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RUDIKE [14]
2 years ago
12

An activity that is relatively short in time (< 10 seconds) and has few repetitions predominately uses the _____________ ener

gy system.
Physics
1 answer:
tensa zangetsu [6.8K]2 years ago
5 0
An activity that is relatively short in time <10 seconds and has few repetitions predominantly uses the ATP/PC energy system. The cellular respiration procedure that changes food energy into ATP which is a form of energy is largely reliant on oxygen obtainability. During exercise the source and request of oxygen obtainable to muscle is unnatural by period and strength and by the individual’s cardiorespiratory suitability level. 
Steps of the ATP-PC system:


1. Primarily, ATP kept in the myosin cross-bridges which is microscopic contractile parts of muscle is broken down to issue energy for muscle shrinkage.  This action consents the by-products of ATP breakdown which are the adenosine diphosphate and one single phosphate all on its own.


2. Phosphocreatine is then broken down by the enzyme creatine kinase into creatine and phosphate.


3. The energy free in the breakdown of PC permits ADP and Pi to rejoin creating more ATP.  This newly made ATP can now be broken down to issue energy to fuel activity. 
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A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? A flat metal washe
jenyasd209 [6]

answer;

The hole in the center of the washer will expand

explanation;

<em>A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? The hole in the center will remain the same size. Changes in the hole cannot be determined without know the composition of the metal. The hole in the center of the washer will expand. The hole in the center of the washer will contract.</em>

this is an example of area expansivity.

coefficient of area expansivity is change in area per area per degree rise in temperature

a=dA/(A*dt)

as the temperature rises , there will be volumetric and area expansivity on the body. volume also increases because of the intermolecular forces of attraction between the molecule is now getting apart.

7 0
2 years ago
A 8.00g sample of substance (substance, molar mass = 152.0 g/mol) was combusted in a bomb calorimeter with a heat capacity of 6.
aleksandrvk [35]

Answer:

ΔH°comb=-5899.5 kJ/mol

Explanation:

First, consider the energy balance:

m_{c} *Cp*(T_{2}-T_{1})=-n_{s} *H_{c} Where m_{c} is the calorimeter mass and n_{s} is the number of moles of the samples; H_{c} is the combustion enthalpy. The energy balance says that the energy that the reaction release is employed in rise the temperature of the calorimeter, which is designed to be adiabatic, so it is suppose that the total energy is employed rising the calorimeter temperature.

The product m_{c} *Cp is the heat capacity, so the balance equation is:

6.21\frac{kJ}{K}*(75-25)=-8.00g*\frac{mol}{152.0g}*H_{c}

So, the enthalpy of combustion can be calculated:

H_{c}=-5899.5\frac{kJ}{mol}

I will be happy to solve any doubt you have.

4 0
2 years ago
A vertical cylinder is divided into two parts by a movable piston of mass m. The piston and cylinder system is well insulated (t
Mekhanik [1.2K]

Answer:

Final temperature will be 438.076 K

Explanation:

We have given temperature T_1=323K

Volume V_1=V\ and\ V_2=\frac{V}{2}

As there is no heat transfer so this is an adiabatic process

For and adiabatic process TV^{\gamma -1}=constant

Here \gamma =1.4

So T_1V_1^{\gamma -1}=T_2V_2^{\gamma -1}

T_2=\left ( \frac{V_1}{V_2} \right )^{\gamma -1}\times T_1

T_2=\left ( \frac{V}{\frac{V}{2}} \right )^{1.4 -1}\times 332=2^{0.4}\times 332=438.076K

4 0
2 years ago
A boat is floating in a small pond. the boat then sinks so that it is completely submerged. what happens to the level of the pon
lukranit [14]
A boat is floating in a small pond. the boat then sinks so that it is completely submerged. what happens to the level of the pond?
It increases!
4 0
2 years ago
A 40-mH ideal inductor is connected in series with a 50 Ω resistor through an ideal 15-V DC power supply and an open switch. If
sergey [27]

Answer:i=300 mA

Explanation:

Given

inductance(L)=40 mH

Resistor(R)=50 \Omega

Voltage(V)=15 V

Time constant(\tau)=\frac{L}{R}

\tau =\frac{40\times 10^{-3}}{50}=8\times 10^{-4}

current i_0=\frac{V}{R}

i_0=\frac{15}{50}=0.3 A

Current as a function of time is given by

i=i_0\left ( 1-e^{-\frac{t}{\tau }}\right )

i=0.3\times 0.9998

i= 299.95 mA

6 0
2 years ago
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