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Romashka-Z-Leto [24]
2 years ago
10

A 60-kg runner raises his center of mass approximately 0.5 m with each step. Although his leg muscles act as a spring, recapturi

ng the energy each time his feet touch down, there’s an average 10% loss with each compression. What must the runner’s additional power output be to account for just this loss, if he averages 0.8 s per stride?
Physics
1 answer:
Darina [25.2K]2 years ago
5 0

Answer: P = 36.75W

The additional power needed to account for the loss is 36.75W.

Explanation:

Given;

Mass of the runner m= 60 kg

Height of the centre of gravity h= 0.5m

Acceleration due to gravity g= 9.8m/s

The potential energy of the body for each step is;

P.E = mgh

P.E = 60 × 9.8 × 0.5

PE = 294J

Since the average loss per compression on the leg is 10%.

Energy loss = 10% (P.E)

E = 10% of 294J

E = 29.4J

To calculate the runner's additional power

given that time per stride is = 0.8s

Power P = Energy/time

P = E/t

P = 29.4J/0.8s

P = 36.75W

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A motorcyclist heading east through a small Iowa town accelerates after he passes a signpost at x=0 marking the city limits. His
adoni [48]

Answer:

1) v_2=23\ m.s^{-1}              &     x_2=43\ m east of sign post

2) x'=55\ m east of sign post

3) x_n=205\ m east of the signpost.

4) v_z=35\ m.s^{-1}

Explanation:

Given:

  • position of motorcyclist on entering the city at the signpost, x_0=0\ m
  • time of observation after being at x=5m east of the signpost, t_m=0\ s
  • constant acceleration of the on entering the city, a=4\ m.s^{-2}
  • distance of the motorcyclist moments later after entering, s_m=5\ m
  • velocity of the motorcyclist moments later after entering, u_m=15\ m.s^{-1}

<u>Now the initial velocity on at the sign board:</u>

u_m^2=u^2+2.a.x_m

where:

u= initial velocity of entering the city at the signpost

Putting respective values:

15^2=u^2+2\times 4\times 5

u=13.6015\ m.s^{-1}

1)

Position at time t_2=2\ s sec.:

Using equation of motion,

x_2=u_m.t_2+\frac{1}{2} a.(t_2)^2+5 because it has already covered 5m before that point

x_2=15\times 2+0.5\times 4\times 2^2+5

x_2=43\ m east of sign post

Velocity at time t_2=2\ s sec.:

v_2=u_m+a.t_2

v_2=15+4\times 2

v_2=23\ m.s^{-1}

2)

Position when the velocity is v'=25\ m.s^{-1}:

using equation of motion,

v'^2=u_m^2+2.a.x'+5

25^2=15^2+2\times 4\times x'+5

x'=55\ m east of sign post

3)

Given that:

acceleration be, a_n=2\ m.s^{-2}

time, t_n=5\ s

Position after the new acceleration and the new given time:

using equation of motion,

x_n=u_m.t_n+\frac{1}{2} a_n.(t_n)^2+5

x_n=15\times 5+0.5\times 2\times 5^2+5

x_n=205\ m east of the signpost.

4)

now time of observation, t_z=5\ s

v_z=u_m+a.t_z

v_z=15+4\times 5

v_z=35\ m.s^{-1}

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Jocko the clown, whose mass is 60-Kg, stands on a skateboard. A 20-Kg ball is thrown at Jocko at 3m/s, and when he catches the b
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Answer:

The speed of the Jocko and the ball move after he catches the ball is 0.75 m/s.

Explanation:

Given that,

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Let V is the speed of Jocko and the ball move after he catches the ball. The momentum of the system remains conserved. Using the conservation of momentum as :

m'v'=(m+m')V\\\\V=\dfrac{m'v'}{(m+m')}\\\\V=\dfrac{20\times 3}{(60+20)}\\\\V=0.75\ m/s

So, the speed of the Jocko and the ball move after he catches the ball is 0.75 m/s.

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For example, the African Savanna has an almost constant temperature all year (see the first figure below).

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The temperature of a temperate grassland (see the second figure below) has a much greater seasonal variation.  

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