Transfer of thermal or heat energy through "touch"

The weightlifter's internal store of energy decreased when he lifted the bar.

Mumz [18]

Answer:

The energy returns to the weightlifter's muscles, where it is dissipated as heat.

Explanation:

The energy returns to the weightlifter's muscles, where it is dissipated as heat. As long as the weightlifter controls the weight's descent, their muscles are acting as an overdamped shock absorber, as if the weight were sitting on a piston containing very thick fluid, slowly compressing it downward (and slightly heating up the fluid in the process). Since muscles are complicated biological systems and not simple pistons, they require metabolic energy to maintain tension throughout the controlled descent, so the weightlifter feels like they're putting energy into the weight, even though the weight's gravitational potential energy is being converted into heat within the lifter's muscles.

5 0

2 years ago

a 25-N net force is applied to a rolling cart and pruduces an acceleration of 5 m/s 2 what’s the cart mass

nignag [31]

Answer:

The mass of the cart is 5 kg

Explanation:

You divide 25 by 5 and get 5. Have a great day! :D

<em>The Equation:</em>

25/5 = 5

8 0

2 years ago

Differentiate between center of mass and center of gravity

ELEN [110]

Answer:

Centre of mass is the point at which the distribution of mass is equal in all directions, and does not depend on gravitational field. Centre of gravity is the point at which the distribution of weight is equal in all directions, and does depend on gravitational field.

Explanation:

google

3 0

3 years ago

Can you find the net force of one vertical force and one horizontal force, such as in the picture below? Yes or no, and why?

den301095 [7]

Answer:

jwuwhisbuebeu said that good morning and private s the

5 0

2 years ago

Read 2 more answers

You’ve made the finals of the science Olympics. As one of your tasks you’re given 1.0 g of copper and asked to make a cylindrica

Pani-rosa [81]

Answer:

Length = 2.92 m

Diameter = 0.11 mm

Explanation:

We have $m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A}$ , where:

$l$ is the length

$m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3$

We divide the first equation by the second equation to get:

$\frac{m}{R} = \frac{d A^2}{\rho}$

$A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8} \ m^2$