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Romashka-Z-Leto [24]
4 years ago
8

A spring has a spring constant of 30000 N/m. How far must it be stretched (in meters) for its potential energy to be 47 J?

Physics
1 answer:
Evgen [1.6K]4 years ago
5 0
The elastic potential energy in a spring is given by:
EPE = 1/2 kx²
x = √(2 x 47 / 30,000)
x = 0.056 m
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which is warmer -30 degrees Celcius or -30 degrees Fahrenheit? And how would i show my work for this question?
Alenkasestr [34]
Convert the temperature scales:

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->TC/5 = TF - 32/9

-30/5 = TF - 32/9

-270 = 5TF - 160

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5TF = -110

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So, -30°C is warmer than -30°F.

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4 years ago
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Given a triangle with sides X = 6.35 cm and Y = 12.25 cm with an angle of 90 degrees between them, find the length of the hypote
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fomenos

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