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hjlf
3 years ago
8

URGENT PLEASE ANSWER

Physics
1 answer:
Anettt [7]3 years ago
6 0

Answer:

3.523 * 10²² N

Explanation:

Newton's Law of Universal Gravitation:

  • \displaystyle \bf{F=G\frac{m_1 m_2}{r^2}
  • F = force of gravity between two objects
  • G = gravitational constant (6.67 * 10⁻¹¹ Nm²kg⁻²)
  • m₁ = mass of object #1
  • m₂ = mass of object #2
  • r = distance between the center of mass of two objects

We are given the mass of the sun (m₁) and its distance from Earth (r). We want to find the value of F.

The mass of Earth (m₂) is 5.972 * 10²⁴ kg and we know the gravitational constant (G).

Let's plug these known values into the equation to solve for F.

  • \displaystyle \bf{F=6.67\cdot10^-^1^1 \Big [ \frac{(1.99\cdot 10^3^0)(5.972\cdot 10^2^4)}{(150\cdot 10^9)^2} \Big ]  

Plugging this into your calculator will output a value of

  • \displaystyle \bf{ F = 3.523028782\cdot 10^2^2

The gravitational force between the sun and the Earth is approximately:

  • \displaystyle \bf{F=3.523\cdot 10^2^2 \ N
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Point A of the circular disk is at the angular position θ = 0 at time t = 0. The disk has angular velocity ω0 = 0.17 rad/s at t
statuscvo [17]

Given that,

Angular velocity = 0.17 rad/s

Angular acceleration = 1.3 rad/s²

Time = 1.7 s

We need to calculate the angular velocity

Using angular equation of motion

\omega=\omega_{0}+\alpha t

Put the value in the equation

\omega=0.17+1.3\times1.7

\omega=2.38(k)\ m/s

We need to calculate the angular displacement

Using angular equation of motion

\theta=\theta_{0}+\omega_{0}t+\dfrac{\alpha t^2}{2}

Put the value in the equation

\theta=0+0.17\times1.7+\dfrac{1.3\times1.7^2}{2}

\theta=2.1675\times\dfrac{180}{\pi}

\theta= 124.18^{\circ}

We need to calculate the velocity at point A

Using equation of motion

v_{A}=v_{0}+\omega\times r

Put the value into the formula

v_{A}=0+2.38(k) \times0.2(\cos(124.18)i+\sin(124.18)j))

v_{A}=0.476\cos(124.18)j+0.476\sin(124.18)i

v_{A}=(-0.267j-0.393i)\ m/s

We need to calculate the acceleration at point A

Using equation of motion

a_{A}=a_{0}+\alpha\times r+\omega\times(\omega\times r)

Put the value in the equation

a_{A}=0+1.3(k)\times0.2(\cos(124.18)i+\sin(124.18)j)+2.38\times2.38\times0.2(\cos(124.18)i+\sin(124.18)j)

a_{A}=0.26\cos(124.18)i+0.26\sin(124.18)j+(2.38)^2\times0.2(\cos(124.18)i+\sin(124.18)j)

a_{A}=-0.146j-0.215i−0.636i+0.937j

a_{A}=0.791j-0.851i

a_{A}=-0.851i+0.791j\ m/s^2

Hence, (a). The velocity at point A is (-0.267j-0.393i)\ m/s

(b). The acceleration at point A is (-0.851i+0.791j)\ m/s^2

3 0
4 years ago
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