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ludmilkaskok [199]
3 years ago
13

The existence of the dwarf planet Pluto was proposed based on irregularities in Neptune's orbit. Pluto was subsequently discover

ed near its predicted position. But it now appears that the discovery was fortuitous, since Pluto is small and the irregularities in Neptune's orbit were not well known. Answer the following to illustrate that Pluto has a minor effect on the orbit of Neptune compared with other planets. (a) Calculate the acceleration of gravity at Neptune due to Pluto when they are 3.2 1012 m apart. The mass of Pluto is 1.3 1022 kg. g1 = m/s2 (b) Calculate the acceleration of gravity at Neptune due to Uranus, when they are 3.2 1012 m apart. The mass of Uranus is 8.68 1025 kg. g2 = m/s2 (c) Give the ratio of this acceleration to that due to Pluto. g2 g1 =
Physics
1 answer:
iris [78.8K]3 years ago
3 0

Answer:

a) g1 = 8.467 10⁻¹⁴ m / s² , b)  g2 = 5.65 10⁻¹⁰ m / s² , c) g2 / g1 = 6.6 10³

Explanation:

a) for this we will use Newton's second law where force is the force of universal gravitation

         F = m a

        G m M / r² = m a

        a = g = G M / r²

By the effect of Pluto

        g1 = 6.67 10⁻¹¹  1.3 10²² / (3.2 10¹²)²

        g1 = 8.467 10⁻¹⁴ m / s²

b) By the effect of Uranus

        g2 = 6.67 10⁻¹¹ 8.68 10²⁵ / (3.2 10¹²)²

        g2 = 5.65 10⁻¹⁰ m / s²

c) the relationship between these accelerations

        g2 / g1 = 5.65 10⁻¹⁰ / 8.47 10⁻¹⁴

        g2 / g1 = 0.66 10⁴

        g2 / g1 = 6.6 10³

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A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a
Elza [17]

Answer:

\dfrac{dz}{dt}=0.65\ ft/s

Explanation:

Given that

x= 150 ft

\dfrac{dy}{dt}= 7\ ft/s

y= 14 ft

From the diagram

z^2=x^2+y^2

When ,x= 150 ft and y= 14 ft

z^2=150^2+14^2

z=\sqrt{150^2+15^2}

z=150.74 ft

z^2=x^2+y^2

By differentiating with respect to time t

2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}

z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}

Here x is constant that is why

\dfrac{dx}{dt}=0

z\dfrac{dz}{dt}= y\dfrac{dy}{dt}

Now by putting the values in the above equation we get

150.74\times \dfrac{dz}{dt}=14\times 7

\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s

\dfrac{dz}{dt}=0.65\ ft/s

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

5 0
3 years ago
A car travels up a hill at a constant speed of 38 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the a
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So time required to return fro hill t=\frac{d}{66}h

Total time t_{total}=\frac{t}{38}+\frac{t}{66}

Total distance = d+d =2d

So average speed=\frac{total\ distance}{total\ time}=\frac{2d}{\frac{d}{38}+\frac{d}{66}}=48.23km/h

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Hole filling fasteners (for example, MS20470 rivets) should not be used in composite structures primarily because of the?
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We can cause delamination.

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If we analyze the force applied by the expanded rod will cause that the matrial will be deteriorated and will cause that the material to delaminate around the edges of the hole and we can cause possible control and no protection to the material.

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