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lilavasa [31]
3 years ago
6

What is the distance from the center of the Moon to the point between Earth and the Moon where the gravitational pulls of Earth

and Moon are equal? The mass of Earth is 5.97 × 1024 kg, the mass of the Moon is 7.35 × 1022 kg, the center-to-center distance between Earth and the Moon is 3.84 × 108 m, and G = 6.67 × 10-11 N ∙ m2/kg2.
Physics
1 answer:
Tanya [424]3 years ago
5 0

Answer:

r = 3.84 \times 10^7 m

Explanation:

Let say the distance at which gravitational force due to both Earth and moon is zero is given by force force balance

so it is given as

F_{moon} = F_{Earth}

\frac{GM_{moon}}{r^2} = \frac{GM_{earth}}{(d - r)^2}

so we have

\frac{M_{moon}}{r^2} = \frac{M_{earth}}{(d - r)^2}

\frac{7.35 \times 10^{22}}{r^2} = \frac{5.97 \times 10^{24}}{(3.84 \times 10^8 - r)^2}

now we have

\frac{2.71}{r} = \frac{24.4}{3.84\times 10^8 - r}

10.4 \times 10^8 - 2.71 r = 24.4 r

now we have

r = 3.84 \times 10^7 m

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2. a) A disc rotates about its axis at speed 25 revolutions per minute and takes 15 s to stop. Calculate the
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The statement shows a case of rotational motion, in which the disc <em>decelerates</em> at <em>constant</em> rate.

i) The angular acceleration of the disc (\alpha), in revolutions per square second, is found by the following kinematic formula:

\alpha = \frac{\omega_{f}-\omega_{o}}{t} (1)

Where:

  • \omega_{o} - Initial angular speed, in revolutions per second.
  • \omega_{f} - Final angular speed, in revolutions per second.
  • t - Time, in seconds.

If we know that \omega_{o} = \frac{5}{12}\,\frac{rev}{s}, \omega_{f} = 0\,\frac{rev}{s} y t = 15\,s, then the angular acceleration of the disc is:

\alpha = \frac{0\,\frac{rev}{s}-\frac{5}{12}\,\frac{rev}{s}}{15\,s}

\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}

The angular acceleration of the disc is \frac{1}{36} radians per square second.

ii) The number of rotations that the disk makes before it stops (\Delta \theta), in revolutions, is determined by the following formula:

\Delta \theta  = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \alpha} (2)

If we know that \omega_{o} = \frac{5}{12}\,\frac{rev}{s}, \omega_{f} = 0\,\frac{rev}{s} y \alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}, then the number of rotations done by the disc is:

\Delta \theta = 3.125\,rev

The disc makes 3.125 revolutions before it stops.

We kindly invite to check this question on rotational motion: brainly.com/question/23933120

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3 years ago
A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V. What are (a) the total energy stored i
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Answer:

(A) Total energy will be equal to 0.044\times 10^{-5}J

(b) Energy density will be equal to 0.0175J/m^3

Explanation:

We have given diameter of the plate d = 2 cm = 0.02 m

So area of the plate A=\pi r^2=3.14\times 0.02^2=0.001256m^2

Distance between the plates d = 0.50 mm = 0.50\times 10^{-3}m

Permitivity of free space \epsilon _0=8.85\times 10^{-12}F/m

Potential difference V =200 volt

Capacitance between the plate is equal to C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 0.001256}{0.50\times 10^{-3}}=0.022\times 10^{-9}F

(a) Total energy stored in the capacitor is equal to

E=\frac{1}{2}CV^2

E=\frac{1}{2}\times 0.022\times 10^{-9}\times 200^2=0.044\times 10^{-5}J

(b) Volume will be equal to V=Ad, here A is area and d is distance between plates

V=0.001256\times 0.02=2.512\times 10^{-5}m^3

So energy density =\frac{Energy}{volume}=\frac{0.044\times 10^{-5}}{2.512\times 10^{-5}}=0.0175J/m^3

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4 years ago
a hydraulic machine can be used to lift extremely heavy objects. why is the fluid in the hydraulic machine a liquid rather than
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Answer:

1) The mass of the student is approximately 73.39 kg

2) The net force on the student is approximately 947.523 N

3) The value the scale will read is approximately 96.59 kg

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The weight of the student = 720 N

The speed at which the elevator is decreasing = 3.1 m/s²

1) The weight of the student = The mass of the student × The acceleration due to gravity

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Substituting the known values gives;

720 N = The mass of the student × 9.81 m/s²

∴ The mass of the student = 720 N/(9.81 m/s²) ≈ 73.39 kg

2) The forces acting on the student are;

i) The force of gravity which is the weight of the student acting downwards

ii) The inertia force of the slowing elevator acting downwards in the same direction as the weight of the student

The net force, F_{net} = The weight of the student + The inertia force of the slowing elevator

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3) The scale will read the mass of the student as follows;

Mass reading of student on the scale = Force on scale/9.81

∴ Mass reading of student on the scale = 947.523/9.81 ≈ 96.59 kg

The value the scale will read = 96.59 kg.

3 0
3 years ago
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