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2 years ago

What important discovery was made by edmond halley and alexis clairaut, using newton's ideas of mathematics and gravity?

Physics

2 answers:

Semmy [17]2 years ago

8 0

They were able to find the orbit of a comet and predict the year of its return.

Effectus [21]2 years ago

8 0

Answer:

Comet's orbit.

Explanation:

Edmond Halley and Alexis Clairaut were one of those Newtonian who helped to validate the results of Isaac Newton.

They correctly explained Comet' motion and how planets gravitational force affect them.

But using Newton's idea of Mathematics and Gravity, they have made two important discoveries:

1. Why celestial bodies have different colors.

2. They were able to find the Comet's orbit and prediction(year) of its return.

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Which subatomic particle identifies the element?

grin007 [14]

A. protons.ekgfsbdbsi

3 0

2 years ago

Read 2 more answers

A 95.0 kg satellite moves on a circular orbit around the Earth at the altitude h=1.20*103 km. Find: a) the gravitational force e

Natali5045456 [20]

Answer:

a) $F = 660.576\,N$ , b) $a_{c} = 6.953\,\frac{m}{s^{2}}$ , c) $v \approx 7255.423\,\frac{m}{s}$ , $\omega = 9.583\times 10^{-4}\,\frac{rad}{s}$ , d) $T \approx 1.821\,h$

Explanation:

a) The gravitational force exerted by the Earth on the satellite is:

$F = G\cdot \frac{m\cdot M}{r^{2}}$

$F = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot \frac{(95\,kg)\cdot (5.972\times 10^{24}\,kg)}{(7.571\times 10^{6}\,m)^{2}}$

$F = 660.576\,N$

b) The centripetal acceleration of the satellite is:

$a_{c} = \frac{660.576\,N}{95\,kg}$

$a_{c} = 6.953\,\frac{m}{s^{2}}$

c) The speed of the satellite is:

$v = \sqrt{a_{c}\cdot R}$

$v = \sqrt{\left(6.953\,\frac{m}{s^{2}} \right)\cdot (7.571\times 10^{6}\,m)}$

$v \approx 7255.423\,\frac{m}{s}$

Likewise, the angular speed is:

$\omega = \frac{7255.423\,\frac{m}{s} }{7.571\times 10^{6}\,m}$

$\omega = 9.583\times 10^{-4}\,\frac{rad}{s}$

d) The period of the satellite's rotation around the Earth is:

$T = \frac{2\pi}{\left(9.583\times 10^{-4}\,\frac{rad}{s} \right)} \cdot \left(\frac{1\,hour}{3600\,s} \right)$

$T \approx 1.821\,h$

6 0

2 years ago

A day on a distant planet observed orbiting a nearby star is 21.5 hr. Also, a year on the planet lasts 69.3 Earth days. In other

serg [7]

Answer:

Part A

The angular speed of rotation of the plane is 8.11781 × 10⁻⁵ rad/s

Part B

The angular speed of orbit of the planet is 1.04938 × 10⁻⁶ rad/s

Explanation:

The parameters of the planet are;

The duration of a day on the distant planet = 21.5 hr.

The duration of a year on the distant planet = 69.3 Earth days

Part A

The duration of a day = The time to make one complete revolution of 2·π radians

∴ The average angular speed about its axis, $\omega_{rotation}$ = Angle turned/Time

∴ $\omega_{rotation}$ = 2·π/(21.5 × 60 × 60) s ≈ 8.11781 × 10⁻⁵ rad/s

The average angular speed of the planet about its own axis, $\omega_{rotation}$ = 8.11781 × 10⁻⁵ rad/s

The angular speed of rotation of the plane $\omega_{rotation}$ = 8.11781 × 10⁻⁵ rad/s

Part B

The time it takes the planet to revolve round the neighboring star once = 69.3 Earth days

Therefore, the average angular speed of the planet around its neighboring star, $\omega _{Star}$ , is given as follows;

$\omega _{Orbit}$ = 2·π/((69.3 × 24 × 60 × 60) s) = 1.04938 × 10⁻⁶ rad/s

The average angular speed of orbit, $\omega _{Orbit}$ = 1.04938 × 10⁻⁶ rad/s

The angular speed of orbit of the planet, $\omega _{Orbit}$ = 1.04938 × 10⁻⁶ rad/s.

3 0

2 years ago

A child is sledding. He starts at the top of the hill with a velocity of zero, and 3 seconds later he is speeding down the hill

Flauer [41]

Magnitude of acceleration = (change of speed) / (time for the change) =

(12 m/s - 0) / (3 sec) =

12/3 = <em>4 m/s²</em>

What's a challenge question ? Have we all passed the event horizon
and been spaghettified without knowing it ?

8 0

2 years ago

A 70.0-cm length of a cylindrical silver wire with a radius of 0.175 mm is extended horizontally between two leads. The potentia

Alla [95]

Answer:

(a) the magnitude and direction of the electric field in the wire is 4.5 N/C towards the left end of the wire.

(b) the resistance of the wire is 0.1154 Ω

(d) the current density in the wire is 4.053 x 10⁸ A/m³

Explanation:

Given;

Length of cylinder = 70cm = 0.7m

radius of cylinder = 1.75 x 10⁻⁴ m

potential V = 3.15 V

resistivity = 1.586 ✕ 10⁻⁸ Ω · m

Part (a) the magnitude and direction of the electric field in the wire

V = E x d

E = V/d

E = 3.15/0.7 = 4.5 N/C towards the left end of the wire.

Part (b) the resistance of the wire

$R = \frac{\rho L}{A} \\\\R =\frac{\rho L}{\pi r^2} =R = \frac{1.586X10^{-8} X 0.7}{\pi (1.75X10^{-4})^2} = 0.1154 ohms$

R = 0.1154 Ω

Part (c) the magnitude and direction of the current in the wire

V = IR

I =V/R

I = 3.15/0.1154

I = 27.3 A towards the left end of the wire

Part (d) the current density in the wire

current density,J = current /volume

volume = πr²h = π x (1.75 × 10⁻⁴)² x 0.7 = 6.7357 x 10⁻⁸ m³

$J = \frac{27.3}{6.7357 X10^{-8}} = 4.053 X10^8 \frac{A}{m^3}$

J = 4.053 x 10⁸ A/m³

3 0

3 years ago