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postnew [5]
3 years ago
12

Determine the Universal Gravitational Constant (G).

Physics
1 answer:
aivan3 [116]3 years ago
8 0

Answer:

G = 6.6675 x 10^{-11} Nm^{2}/kg^{2}

Explanation:

From Newton's law of universal gravitation,

F = \frac{GMm} {r^{2} }  .................. 1

Where F is the force of attraction, G is the universal gravitation constant, M is the mass of a massive object e.g earth, m is the mass of the smaller object e.g moon, r is the radius of the massive object.

Newton's second law states;

F = mg  ...................  2

F is the force, m is the mass of the object and g is the gravitational force on the object.

Equating equations 1 and 2 gives;

mg = \frac{GMm} {r^{2} }

So that;

g = \frac{GM}{r^{2} }

⇒ G = \frac{gr^{2} }{M}

where: g = 9.81 m/s^{2}, r = 6371 km, M = 5.972 x 10^{24} kg.

G = \frac{9.81*(6371000)^{2} }{5.972*10^{24} }

  = \frac{3.98184*10^{14} }{5.972*10^{24} }

  = 6.6675 x 10^{-11} Nm^{2}/kg^{2}

The Universal Gravitational Constant (G) is 6.6675 x 10^{-11} Nm^{2}/kg^{2}.

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A 34-kg child runs with a speed of 2.8 m/s tangential to the rim of a stationary merrygo-round. The merry-go-round has a momentu
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Answer: 0.43\ rad/s

Explanation:

Given

Mass of child m=34\ kg

speed of child is v=2.8\ m/s

Moment of inertia of merry go round is I=510\ kg.m^2

radius r=2.31\ m

Conserving the angular momentum

\Rightarrow mvr=I\omega \\\Rightarrow 34\times 2.8\times 2.31=510\times \omega\\\\\Rightarrow \omega=\dfrac{219.912}{510}\\\Rightarrow \omega=0.43\ rad/s

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3 years ago
Question 3 A woman sitting in a stationary car notices a man cycling past her at 40 km/hr. Five seconds after he passes her car,
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The shortest possible distance in which she catches up with the man is 129.7 m.

The given parameters:

speed of the man, v₁ = 40 km/hr

speed of the woman, v₂ = 70 km/hr

time when the woman started moving, t₁ = 5 seconds

To find:

  • the shortest distance when she catches up with the man

Convert the given speed in km/h to m/s

  • 3.6 km/h = 1 m/s
  • 40 km/h = 40/3.6 = 11.11 m/s
  • 70 km/h = 70/3.6 = 19.44 m/s

Calculate the distance moved by the man before she started following him.

Distance moved by the man = speed x time

                                                = 11.11 m/s  x  5 s

                                                = 55.55 m

Let the time the woman catches up with the man = t

Apply the following simple principle to calculate the time.

The man is moving forward and the woman is moving forward to close the gap between them, thus both speeds are working against each other.

(v_2-v_1)t = 55.55 \ m

  • the difference in the speed is because they are working against each other.

(19.44 - 11.11)t = 55.55\\\\8.33t = 55.55\\\\t = \frac{55.55}{8.33} \\\\t = 6.67 \ s

The time when the woman catches up with the man = 6.67 s

The shortest distance at which the woman catches up with the man is calculated as;

Shortest distance = woman's speed x time when she catches up

                               = 19.44 m/s    x  6.67 s

                               = 129.7 m

Thus, the shortest possible distance in which she catches up with the man is 129.7 m.

Learn more here: https://brainly.in/question/9541933?tbs_match=3

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Answer:

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