(1.1-0)\\\Rightarrow v_i=10.791\ m/s

The speed that the ball had just after it left the hand is 10.791 m/s

As the energy of the system is conserved

K_i=U\\\Rightarrow \dfrac{1}{2}mv_i^2=mgh\\\Rightarrow h=\dfrac{v_i^2}{2g}\\\Rightarrow h=\dfrac{10.791^2}{2\times 9.81}\\\Rightarrow h=5.93505\ m

The maximum height above your hand reached by the ball is 5.93505 m

5 0
2 years ago
A mechanic uses a hydraulic car jack to lift the front end of a car to change the oil. The jack used exerts 8,915 N of force fro
Vaselesa [24]

Answer:

63.05 cm²

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The law says that in an incomprehensible , non-viscous fluid the pressure applied will transmit through out the fluid without a change.

So, Pressure on larger piston = pressure on smaller piston.

      \frac{444}{3.14}  = \frac{8915}{A}

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1 year ago
You ride a roller coaster with a loop-the-loop. Compare the normal force that the seat exerts on you to the force that Earth exe
timofeeve [1]

Answer:

N = mg + \frac{mv^2}{R}

Explanation:

At the bottom of the loop, the normal force is opposite to my weight.

I am making a circular motion. So,

F_{net} = \frac{mv^2}{R}

The relationship between the normal force, my weight, my speed and the radius of the loop is

N - mg = \frac{mv^2}{R}\\mg = N - \frac{mv^2}{R}\\ N = mg + \frac{mv^2}{R}

Here, my weight (mg) is constant. But the normal force is inversely proportional to my speed.

If my speed is zero, the normal force would be maximum and equal to my weight. If my speed is to much, then the normal force would be equally high too.

4 0
2 years ago
A small airplane is sitting at rest on the ground. Its center of gravity is 2.58 m behind the nose of the airplane, the front wh
otez555 [7]

Answer:

The percentage of the weight supported by the front wheel is  A= 19.82 %

Explanation:

From the question we are told that

   The center of gravity of the plane to its nose  is  z = 2.58 m

    The distance of the front wheel of the plane to  its nose is l = 0.800\ m

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At equilibrium  the Torque about the nose of the airplane is mathematically represented as

          mg (z- l) -  G_B *(e - l) = 0

Where m is the mass of the airplane

          G_B is the weight of the airplane supported by the main wheel  

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             G_B =\frac{mg (z-l)}{(e - l)}

Substituting values

            G_B =\frac{mg (2.58 -0.8 )}{(3.02  - 0.80)}

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Now the weight supported at the frontal wheel is mathematically evaluated as

           G_F = mg - G_B

Substituting values      

       G_F = mg - 0.8018mg    

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Now the weight of the airplane is  =  mg

Thus percentage of this weight supported by the front wheel is  A = 0. 1982 *100 = 19.82 %

7 0
2 years ago
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Answer:

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