Ask question

Ask question

All categories

3 years ago

5

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

died the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters tendons stretch 41 mm, while nonathletes' stretch only 33 mm. The spring constant for the tendon is the same for both groups, 33 N/mm. What is the difference in maximum stored energy between the sprinters and the nonathlethes?

1 answer:

Anni [7]3 years ago

7 0

Answer:

ΔE = 9.7mJ

Explanation:

See attachment below.

You might be interested in

If I = 2.0 A in the circuit segment shown below, what is the potential difference VB - VA?

Tom [10]

Answer:

10 V

Explanation:

The potential difference between two points is the amount of work required to carry a unit charge from one point to the other point. This would result in a potential difference between this two points.

The difference between the potential across two points B and A is $V_{BA}=V_B-V_A$

From the image attached:

$V_B-V_A=10-10I+20\\\\But\ I = 2A,hence:\\\\V_B-V_A=10-10(2)+20\\\\V_B-V_A=10-20+20\\\\V_B-V_A=10\ V$

7 0

2 years ago

Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg

vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

$\omega = \frac{v}{R}$

Where,

$\omega =$ Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

$\alpha = \frac{\omega}{t}$

Where

$\alpha =$ Angular acceleration

$\omega =$ Angular velocity

t = Time

Our values are

$v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})$

$v = 16.67m/s$

$r = 0.25m$

$t=6s$

Replacing at the previous equation we have that the angular velocity is

$\omega = \frac{v}{R}$

$\omega = \frac{ 16.67}{0.25}$

$\omega = 66.67rad/s$

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

$\alpha = \frac{\omega}{t}$

$\alpha = \frac{66.67}{6}$

$\alpha = 11.11rad/s^2$

Therefore the angular acceleration of a point on the outer edge of the tires is $11.11rad/s^2$

5 0

2 years ago

What is the impulse

vichka [17]

Impulse is the integral of a force, F.
Hope this helps.
(Please mark this brainliest, I would really appreciate it) Thanks!

5 0

3 years ago

A garden hose having with an internal diameter of 1.1 cm is connected to a (stationary) lawn sprinkler that consists merely of a

trapecia [35]

Answer:

Water leaves the sprinkler at a speed of 2.322 m/sec

Explanation:

We have given internal diameter of the garden $d_1=1.1cm$

Speed of water in the hose is $v_1=0.95m/sec$

Number of holes n = 22

Diameter of each holes $d_2=15cm$

According to continuity equation $A_1v_1=A_2v_2$

$d_1^2\times v_1=22\times d_2^2v_2$

$1.1^2\times 0.95=22\times 0.15^2\times v_2$

$v_2=2.322m/sec$

So water leaves the sprinkler at a speed of 2.322 m/sec

6 0

3 years ago

Need Help ASAP!!<br> (Picture)

olchik [2.2K]

Answer:225000000000

Explanation:

5 0

3 years ago