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Sergeeva-Olga [200]
3 years ago
11

A potential energy function for a system in which a two-dimensional force acts is of the form U = 3x6y − 5x. Find the force that

acts at the point (x, y). (Use the following as necessary: x and y.)
Physics
1 answer:
liq [111]3 years ago
7 0

Answer:

Force, F=((-18x^5y+5)i+(-3x^6)j)\ N

Explanation:

Given that,

A potential energy function for a system in which a two-dimensional force acts is of the form of :

U=3x^6y-5x

We need to find the force that acts at the point (x, y). The force in 2 dimensional with components is given by :

F=(\dfrac{-dU}{dx},\dfrac{-dU}{dy})\\\\F=(\dfrac{-d(3x^6y-5x)}{dx},\dfrac{d(3x^6y-5x)}{dy})\\\\F=(-(18x^5y-5)),(-(3x^6))\\\\F=((-18x^5y+5)i+(-3x^6)j)\ N

So, the force acting at the point (x,y) is ((-18x^5y+5)i+(-3x^6)j)\ N. Hence, this is the required solution.

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Calculate the force of gravity between planet X and planet y if both planets are 3.75 X 10^11 m apart, planet X has a mass of 1.
GenaCL600 [577]

So, the force of gravity that the asteroid and the planet have on each other approximately \boxed{\sf{2.9 \times 10^{17} \: N}}

<h3>Introduction</h3>

Hi ! Now, I will help to discuss about the gravitational force between two objects. The force of gravity is not affected by the radius of an object, but radius between two object. Moreover, if the object is a planet, the radius of the planet is only to calculate the "gravitational acceleration" on the planet itself,does not determine the gravitational force between the two planets. For the gravitational force between two objects, it can be calculated using the following formula :

\boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}}

With the following condition :

  • F = gravitational force (N)
  • G = gravity constant ≈ \sf{6.67 \times 10^{-11}} N.m²/kg²
  • \sf{m_1} = mass of the first object (kg)
  • \sf{m_2} = mass of the second object (kg)
  • r = distance between two objects (m)

<h3>Problem Solving</h3>

We know that :

  • G = gravity constant ≈ \sf{6.67 \times 10^{-11}} N.m²/kg²
  • \sf{m_X} = mass of the planet X = \sf{1.55 \times 10^{22}} kg.
  • \sf{m_Y} = mass of the planet Y = \sf{3.95 \times 10^{28}} kg.
  • r = distance between two objects = \sf{3.75 \times 10^{11}} m.

What was asked :

  • F = gravitational force = ... N

Step by step :

\sf{F = G \times \frac{m_X \times m_Y}{r^2}}

\sf{F = 6.67 \cdot 10^{-11} \times \frac{1.55 \cdot 10^{22} \cdot 3.95 \times 10^{28}}{(3.75 \times 10^{11})^2}}

\sf{F \approx \frac{40.84 \times 10^{-11 + 22 + 28}}{14.0625 \times 10^{22}}}

\sf{F \approx 2.9 \times 10^{39 - 22}}

\sf{F \approx 2.9 \times 10^{17} \: N}

<h3>Conclusion</h3>

So, the force of gravity that the asteroid and the planet have on each other approximately

\boxed{\sf{2.9 \times 10^{17} \: N}}

<h3>See More</h3>
  • Gravity is a thing has depends on ... brainly.com/question/26485200
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Assume n1 and n2 are two adjacent energy levels of an atom. The emission of radiation with the longest wavelength would occur fo
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Answer:

Explanation:

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attashe74 [19]

<u>Answer:</u>

The amount of the lighted side of the moon you can see is the same during "how much of the sunlit side of the moon faces Earth".

<u>Explanation:</u>

The Moon is in sequential rotation with Earth, and thus displays the Sun, the close side, always on the same side. Thanks to libration, Earth can display slightly greater than half (nearly 59 per cent) of the entire lunar surface.

The side of the Moon facing Earth is considered the near side, and the far side is called the reverse. The far side is often referred to as the "dark side" inaccurately but it is actually highlighted as often as the near side: once every 29.5 Earth days. During the New Moon the near side becomes blurred.

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