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Oksana_A [137]
2 years ago
11

A spring whose stiffness is 980 N/m has a relaxed length of 0. 50 m. If the length of the spring changes from 0. 25 m to 0. 81 m

, what is the change in the potential energy of the spring
Physics
1 answer:
Gelneren [198K]2 years ago
8 0

151.9j

Explanation:

PE=1/2kx^2

PE=1/2(980)(.50)= 245j

PE=(1/2)(980)(.81)= 396.9j

396.9- 245= 151.9j

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In the two-slit experiment, for the condition of bright fringes, the value of m = +2 corresponds to a path difference of λ.
melisa1 [442]

Answer:

False

Explanation:

We know that if path difference is even multiple of wavelength then bright fringes are formed and if path difference is odd multiple of wavelength then dark fringes are formed .

For bright fringes

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If m = 2 then the path difference will be

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therefore the above statement if false.

False

7 0
3 years ago
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Novosadov [1.4K]

Answer:

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4 years ago
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musickatia [10]

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6 0
3 years ago
A 4 kg billiard ball moving on a horizontal surface has a speed of 16 m/s when it strikes a horizontal coiled spring is brought
ruslelena [56]

Answer:

spring constant of the spring is 1820.44 N/m

Explanation:

given data

ball mass = 4 kg

speed = 16 m/s

distance = 0.75 m

to find out

spring constant of the spring

solution

we know that kinetic energy of ball = energy store in spring as compression

so we can express it as

0.5 × m × v² = 0.5 × k × x²    ....................1

so put here value we get spring constant k

m × v² =  k × x²

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solve it we get

k  =  1820.44 N/m

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