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valentina_108 [34]
3 years ago
14

A hockey stick stores 4.2 J of potential energy when it is bent 3.1 cm. Treating the hockey stick as a spring, what is its sprin

g constant
Physics
1 answer:
storchak [24]3 years ago
3 0

Answer:

Spring constant, k = 8740.89 N/m

Explanation:

Given the following data;

Potential energy = 4.2 J

Extension, e = 3.1 cm to meters = 3.1/100 = 0.031 m

To find the spring constant;

Mathematically, the potential energy stored in a spring is given by the formula;

P.E = \frac {1}{2}Kx^{2}

Where;

P.E is potential energy.

K is the spring constant.

x is the extension of the spring.

Substituting into the equation, we have;

4.2 = \frac {1}{2}*k*0.031^{2}

Cross-multiplying, we have;

8.4 = k*0.000961

k = \frac {8.4}{0.000961}

Spring constant, k = 8740.89 N/m

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The effective height of the water for Smith's house will be 24.61m.

<h3>How to calculate the height?</h3>

Based on the information given, the volume of the water in sphere will be:

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The effective height of the water will be:

= 18.0 + 2(5.18)

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The gauge pressure at Faucet of Jones house will be:

= pgh

= 1000(9.8)(28.36)

= 277.9kPa

The effective height of the water for Smith's house will be:

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A woman stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.90
kiruha [24]

Answer:

1 m/s² down

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PLEASE HELP ME WITH THIS
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Answer:

i think it might be c but i know its not A or B and im

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What is the net force on a car with a mass of 1000 kg if its<br> acceleration is 35 m/s^2?
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A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 19
Hunter-Best [27]

The car will take 300 m before it stops due to applying break.

<h3>What's the relation between initial velocity, final velocity, acceleration and distance?</h3>
  • As per Newton's equation of motion, V² - U² = 2aS
  • V= final velocity velocity of the object, U = initial velocity velocity of the object, a= acceleration, S = distance covered by the object
  • Here, U = 60 ft/sec, V = 0 m/s, a= -6 ft/sec²
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=> -3600 = -12S

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Thus, we can conclude that the distance covered by the car is 300 m before it stopped.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

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