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choli [55]
2 years ago
15

The concept that large masses warp the shape of spacetime and thus change the path of an object moving through that spacetime ap

plies to all objects. As such, Einstein’s theory of general relativity predicts that light will also curve around massive objects as it travels through space. Does Newton’s law of gravity agree with that prediction? Note that the mass of a photon is zero.
Physics
1 answer:
mrs_skeptik [129]2 years ago
4 0

Answer: Newton's law of gravity theory disagrees with Einstein's theory. The last one were named Theory of General Relativity and was discovered and proposed by the physicist Einstein in the year of 1915.

Explanation: Until the beginning of 20 century, the physics were ruled by Isaac Newton's ideas. He believed that the gravity was a force caused by the objects mass on the space, made them to be draw towards each other. Newton thought that the greater the mass of the object, the more intense was its attraction, which would justify the planet's moovements around the sun and how the gravity between them maintain the planets on solar orbit. Concluding, he believed gravity was a immediate force of action, regardless of the distance of the bodies.

Contrary to Newton's law of gravity, in 1915, the physicist Einstein created the Theory of General Relativity, wich discovered that gravity was, in fact, the deformation caused by the attraction of massive celestial bodies. This deformation, related to the Sun, for example, creates a curvature on the space-time and this curvature are followed by the other planets.

So, we can conclude that Newton's law of gravity disagree's with the Theory of General Relativity, once the first believes that light force of attraction are transmitted instantly and, as Einstein already prooved, the gravity influency propagates in the speed of light.

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svetoff [14.1K]

One of the brightest nebulae in the night sky, the Orion Nebula may be seen with the unaided eye. The Trapezium is a young open cluster of four main stars in this magnitude 4 interstellar cloud of ionized atomic hydrogen.

<h3>What is the source of the Orion Nebula's crimson glow?</h3>
  • The hydrogen gas in the Orion Nebula, which is powered by radiation from young stars, gives off a crimson tint. The nebula's blue-violet regions are reflecting radiation from bright, blue-white O-type stars while the red areas are emitting light.
  • The Orion Nebula is one of many massive clouds of gas and dust in our Milky Way galaxy, say contemporary astronomers, and is one of the largest. It is approximately 1,300 light years away from Earth. This enormous hazy cocoon, which measures approximately 30 to 40 light-years in diameter, is generating potentially a thousand stars.  

To learn more about Orion nebula refer to:

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3 0
11 months ago
In any energy transformation, there is always some energy that gets wasted as non-useful heat.
Nady [450]
It is a completely false statement that in <span>any energy transformation, there is always some energy that gets wasted as non-useful heat. The correct option among the two options that are given in the question is the second option. I hope that this is the answer that has actually come to your desired help.</span>
8 0
3 years ago
Read 2 more answers
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
2 years ago
A 1200-kilogram automobile in motion strikes a 0.0001-kilogram insect. As a result, the insect is accelerated at a rate of 100 m
nadya68 [22]

Answer:

force = 1 × 10^{-2} N

Explanation:

given data

automobile mass = 1200 kg

insect mass = 0.0001 kg

insect accelerated = 100 m/s²

to find out

magnitude of the force the insect exerts on the car

solution

we get here force the insect exerts that is express as

force = mass × acceleration    ............1

put here value we get

force = 0.0001 × 100 m/s²

force = 1 × 10^{-2} N

6 0
2 years ago
Read 2 more answers
Calculate the amount of heat (in kj) required to raise the temperature of a 79.0 g sample of ethanol from 298.0 k to 385.0 k. th
earnstyle [38]
We are given with the specific heat capacity of ethanol, the mass of the sample and the temperature change to determine the total amount of heat to raise the temperature. The formula to be followed is H = mCpΔT. Upon subsituting, H = 79 g * 2.42 J/gC *(385-298)C = 16.63 kJ 
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3 years ago
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