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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. 
You might be interested in
What is the threshold frequency for sodium metal if a photon with frequency 6.66 × 1014 s−1 ejects a photon with 7.74 × 10−20 J
FrozenT [24]

Answer:

5.5 × 10^14 Hz or s^-1

no orange light has less frequency so no photoelectric effect

Explanation:

hf = hf0 + K.E

HERE h is Planck 's constant having value 6.63 × 10 ^-34 J s

f is frequency of incident photon and f0 is threshold frequency

hf0 = hf- k.E

6.63 × 10 ^-34 × f0 = 6.63 × 10 ^-34× 6.66 × 10^14 - 7.74× 10^-20

6.63 × 10 ^-34 × f0 = 3.64158×10^-19

                           f0 = 3.64158×10^-19/ 6.63 × 10 ^-34

                           f0 = 5.4925 × 10^14

                            f0 =5.5 × 10^14 Hz or s^-1

frequency of orange light is 4.82 × 10^14 Hz which is less than threshold frequency hence photo electric effect will not be observed for orange light

8 0
2 years ago
A 15.0-gram lead ball at 25.0°C was heated with 40.5 joules of heat. Given the specific heat of lead is 0.128 J/g∙°C, what is th
mr Goodwill [35]

Answer:

T=4985.5^{\circ}K

Explanation:

The equation that relates heat Q with the temperature change T-T_0 of a substance of mass <em>m </em>and specific heat <em>c </em>is Q=mc(T-T_0).

We want to calculate the final temperature <em>T, </em>so we have:

T=\frac{Q}{mc}+T_0

Which for our values means (in this case we do not need to convert the mass to Kg since <em>c</em> is given in g also and they cancel out, but we add 273^{\circ} to our temperature in ^{\circ}C to have it in ^{\circ}K as it must be):

T=\frac{Q}{mc}+T_0=\frac{40.5J}{(15g)(0.128J/g^{\circ}C)}+(298^{\circ}K)=4985.5^{\circ}K

3 0
1 year ago
Hippos spend much of their lives in water, but amazingly, they don’t swim. manatees, They have, like little very body fat. The d
kenny6666 [7]

Answer:

428.59 N

Explanation:

Buoyant force, B=Vg\rho where V is volume, g is gravitational constant and \rho is density

B+F_{upward}=mg where F_{upward} is upward force

Vg\rho_{w}+F_{upward}=mg

F_{upward}=mg- Vg\rho_{w}

F_{upward}=g(mg- V\rho_{w})=g(m-m\frac {\rho_{w}{\rho_{hippo}} where \rho_{hippo} is the density of hippo

F_{upward}=mg(1-\frac {\rho_{w}}{\rho_{hippo}})

Using g as 9.81

F_{upward}=1500*9.81*(1-1000/1030)= 428.5922 N

Therefore, the upward force=428.59 N

3 0
2 years ago
A cliff diver running 3.60 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2.00 s later
mart [117]

Explanation:

It is given that,

The horizontal speed of a cliff diver, v_x=3.6\ m/s

It reaches the water below 2.00 s later, t = 2 s

Let d_x is the distance where the diver hit the water. It can be calculated as follows :

d_x=v_x\times t\\\\=3.6\times 2\\\\=7.2\ m

Let d_y is the height of the cliff. It can be calculated using second equation of motion as follows :

d_y=u_yt+\dfrac{1}{2}gt^2\\\\d_y=\dfrac{1}{2}\times 9.8\times 2^2\\\\=19.6\ m

So, the cliff is 19.6 m high and it will hit the water at a distance of 19.6 m.

8 0
1 year ago
one horsepower is a unit of power equal to 746w. how much energy can a 150-horsepower engine transform in 10.0s?
Dafna1 [17]

1 watt = 1 joule/second

1 horsepower = 746 watts = 746 joule/second

   (150 horsepower) x (746 watt/HP) x (1 joule/sec  /  watt) x (10 sec)

=  (150 x 746 x 1 x 10)  joule  =  1,119,000 joules .   
if correct plz mark brainly
8 0
1 year ago
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