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3 years ago

Point masses m1 m2 are placed at opposite ends

Physics

1 answer:

tiny-mole [99]3 years ago

7 0

(a) $x = \frac{m_2L}{m_1+m_2}$

<u>Explanation:</u>

Given:

Moment of Inertia of m₁ about the axis, I₁ = m₁x²

Moment of Inertia of m₂ about the axis. I₂ = m₂ (L - x)²

Kinetic energy is rotational.

Total kinetic energy is $E = \frac{1}{2} I_1w_0^2 + \frac{1}{2}I_2w_0^2 = \frac{1}{2} w_0^2(m_1x^2 + m_2(L-x)^2)$

Work done is change in kinetic energy.

To minimize E, differentiate wrt x and equate to zero.

$m_1x - m_2(L-x) = 0\\\\x = \frac{m_2L}{m_1+m_2}$

Alternatively, work done is minimum when the axis passes through the center of mass.

Center of mass is at $\frac{m_2L}{m_1 + m_2}$

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What is the rotational kinetic energy of the Earth about the Sun? Assume the earth is a uniform sphere, mass of the Earth is 5.9

Thepotemich [5.8K]

Answer:

2.66x10^33 J

Explanation:

In order to do this, we first need to know the expression for kinetic energy:

E = 1/2 I*w² (1)

I is moment of innertia

w is angular speed.

Moment of Innertia can be calculated using the following expression:

I = 2/5 M*R² (2)

M is mass of earth, R is radius of earth

Replacing the data in expression (2) we have:

I = 2/5 * 5.97x10^24 * (6.37x10^6)²

I = 9.69x10^37 kg m²

Next, we need to calculate the angular speed of Earth over it's axis, this is easy, because we know the Earth rotates over it's own axis once a day, 24 hours (86400 s), and assuming Earth is a perfect sphere, we can calculate the speed:

w = 2π / 86400 = 7.27x10^-5 rad/s

next thing we need to do is calculate the rotational kinetic energy of earth on it's axis, using equation (1) so:

E = 1/2 * 9.69x10^37 * (7.27x10^-5)²

E = 2.56x10^29 J

Now that we have this value, we can finally calculate the rotational kinetic energy of earth about the sun. For that, we need to calculate again the angular speed of earth about the sun. The Earth rotates around the sun once a year, or 365 days, which is 3.1536x10^7 s, so the angular speed would be:

w = 2π/3.1536x10^7 = 1.99x10^-7 rad/s

finally the energy is the combination of the sun and earth so:

K = 1/2 (Ie + Me*Rorb²)wo²

I is innertia for earth

Me mass of earth

Rorb RAdius of orbit around the sun

wo is angular speed around the sun

Replacing the data we finally have:

K = 1/2 [9.69x10^37 + 5.97x10^24 * (1.5x10^11)²]*(1.99x10^-7)²

K = 2.66x10^33 J

4 0

3 years ago

Two Carnot air conditioners, A and B, are removing heat from different rooms. The outside temperature is the same for both rooms

amid [387]

Answer:

a) Work required for air conditioner A = 354.7 J

b) Work required for air conditioner B = 310.3 J

c) The magnitude of the heat deposited outside for conditioner A = 4684.7 J

d) The magnitude of the heat deposited outside for conditioner B = 4640.3 J

Explanation:

In a carnot air conditioner, it operates like a reverse carnot engine; i.e. it removes heat from the cold reservoir (making it colder) and dumps the heat in the hot reservoir (making it hotter)

For a Carnot air conditioner,

Q꜀ is the heat removed from the colder reservoir = 4330 J for both cases

T꜀ is the temperature of the colder reservoir (temperature of the rooms) = 293 K and 296 K for both cases to be considered.

Qₕ is the heat deposited in the warmer reservoir = ? for both cases

Tₕ is the temperature of the hot reservoir (temperature of outside) = 317 K for both cases.

For Carnot air conditioners,

Qₕ = W + Q꜀ (eqn 1)

And

(Qₕ/Tₕ) - (Q꜀/T꜀) = 0 (eqn 2)

Making Qₕ the subject of formula in eqn 2

Qₕ = Tₕ (Q꜀/T꜀)

Substituting this into eqn 1

Tₕ (Q꜀/T꜀) = W + Q꜀

Q꜀ (Tₕ/T꜀) - Q꜀ = W

Q꜀ [(Tₕ - T꜀)/T꜀ ] = W

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ]

For the air conditioner A,

T꜀ = 293 K, Tₕ = 317 K, Q꜀ = 4330 J, W = ?

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ] = 4330 [ (317 - 293)/293] = 354.7 J

For the air conditioner B,

T꜀ = 296 K, Tₕ = 317 K, Q꜀ = 4330 J, W = ?

W = Q꜀ [ (Tₕ - T꜀)/T꜀ ] = 4330 [ (317 - 296)/296] = 310.3 J

c) Qₕ = W + Q꜀

For conditioner A,

Qₕ = 354.7 + 4330 = 4684.7 J

For conditioner B,

Qₕ = 310.3 + 4330 = 4640.3 J

8 0

3 years ago

A 1,000 kg ball traveling at 5 m/s would have

jonny [76]

Answer:

15 because 5×5×5 is the same thing as 5×3 which equals to 15

6 0

3 years ago

Read 2 more answers

One mole of a substance contains 6.02 × 1023 protons and an equal number of electrons. If the protons could somehow be separated

Aleksandr [31]

Explanation:

it is almost zero .this is because the distance and the electrostatic force are inversely proportional

8 0

2 years ago

Calculate the speed of a proton after it accelerates from rest through a potential difference of 350 V.

AVprozaik [17]

The speed of a proton after it accelerates from rest through a potential difference of 350 V is $25.86 \times 10^4 ~m/s$ .

Initial velocity of the proton $u = 0$

Given potential difference $\Delta V = 350V$

let's assume that the speed of the proton is $v$ ,

Since the proton is accelerating through a potential difference, proton's potential energy will change with time. The potential energy of a particle of charge $q$ when accelerated with a potential difference $\Delta V$ is,

$U = q \Delta V$

Due to Work-Energy Theorem and Conservation of Energy - <em>If there is no non-conservative force acting on a particle then loss in Potential energy P.E must be equal to gain in Kinetic Energy K.E</em> i.e

$\Delta K = \Delta V$

If the initial and final velocity of the proton is $u$ and $v$ respectively then,

change in Kinetic Energy $\implies \Delta K = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}mv^2 - 0$

change in Potential Energy $\implies \Delta U = q\Delta V$

from conservation of energy,

$v= \sqrt{\frac{2q\Delta V}{m}}$

so, $v = \sqrt{\frac{2\times 350 \times 1.6\times 10^{-19}}{1.67 \times 10^{-27}}$

$= 25.86 \times 10^4 ~m/s$

To read more about the conservation of energy, please go to brainly.com/question/14668053

7 0

1 year ago