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Romashka-Z-Leto [24]
3 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]3 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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Answer:

option B

Explanation:

given,

height of building = 0.1 km

ball strikes horizontally to ground at = 65 m    

speed at which the ball strike = ?

vertical velocity  = 0 m/s

time at which the ball strike

s = \dfrac{1}{2}gt^2

t = \sqrt{\dfrac{2s}{g}}

t = \sqrt{\dfrac{2\times 100}{9.8}}

t = 4.53 s

vertical velocity at the time  4.53 s = g × t = 9.8 × 4.53 = 44.39 m/s

horizontal velocity = \dfrac{65}{4.53} =14.35 m/s

speed of the ball = \sqrt{44.39^2+14.35^2}

                             = 46.65 m/s

hence, the speed of the ball just before it strike the ground = 47 m/s

The correct answer is option B

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2 years ago
The heat of fusion for water is the amount of energy needed for water to
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Clois what is the weight of a body in the earth, if its weig is 5Nin moon? <br>​
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Explanation:

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2 years ago
if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu
RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

F = \frac{k|q_1||q_2|}{r^2}

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N

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3 years ago
Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire
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To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

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Replacing,

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We have that,

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t = 1403.16s

Therefore the Rocket should be fired around to 1403.16s

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