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Nuetrik [128]
3 years ago
9

How long does creatine phosphate provide every

Physics
1 answer:
Tems11 [23]3 years ago
6 0
Creatine phosphate can supply the energy needs of a working muscle at a very high rate, but only for about 8–10 seconds.
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A ball is tossed int the air and rises to a height of 12m. How long is the ball in the air?
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1.5 seconds that its in the air
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HELP WILL GIVE BRAINLIEST IF CORRECT
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It’s B i literally jus learned this
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You put a room temperature metal spoon into a bowl of hot soup. What will happen to
Semenov [28]

Answer:

the soup will transfer its thermal energy to the spoon

Hope this helps! <u><em>Can you please mark me brainliest? (also can you guys answer my recent questions? thanks ^^ )</em></u>

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7 0
3 years ago
A mass 25kg is attached to a spring which is held stretched a distance 120 cm by a force F=300N (Fig. 7-28), and then released.
azamat

Explanation:

Spring is stretched by force f to distance

"X"

now here by force balance we can say

f = kx

k = !

now here we will we say that energy stored in the spring will convert into kinetic energy

kx² = mv²

(x² = mv²

now solving above equation we will have

PART 2)

now for half of the extension again we can use energy conservation

ka²-k(x/2)² = 1 {mv²

¾fx=mv²

now the speed is given as

3 fr 4m V=

4 0
2 years ago
A particular heat engine has a mechanical power output of 4.00 kW and an efficiency of 26.0%. The engine expels 8.55 103 J of ex
Ivahew [28]

To develop the problem we will start by finding the energy taken by each cycle through the efficiency of the motor and the exhausted energy. Later the work will be found for the conservation of energy in which this is equivalent to the difference between the two calculated energy values. Finally the estimated time will be calculated with the work and the power given,

\text{Efficiency of the heat engine} = \eta = 26\% = 0.26

\text{Energy taken in by the heat engine during each cycle} = Q_h

\text{Energy exhausted by the heat engine in each cycle} = Q_c = 8.55*10^3 J

\eta = 1 - \frac{Q_{c}}{Q_{h}}

0.26 = 1 - \frac{8.55\ast 10^{3}}{Q_{h}}

\frac{8.55* 10^{3}}{Q_{h}} = 0.74

Q_h = \frac{8.55*10^3}{0.74}

Q_h = 11.554*10^3J

PART A)

Work done by the heat engine in each cycle = W

W = Q_h-Q_c

W = 11.554*10^3J-8.55*10^3J

W = 3004J

According to the value given we have that,

P = 4.0kW

P = 4000W

Power is defined as the variation of energy as a function of time therefore,

P = \frac{W}{t}

4000W = \frac{3004J}{t}

t = \frac{3004}{4000}

t = 0.75s

Therefore the interval for each cycle is 0.75s

5 0
3 years ago
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