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Maurinko [17]
3 years ago
6

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have st

udied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 40 mm, while nonathletes' stretch only 32 mm. The spring constant for the tendon is the same for both groups, 32 N/mm. What is the difference in maximum stored energy between the sprinters and the nonathlethes?
Physics
1 answer:
FinnZ [79.3K]3 years ago
8 0

Answer:

Explanation:

From the given information:

The difference in the maximum energy stored is can be determined by finding the difference in the maximum stored energy in the sprinters and that of the non-athlete:

\Delta U = \dfrac{1}{2}k(x_2^2 - x_1^2)

\Delta U = \dfrac{1}{2} (32 \ N/mm) (\dfrac{1 \ mm}{10^{-3} \ m})  ((40\times 10^{-3})^2 - (32\times 10^{-3})^2)

\Delta U =16000 \times  (5.76\times 10^{-4})

\mathbf{\Delta U =9.216\ J}

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Si se deja caer un carrito de una pista de coches sin friccion y su altura inicial es de 1.4 metros, cual es la velocidad maxima
Svet_ta [14]

Answer:

5.241\ \text{m/s}

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m = Masa del coche

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h = Altura = 1.4\ \text{m}

v = Velocidad del automóvil en la parte inferior de la pista

Aquí asumimos que el automóvil desciende verticalmente. La energía potencial del automóvil se completará convertida en energía cinética en la parte inferior de la pista ya que no hay pérdida de energía.

mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.4}\\\Rightarrow v=5.241\ \text{m/s}

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8 0
3 years ago
A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following case
Murljashka [212]

Answer:

The answer is below

Explanation:

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the

cable for the following cases:

a. The load moves downward at a constant velocity

b. The load accelerates downward at a rate 0.4 m/s??

C. The load accelerates upward at a rate 0.4 m/s??

Solution:

Acceleration due to gravity (g) = 10 m/s²

a) Given that the mass of the crane (m) is 140 kg. If the load moves downward, the tension (T) is given by:

mg - T = ma

Since the load has a constant velocity, hence acceleration (a) = 0. Therefore:

mg - T = m(0)

mg - T = 0

T = mg

T = 140(10) = 1400 N

T = 1400 N

b)  If the load moves downward, the tension (T) is given by:

mg - T = ma

T = mg - ma = m(g - a)

T = 140(10 - 0.4) = 140(9.96) = 134.4

T = 134.4 N

c)  If the load moves upward, the tension (T) is given by:

T - mg = ma

T = ma + mg = m(a + g)

T = 140(0.4 + 10) = 140(10.4)

T = 145.6 N

2) To find the distance (s) if the load move from rest (u= 0) and accelerates for 20 seconds (t = 20). We use:

s = ut + (1/2)gt²

s = 0(20) + (1/2)(10)(20)²

s = 2000 m

7 0
3 years ago
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