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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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How can an individual benefit from participating in the recommended amount of physical activity?
Orlov [11]
It will help them stay in better shape resulting in better health, being more physically active, and staying happier as a result.
3 0
3 years ago
A football player kicks a football downfield. The height of the football increases until it reaches a maximum height of 15 yards
Trava [24]

Answer:

kick 1 has travelled 15 + 15 = 30 yards before hitting the ground

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards  

Explanation:

1st kick travelled 15 yards to reach maximum height of 8 yards

so, it has travelled 15 + 15 = 30 yards before hitting the ground

2nd kick is given by the equation

y (x) = -0.032x(x - 50)

Y = 1.6 X - 0.032x^2

we know that maximum height occurs is given as

x = -\frac{b}{2a}

y =- \frac{1.6}{2(-0.032)} = 25

and maximum height is

y = 1.6\times 25 - 0.032\times 25^2

y = 20

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

8 0
3 years ago
Read 2 more answers
A camera lens focuses on an object 75.0 cm from the lensThe image forms 3.50 cm behind the lens. What is the magnification of th
N76 [4]

Answer:

7/150

Explanation:

The following data were obtained from the question:

Object distance (u) = 75cm

Image distance (v) = 3.5cm

Magnification (M) =..?

Magnification is simply defined as:

Magnification (M) = Image distance (v)/ object distance (u)

M = v /u

With the above formula, we can obtain the magnification of the image as follow:

M = v/u

M = 3.5/75

M = 7/150

Therefore, the magnification of the image is 7/150.

6 0
3 years ago
A 5.31 kg object is swung in a vertical circular path on a string 2.99 m long. The acceleration of gravity is 9.8 m/s 2 . If the
11111nata11111 [884]

Answer:

T = 120.3 N

Explanation:

Since, the tension in the rope is acting against both the centripetal force and the weight of the stone. As both act downward towards center of the circle and tension acts towards point of support that is upward. So, tension will be equal to the sum of centripetal force and weight of the stone:

Tension = Centripetal Force + Weight of Stone

T = mv²/r + mg

where,

m = mass of stone = 5.31 kg

r = radius of circle = length of string = 2.99 m

g = 9.8 m/s²

Therefore,

T = (5.31 kg)(6.2 m/s)²/(2.99 m) + (5.31 kg)(9.8 m/s²)

T = 68.27 N + 52.03 N

<u>T = 120.3 N</u>

4 0
2 years ago
A brick is dropped from a high scaffold. a. What is its velocity after 4.0s ?
Ilia_Sergeevich [38]

Answer:

A: 1.962

B: 3.924

Explanation:

g = G *M /R^2

g = 9.807*M/R^2 the gravitational constant of ground level on earth is about 9.807

g = 9.807*5lbs/R^2 the average brick is about 5 pounds.

g = 9.807*5*10^2.   I'm assuming the height is around ten feet to help you out.

with these numbers plugged in you get an acceleration of 0.4905 a final velocity after 4 seconds 1.962. It's height fallen after 4 seconds is 3.924.

( M = whatever the brick weighs it's not specified in the question)

(R = the distance from the ground or how high the scaffold is)

(hopefully you can just plug your numbers in there hope this helps)

6 0
3 years ago
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