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NARA [144]
3 years ago
11

A mass of 1.03 kilograms is placed on a horizontal frictionless surface against an uncompressed spring with spring constant 861.

1 N/m. The mass is pushed against the spring until the spring is compressed a distance 0.36 m and then released. How high (vertically) in m does the mass rise from the original height before it stops (momentarily).
Physics
1 answer:
Lelechka [254]3 years ago
8 0

Answer:

5.522 m

Explanation:

Data provided:

Mass, m = 1.03 kg

spring constant, k = 861.1 N/m

Distance by which the spring is compressed, x = 0.36

Thus,

the energy stored in the spring = \frac{1}{2}kx^2

on substituting the values, we get

the energy stored in the spring =  \frac{1}{2}\times861.1\times0.36^2

now,

by the conservation of energy, we have

Potential energy gained by the mass =  Energy gained by the spring

or

mgh = \frac{1}{2}\times861.1\times0.36^2

where,

g is the acceleration due to the gravity

h is the maximum height reached by the mass before falling

on substituting the values in the above relation, we get

1.03 × 9.81 × h = \frac{1}{2}\times861.1\times0.36^2

or

h = 5.522 m

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Answer:

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The moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is
sleet_krkn [62]

Answer:

I = I₀ + M(L/2)²

Explanation:

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Speed is 300,000 km/sec or 300,000,000 m/s and the wavelength is 300,000  km or 300,000,000 m.
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