U=1/2kx2
This image sums it up
Momentum will be conserved in one dimension in the explosion.
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Given that the fragment a acquires three
times the kinetic energy of the fragment b.
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P</span><span><span>initial </span><span>= p</span></span>final ⇒ 0 =mₐv⁰ₐ+mьv⁰ь= 0 ⇒ v⁰ь = -mₐv⁰ₐ/mь
KE= 3KEь
⇒1/2 mₐv⁰ₐ² = 3 (1/2mьv⁰ь²)
</span><span>
⇒1/2 mₐv⁰ₐ² = 3/2 mь(-mₐv⁰ₐ/mь)²
⇒1/2 mₐv⁰ₐ² = 3/2 mь(mₐ²v⁰ₐ²/mь²)
</span>
⇒1/2 x 2/3 = mₐ/mь= 1/3
<span>
<span>
Thus the ratio
of the masses of the fragments is 1:3.
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Answer:
Acceleration due to gravity is 20 
So option (E) will be correct answer
Explanation:
We have given length of the pendulum l = 2 m
Time period of the pendulum T = 2 sec
We have to find acceleration due to gravity g
We know that time period of pendulum is given by
Squaring both side
So acceleration due to gravity is 20
So option (E) will be correct answer.
Answer:
Explanation:
95.0 km/hr = 26.39 m/s
65 km/hr = 18.06 m/s
Circumference of a tire is 0.9π m
77 revolutions is a distance of
77(0.9π) = 69.3π m
v² = u² + 2as
a = (v² - u²) / 2s
a = (18.06² - 26.39²) / (2(69.3π))
a = -0.85 m/s²
s = (v² - u²) / 2a
s = (0² - 26.39²) / 2(-0.85)
s = 409 m
Distance from the sun.
<span>The third law of planetary motion states that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit</span>. The semi-major axis is the distance from the sun to the epicenter of the ellipse (which would be the planet in question). So, the revolutionary period is directly related to the distance of the planet from the sun.
