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3 years ago

A 5 kg mass is oscillating on a spring with a time period of 2.8 seconds. What is the spring constant k of the spring?

Physics

1 answer:

katrin [286]3 years ago

5 0

Answer:

Explanation:

<u>Simple Harmonic Oscillator</u>

It consists of a weight attached to one end of a spring being allowed to move forth and back.

If m is the mass of the weight and k is the constant of the spring, the period of the oscillation is given by:

$\displaystyle T=2\pi {\sqrt {\frac {m}{k}}}$

If the period is known, we can find the value of the constant by solving for k:

$\displaystyle k=m\left(\frac{2\pi}{T}\right)^2$

Substituting the given values m=5 Kg and T=2.8 seconds:

$\displaystyle k=5 \left(\frac{2\pi}{2.8}\right)^2$

k = 25.18 N/m

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3 years ago

A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo

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Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

$E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2$

$E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2$

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

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3 years ago

Why does the solar system consist of small planets orbiting close to sun and larger on more distant

ahrayia [7]

Answer:

The inner planets are smaller and rockier

Explanation:

Astronomers divide the planets into two groups in Solar system, the inner planets and outer planets. The inner planets are smaller and rockier and it is closer to the sun. The outer planets are larger , further far away and made of gas

The inner planets are Mercury, Venus , Earth and Mars. The outer planets Jupiter , Saturn , Uranus and Neptune comes after an asteroid belt. In some other planetary systems the gas are close to the sun.

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2 years ago

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has

Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = $\sqrt{6gL}$

Hence, v0 = 6.86 m/s

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3 years ago

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vovikov84 [41]

Answer:

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Explanation:

The expression they give for the strength of the tide is

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They ask us to perform the calculation

F = 2 6.67 10⁻¹¹ 135 5.98 10²⁴ 13 / (6.79 10⁶)³

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