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krok68 [10]
3 years ago
11

Block A is also connected to a horizontally-mounted spring with a spring constant of 281 J/m2. What is the angular frequency (in

rad/s) of simple harmonic oscillations of this system?
Physics
1 answer:
Svetradugi [14.3K]3 years ago
5 0

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, the formula to be used here is

ω = 2π/T

Where ω is the angular frequency (in rad/s)

T is the period - the time taken for Block A to complete one oscillation and return to it's original position.

To solve for this period T, the formula below should be used

T = 2π√m/k

where m is the mass of the object (Block A) and k is the spring constant (281 J/m²)

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Hence, the correct option is a.

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As a moon follows its orbit around a planet, the maximum gravitational force exerted on the moon by the planet exceeds the minim
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F = GmM / r²

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                      1/rmin² = 0.127(1/rmax²) + 1/rmax²

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Read 2 more answers
A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length ℓ and of negligible
mr Goodwill [35]

Answer:

a)

mv l

b)

\frac{M }{(M + m)}

Explanation:

Complete question statement is as follows :

A wooden block of mass M resting on a friction less, horizontal surface is attached to a rigid rod of length ℓ and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.

(a) What is the angular momentum of the bullet–block system about a vertical axis through the pivot? (Use any variable or symbol stated above as necessary.)

(b) What fraction of the original kinetic energy of the bullet is converted into internal energy in the system during the collision? (Use any variable or symbol stated above as necessary.)

a)

m = mass of the bullet

v = velocity of the bullet before collision

r = distance of the line of motion of bullet from pivot = l

L = Angular momentum of the bullet-block system

Angular momentum of the bullet-block system is given as

L = m v r

L = mv l

b)

V = final velocity of bullet block combination

Using conservation of momentum

Angular momentum of bullet block combination = Angular momentum of bullet

(M + m) V l = m v l\\V =\frac{mv}{(M + m)}

K_{o} = Initial kinetic energy of the bullet

Initial kinetic energy of the bullet is given as

K_{o} = (0.5) m v^{2}

K_{f} = Final kinetic energy of bullet block combination

Final kinetic energy of bullet block combination is given as

K_{f} = (0.5) (M + m) V^{2}

Fraction of original kinetic energylost is given as

Fraction = \frac{(K_{o} - K_{f})}{K_{o}} = \frac{((0.5) m v^{2} - (0.5) (M + m) V^{2})}{(0.5) m v^{2}}

Fraction = \frac{(m v^{2} - (M + m) (\frac{mv}{(M + m)})^{2})}{m v^{2}} = \frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}

Fraction = \frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}\\ \frac{M }{(M + m)}

6 0
3 years ago
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