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Vladimir [108]
2 years ago
12

A spring whose spring constant is 260 lbf/in has an initial force of 100 lbf acting on it. Determine the work, in Btu, required

to compress it another 1 in.
Physics
1 answer:
Readme [11.4K]2 years ago
7 0

Answer:

A spring whose spring constant is 200 lbf/in has an initial force of 100 lbf acting on it. Determine the work, in Btu, required to compress it another 1 inch.

Step 1 of 4

The force at any point during the deflection of the spring is given by,

where is the initial force

and x is the deflection as measured from the point where the initial force occurred.

The work required to compress the spring is

Therefore work required to compress the spring is

The work required to compress the spring in Btu is calculated by

Where 1Btu =778

The work required to compress the spring,  

eman Asked on February 19, 2018 in thermal fluid Sciences 4th solutions.

Explanation:

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A, a spark because if you touch something after rubbing your hand on carpet or something you get a spark
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What constant acceleration is required to increase the speed of a car from 26 mi/h to 51 mi/h in 5 seconds
mr_godi [17]

2.25 m/s² of acceleration is required to increase the speed of a car from 26 mi/h to 51 mi/h in 5 seconds.

To find the answer, we need to know about the acceleration.

<h3>What is acceleration?</h3>
  • Acceleration is given as the ratio of velocity to time.
  • Mathematically, acceleration= velocity/time.
<h3>What is the acceleration required to increase the speed of a car from 26 mi/h to 51 mi/h in 5 seconds?</h3>
  • Here change in velocity of the car is 51-26= 25 mi/h.
  • As 1 mi/h = 0.45 m/s. So 25mi/h = 11.25 m/s.
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Thus, we can conclude that the constant acceleration is 2.25 m/s².

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8 0
1 year ago
Oscilloscopes have parallel metal plates inside them to deflect the electron beam. These plates are called the deflecting plates
alexdok [17]

Answer:

1.6 pF

Explanation:

The capacitance of a parallel-plate capacitor in air is given by:

C=\frac{\epsilon_0 A}{d}

where

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d is the separation between the plates

In this problem we have:

- Separation between the plates: d = 5.00 mm = 0.005 m

- Area of the plates: A = 3.00 cm \cdot 3.00 cm = 9.00 cm^2 = 9\cdot 10^{-4}m^2

Therefore, the capacitance is

C=\frac{(8.85\cdot 10^{-12}F/m)(9\cdot 10^{-4} m^2)}{0.005 m}=1.6\cdot 10^{-12} F=1.6 pF

8 0
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Answer:

Explanation:

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DR= 5729.57795

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R = 5729.57795/4

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0.08 + 0.13 = V^2 / (32*1432.4)

V^2 = 9625.728 or V = 98 ft/sec

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7 0
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lisov135 [29]

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