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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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Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i
SCORPION-xisa [38]

Answer:

The value is  r =  5.077 \  m

Explanation:

From the question we are told that

   The  Coulomb constant is  k =  9.0 *10^{9} \  N\cdot  m^2  /C^2

   The  charge on the electron/proton  is  e =  1.6*10^{-19} \  C

    The  mass of proton m_{proton} =  1.67*10^{-27} \  kg

    The  mass of  electron is  m_{electron } =  9.11 *10^{-31} \ kg

Generally for the electron to be held up by the force gravity

   Then    

       Electric force on the electron  =  The  gravitational Force

i.e  

            m_{electron} *  g  = \frac{ k *  e^2  }{r^2 }

         \frac{9*10^9 *  (1.60 *10^{-19})^2  }{r^2 }  =     9.11 *10^{-31 }  *  9.81

         r =  \sqrt{25.78}

         r =  5.077  \  m

7 0
2 years ago
A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the
olga2289 [7]

Answer:

a) F = 30 N, b)   I = 12 N s , c)  I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

     p₀ = m v

The speed can be found by kinematics

     v = v₀ - g t

     v = 0 - 10 0.4

     v = -4.0 m / s

Final after division

     pf = m₁ v₁f + m₂ v₂f

    p₀ = pf

    M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

     F = mg

     F = 3 10 = 30 N

b) The expression for momentum is

     I = Ft

Before the explosion the only force that acts is the weight

    I = mg t

    I = 3 10 0.4

    I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

    M v = m₁ v₁f + m₂ v₂f

    v₂f = (M v - m₁ v₁f) / m₂

    v₂f = (3 (-4) - 1 6) / 2

   v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

    I = F t = Dp

   F t = pf –po)

   F t= [m₁ v₁f + m₂ v₂f]

   

   I = [1 6 + 2 (-9) -0]

   I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

 ΔI =If – I₀  

ΔI = - 12 – (-12)

ΔI = -0 N s

3 0
2 years ago
A 4.0-mF capacitor initially charged to 50 V and a 6.0-mF capacitor charged to 30 V are connected to each other with the positiv
Juli2301 [7.4K]

Answer:

<em>The final charge on the 6.0 mF capacitor would be 12 mC</em>

Explanation:

The initial charge on 4 mF capacitor  = 4 mf  x 50 V = 200 mC

The initial Charge on 6 mF capacitor  = 6 mf x 30 V =180 mC

Since the negative ends are joined together  the total charge on both capacity would be;

q = q_{1} -q_{2}

q = 200 - 180

q = 20 mC

In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage

q = (4 x V) + (6 x V)

20 = 10 V

V = 2 V

For the final charge on 6.0 mF;

q = CV

q = 6.0 mF x 2 V

q =  12 mC

Therefore the final charge on the 6.0 mF capacitor would be 12 mC

5 0
2 years ago
Read 2 more answers
Which statement best explains why an object appears green in sunlight?
Helen [10]

Answer:

This question is incomplete, the options are:

A) The object absorbs most white light and refracts most green light.

B) The object refracts most white light and absorbs most green light.

C) More green light is absorbed while more red and blue light is reflected.

D) More green light is reflected while more red and blue light is absorbed.

The answer is D.

Explanation:

Light is an electromagnetic wave that contains different colours at different wavelength. The colour of light that is seen depends on the wavelength of light that is REFLECTED, while other wavelengths of light are ABSORBED. This feature is dependent on the properties of each object that received the sunlight.

For example, an object will appear GREEN because it has properties that enables it to REFLECT most of the GREEN LIGHT but absorbs most of the RED AND BLUE LIGHT in the sunlight passing through the object.

3 0
1 year ago
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A car traveling at 91.0 km/h approaches the turn off for a restaurant 30.0 m ahead. If the driver slams on the brakes with the a
eduard

Answer: 49.92 m

Explanation:

In this situation the following equation will be useful:

V^{2}=V_{o}^{2} +2 a d

Where:

V=0 m/s is the final velocity of the car, when it finally stops

V_{o}=91 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=25.27 m/s is the initial velocity of the car

a=-6.4 m/s^{2} is the constant acceleration of the car after the driver slams on the brakes

d is the stopping distance

Isolating d:

d=\frac{-V_{o}^{2}}{2a}

d=\frac{-(25.27 m/s)^{2}}{2(-6.4 m/s^{2})}

d=41.919 m \approx 41.92 m

7 0
2 years ago
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