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3 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]3 years ago

8 0

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

Effectus [21]3 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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The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a

likoan [24]

Answer:

14869817.395 m

Explanation:

$\theta$ =22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

$22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians$

From Rayleigh Criterion

$\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m$

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0

3 years ago

A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force of friction on the cart

Ber [7]

Newton's second law states that the resultant of the forces applied to an object is equal to the product between the object's mass and its acceleration:
$\sum F = ma$
where in our problem, m is the mass the (child+cart) and a is the acceleration of the system.

We are only concerned about what it happens on the horizontal axis, so there are two forces acting on the cart+child system: the force F of the man pushing it, and the frictional force $F_f$ acting in the opposite direction. So Newton's second law can be rewritten as
$F-F_a = ma$
or
$F=ma + F_f$

since the frictional force is 15 N and we want to achieve an acceleration of $a=1.50 m/s^2$ , we can substitute these values to find what is the force the man needs:
$F=(30 kg)(1.5 m/s^2)+15 N=60 N$

8 0

3 years ago

Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father

Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity $\omega_o = 0$

angular acceleration $\alpha$ = 4.44 rad/s²

Using the formula:

$\omega = \omega_o+ \alpha t$

Making t the subject of the formula:

$t= \dfrac{\omega- \omega_o}{ \alpha }$

where;

$\omega = 1.53 \ rad/s^2$

∴

$t= \dfrac{1.53-0}{4.44 }$

t = 0.345 s

Using the formula:

$\omega ^2 = \omega _o^2 + 2 \alpha \theta$

here;

$\theta$ = angular displacement

∴

$\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }$

$\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }$

$\theta =0.264 \ rad$

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

$x = \dfrac{0.264 \times 1}{2 \pi}$

x = 0.042 revolutions

Here; force = 270 N

radius = 1.20 m

The torque = F * r

$\tau = 270 \times 1.20 \\ \\ \tau = 324 \ Nm$

However;

From the moment of inertia;

$Torque( \tau) = I \alpha \\ \\ Since( I \alpha) = 324 \ Nm. \\ \\ Then; \\ \\ \alpha= \dfrac{324}{I}$

given that;

I = 84.4 kg.m²

$\alpha= \dfrac{324}{84.4} \\ \\ \alpha=3.84 \ rad/s^2$

For re-tardation; $\alpha=-3.84 \ rad/s^2$

Using the equation

$t= \dfrac{\omega- \omega_o}{ \alpha }$

$t= \dfrac{0-1.53}{ -3.84 }$

$t= \dfrac{1.53}{ 3.84 }$

t = 0.398s

The required time it takes= 0.398s

5 0

2 years ago

A steam engine takes in superheated steam at 270 °C and discharges condensed steam from its cylinder at 50 °C. The engine has an

Vanyuwa [196]

Answer:

b) 20 kJ

Explanation:

Efficiency of carnot engine = (T₁ - T₂ ) / T₁ Where T₁ is temperature of hot source and T₂ is temperature of sink .

T₁ = 270 + 273 = 543K

T₂ = 50 + 273 = 323 K

Putting the given values of temperatures

efficiency = (543 - 323) / 543

= .405

heat input = 50 KJ

efficiency = output work / input heat energy

.405 = output work / 50

output work = 20.25 KJ.

= 20 KJ .

6 0

3 years ago

Four point charges are placed at the corners of a square. Each charge has the identical value +Q. The length of the diagonal of

kvv77 [185]

Answer: $V_{net}=4\frac{kQ}{a}$

Explanation:

Given

charge on each Particle is Q

Length of diagonal of the square is 2a

therefore distance between center and each charge is $\frac{2a}{2}=a$

Electric Potential of charged Particle is given by

For First Charge

$V_1=\frac{kQ}{a}$

$V_2=\frac{kQ}{a}$

$V_3=\frac{kQ}{a}$

$V_4=\frac{kQ}{a}$

total Electric Potential At center is given by

$V_{net}=V_1+V_2+V_3+V_4$

$V_{net}=4\times \frac{kQ}{a}$

$V_{net}=4\frac{kQ}{a}$

7 0

3 years ago