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Elanso [62]
3 years ago
15

A mass m1=1.5 kg rests on a 30 degree ramp with a coefficient of kinetic friction = 0.40. Mass m1 is tied to another mass m with

a string which runs over a frictionless pulley. Mass m is hanging above the ground. The acceleration of masses is measured 2.68 m/s2. What is m?
Physics
1 answer:
soldier1979 [14.2K]3 years ago
6 0

Answer:

m = 2.31 Kg

Explanation:

given,

m₁ is the mass rest on the inclined plane = 1.5 Kg

inclination of plane = 30°

Kinetic friction = μ = 0.4

acceleration of the body = 2.68 m/s²

mass m is hanging  = ?

using equation to solve

T - m₁g sin θ - μ m₁g cos θ = m₁ a

on the block on inclined plane there will be acting tension T on the string,

mg sin θ will be the force acting opposite to the tension on the string and a frictional force will be acting which will oppose moment which will be equal to  μ m₁g cos θ

T = m₁(g sin θ +  μ g cos θ + a )

T = 1.5 (9.8 x sin 30° +  0.4 x 9.8 x cos 30°+ 2.68 )

T = 16.46 N

now, forces on the other side of pulley

m g - T = m a

m (g - a ) = T

m = \dfrac{16.46}{9.8-2.68}

m = 2.31 Kg

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The Earth's rotational kinetic energy is the kinetic Energy that the Earth

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The rotational kinetic energy of the Earth is approximately <u>3.331 × 10³⁶ J</u>

Reasons:

<em>The parameters required for the question are;  </em>

<em>Mass of the Earth, M = </em><em>5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = </em><em>6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = </em><em>24.0 hrs</em><em>.</em>

The \ moment  \ of \  inertia \  of \  uniform \  sphere \  is \ I =   \mathbf{\dfrac{2}{5} \cdot M \cdot R^2}

Which gives;

\mathbf{I_{Earth}} =   \dfrac{2}{5} \times 5.97 \times 10 ^{24} \cdot \left(6.38 \times 10^6 \right)^2 = 9.7202107 \times 10^{37}

\mathrm{The \ rotational \  kinetic  \ energy \  is} \   E_{rotational} = \mathbf{\dfrac{1}{2} \cdot I \cdot \omega^2}

\mathrm{The \ angular \ speed, \ \omega} = \mathbf{\dfrac{2 \dcdot \pi}{T}}

Therefore;

\omega = \dfrac{2 \cdot \pi}{24}  = \dfrac{\pi}{24}

Which gives;

\mathbf{E_{rotational}} = \dfrac{1}{2} \times  9.7202107 \times 10^{37} \times  \left(  \dfrac{\pi}{12} \right)^2 = 3.331 \times 10^{36}

The rotational kinetic energy of the Earth, E_{rotational} = <u>3.331 × 10³⁶ Joules</u>

Learn more here:

brainly.com/question/13623190

<em>The moment of inertia from part A  of the question (obtained online) is that of the Earth approximated to a perfect sphere</em>.

<em>Mass of the Earth, M = 5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = 6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = 24.0 hrs</em>

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am I right or am I right?

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