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RUDIKE [14]
3 years ago
9

(8c7p26) During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with lar

ge catapults that are made with surgical hose mounted on a window frame. A balloon filled with dyed water is placed in a pouch attached to the hose, which is then stretched through the width of the room. Assume that the stretching of the hose obeys Hooke's law and has a spring constant of 102 N/m. If the hose is stretched by 4.8 m and then released, how much work does the force from the hose do on the balloon in the pouch by the time the hose reaches its relaxed length
Physics
1 answer:
likoan [24]3 years ago
7 0

Answer: 1175 J

Explanation:

Hooke's Law states that "the strain in a solid is proportional to the applied stress within the elastic limit of that solid."

Given

Spring constant, k = 102 N/m

Extension of the hose, x = 4.8 m

from the question, x(f) = 0 and x(i) = maximum elongation = 4.8 m

Work done =

W = 1/2 k [x(i)² - x(f)²]

Since x(f) = 0, then

W = 1/2 k x(i)²

W = 1/2 * 102 * 4.8²

W = 1/2 * 102 * 23.04

W = 1/2 * 2350.08

W = 1175.04

W = 1175 J

Therefore, the hose does a work of exactly 1175 J on the balloon

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6 0
3 years ago
Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250x10-7C (charges)
SIZIF [17.4K]

The electric field at one corner of a square is 1614217 N/C.

Explanation:

The distance between x and y direction diagonals.

As per the given details the distance between diagonals is calculated as

0.5² + 0.5² = c²  =>  c = 0.707 m

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In order to find the electric charge towards x direction

we use e = kq/r² formula

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e = (9 x 10^{9})(250 x 10^{-7}) / (0.5)²

e = 9 x 10^{5} N/C

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e = kq/r²

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e = kq/r²

e = (9 x 10^{9})(250 x 10^{-7}) / (0.5)²

e = 9 x 10^{5} N/C

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e = kq/r²

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4 0
3 years ago
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Unknown:

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Velocity brainly.com/question/10883914

#learnwithBrainly

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