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____ [38]
3 years ago
14

A Sense of Proportion: Saturn is about 60,000 km in radius, and its rings are only about 0.01 km thick with ripples 100m high. D

esign a really big model with Saturn 60 inches in radius (10 feet in diameter). How thick must the rings be in your model and how high can the ripples be
Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
0 0

Answer:

The thickness of ring in model is 0.00001 inch and the height of ring is 0.0001 inch.

Explanation:

The radius of Saturn = 60000 km

The thickness of rings = 0.01 km

The height of ring ripples = 100 m or 0.1 km

Now design a new model with a radius of 60 inches.

Therefore, the  radius of Saturn (600000 km)  = radius of the model (60 inches)

Here, one inch of the model will be equal to 1000 km of Saturn.

1 km of Saturn = 0.001 inch of model

Thickness of ring = 0.01 km

Therefore, thickness of ring in model = 0.01 x 0.001 = 0.00001 inch

The height of Saturn ring = 0.1 km

Therefore, the height of ring in model = 0.1 x 0.001 = 0.0001 inch.

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5 0
3 years ago
it takes 90 j of work to stretch a spring 0.2 m from its equilibrium position. How muc work is needed to stretch it an additiona
Vinvika [58]

Work needed: 720 J

Explanation:

The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the stretching of the spring from the equilibrium position

In this problem, we have

E = 90 J (work done to stretch the spring)

x = 0.2 m (stretching)

Therefore, the spring constant is

k=\frac{2E}{x^2}=\frac{2(90)}{(0.2)^2}=4500 N/m

Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of

x = 0.2 + 0.4 = 0.6 m

Substituting,

E'=\frac{1}{2}kx^2=\frac{1}{2}(4500)(0.6)^2=810 J

Therefore, the additional work needed is

\Delta E=E'-E=810-90=720 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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