Ask question

Ask question

All categories

3 years ago

10

After Pedro had breakfast, he went for a run. Identify at least three energy transformations that had to take place to allow Ped

ro to complete his run. Then use evidence to support the claim that energy in this system was conserved as it was transformed.

1 answer:

FinnZ [79.3K]3 years ago

3 0

Answer:

1) Chemical to mechanical (calories -> muscle contractions)

2) mechanical to mechanical (muscle -> motion)

3) mechanical to thermal (motion -> heat from the body causes sweating)

The potential energy is stored in the body chemically (via food). That potential energy is then released when the body is put in motion.

Explanation:

You might be interested in

How does the centre of gravity vary with different objects?

Aliun [14]

Answer:

The point through which the Whole mass of the body acts, irrespective of the position of the body, is known as centre of gravity (briefly written as c.g.). ... Mass affects the centre of gravity of the object.

Explanation:

5 0

3 years ago

Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter

alukav5142 [94]

Answer: 0.258

Explanation:

The resistance $R$ of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. For aluminium is $\rho_{Al}=2.65(10)^{-8}m\Omega$ and for copper is $\rho_{Cu}=1.68(10)^{-8}m\Omega$

$l$ is the length of the wire, which in the case of aluminium is $l_{Al}=12m$ , and in the case of copper is $l_{Cu}=30m$

$s$ is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

$s=\pi{(\frac{d}{2})}^{2}$ (2) Where $d$ is the diameter of the circumference.

For aluminium wire the diameter is $d_{Al}=2.5mm=0.0025m$ and for copper is $d_{Cu}=1.6mm=0.0016m$

So, in this problem we have two transversal areas:

<u>For aluminium:</u>

$s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}$

$s_{Al}=0.000004908m^{2}$ (3)

<u>For copper:</u>

$s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}$

$s_{Cu}=0.00000201m^{2}$ (4)

Now we have to calculate the resistance for each wire:

<u>Aluminium wire:</u>

$R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}$ (5)

$R_{Al}=0.0647\Omega$ (6) Resistance of aluminium wire

<u>Copper wire:</u>

$R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}$ (6)

$R_{Cu}=0.250\Omega$ (7) Resistance of copper wire

At this point we are able to calculate the ratio of the resistance of both wires:

$Ratio=\frac{R_{Al}}{R_{Cu}}$ (8)

$\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}$ (9)

Finally:

$\frac{R_{Al}}{R_{Cu}}=0.258$ This is the ratio

3 0

4 years ago

O prevent tank rupture during deep-space travel, an engineering team is studying the effect of temperature on gases confined to

Ira Lisetskai [31]

To calculate for the pressure of the system, we need an equation that would relate the number of moles (n), pressure (P), and temperature (T) with volume (V). There are a number of equations that would relate these values however most are very complex equations. For simplification, we assume the gas is an ideal gas. So, we use PV = nRT.<span>

PV = nRT where R is the universal gas constant
P = nRT / V</span>

<span>P = 3.40 mol ( 0.08205 L-atm / mol-K ) (251 + 273.15 K) / 1.75 L </span>

<span>P = 83.56 atm</span>

<span>
</span>

<span>Therefore, the pressure of the gas at the given conditions of volume and temperature would be 83.56.</span>

7 0

3 years ago

Based on its location on the periodic table, an element that is not naturally occurring is?

Scorpion4ik [409]

this is a link to a web sight with a diagram to help you

https://www.google.com/url?sa=i&source=images&cd=&ved=2ahUKEwjw4v7knNDgAhVyUN8KHWgWD-wQjRx6BAgBEAU&url=%2Furl%3Fsa%3Di%26source%3Dimages%26cd%3D%26ved%3D%26url%3Dhttps%253A%252F%252Ffuturism.com%252Fwhere-do-all-the-elements-come-from%26psig%3DAOvVaw19_FOCuWs_nMsyY1YT0Da-%26ust%3D1550955269844922&psig=AOvVaw19_FOCuWs_nMsyY1YT0Da-&ust=1550955269844922

7 0

4 years ago

Which of the following is NOT an exercise myth?

galben [10]

I believe it would be A, (B) just because a women trains doesn't mean she'll bulk, the may become thinner, or more tone. No to C because You have to change up and make workouts more difficult to help improve your body, and no to D because a workout doesn't have to be strenuous, you can be careful and not crazy about working out and still help your body. hope this helped

8 0

3 years ago