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RUDIKE [14]
3 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]3 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 tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly
never [62]

Answer:

1) Vf = 3.36 m/s

2) v = 6.86 m/s

3) s = 92.3 m

4) a = - 0.23 m/s²

Explanation:

1)

We use first equation of motion in this case:

Vf = Vi + at

where,

Vf = Final velocity = ?

Vi = Initial velocity = 0 m/s

a = acceleration = 1.4 m/s²

t = time = 2.4 s

Therefore,

Vf = 0 m/s + (1.4 m/s²)(2.4 s)

<u>Vf = 3.36 m/s</u>

2)

We again use first equation of motion but with t= 4.9 s now:

Vf = 0 m/s + (1.4 m/s²)(4.9 s)

Vf = 6.86 m/s

Now, this velocity remained constant for net 11 seconds. Hence, the velocity of hare after 8.9 s is:

<u>v = 6.86 m/s</u>

3)

First we use second equation of motion to find distance covered in accelerated motion:

s₁ = Vi t + (0.5)at²

s₁ = (0 m/s)(4.9 s) + (0.5)(1.4 m/s²)(4.9 s)²

s₁ = 16.8 m

Now, we calculate the distance covered in uniform motion:

s₂ = vt

s₂ = (6.86 m/s)(11 s)

s₂ = 75.5 m

Now, total distance covered before slowing down is given as:

s = s₁ + s₂ = 16.8 m + 75.5 m

<u>s = 92.3 m</u>

3)

Using third equation of motion for the decelerated motion:

2as = Vf² - Vi²

where,

a = deceleration = ?

s = distance covered = 102 m

Vf = Final Speed = 0 m/s

Vi = Initial Speed = 6.86 m/s

Therefore,

(2)(a)(102 m) = (0 m/s)² - (6.86 m/s)²

a = - (47.06 m²/s²)/(204 m)

<u>a = - 0.23 m/s²</u>

3 0
3 years ago
A horizontal force of 300.0 N is used to push a 145-kg mass 30.0 m horizontally in 3.00 s. Calculate the power developed.
Svetradugi [14.3K]

Answer:

3 * 10³J/s

Explanation:

Given :

Force applied, F = 300 N

Distance, d = 30 m

Time, t = 3 seconds

Power, P = Workdone / time

Recall :

Workdone = Force * distance

Workdone = 300 N * 30 m = 9000 Nm

Workdone = 9 * 10³ J

Power = (9 * 10³ J) / 3s

Power = 3 * 10³J/s

4 0
3 years ago
Match the microscopes with their useable
umka21 [38]
Microscopes with life science and physics
6 0
3 years ago
Read 2 more answers
The head of a grass string trimmer has 100 g of cord wound in a light, cylindrical spool with inside diameter 3.00 cm and outsid
faust18 [17]

Answer:

a).11.546J

b).2.957kW

Explanation:

Using Inertia and tangential velocity

a).

w=2250*2\pi *\frac{1}{60}\\ w=235.61

I=\frac{1}{2}*m*((\frac{d_{i} }{2})^{2} +(\frac{d_{e} }{2})^{2})\\m=100g *\frac{ikg}{1000g}=0.1kg\\ d_{i}=3cm*\frac{1m}{100cm}=0.03m \\ d_{e}=18cm*\frac{1m}{100cm}=0.18m\\I=\frac{1}{2}*0.1kg*((\frac{0.03m}{2})^{2} +(\frac{0.18m}{2})^{2})\\I=0.41625x10^{-3}kg*m^{2}

Now using Inertia an w

E=\frac{1}{2}*I*(w)^{2} \\ E=\frac{1}{2}*0.416x10^{-3}*(235.61)^{2} \\E=11.54J

average power=\frac{11.4J}{0.230s}=50.2 W

b).

power=t*w

P=11.5465*0.25*235.61

P=2.957 kW

4 0
3 years ago
Read 2 more answers
HELP
Digiron [165]
Speed is the answer
4 0
2 years ago
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