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Orlov [11]
3 years ago
9

Jupiter and Saturn each have the same three basic cloud layers, but the spacing of the layers differs on the two planets. Why is

this?
a. On Saturn, the "rain" of helium condensate has forced the lower layers downward, so they are more separated there.
b. Jupiter's greater gravity has compressed the layers, so they are closer together there.
c. Jupiter's greater rotation rate has flung the layers outward, so they are separated more there.
d. Jupiter's warmer temperature has expanded the atmosphere, so they are more separated there.
Physics
1 answer:
-Dominant- [34]3 years ago
3 0

Answer:

b. Jupiter’s greater gravity has compressed the layers, so they are closer together there.

Explanation:

The value for Jupiter mass is 1.8981×10²⁷kg, while the mass of Saturn is 5.6832×10²⁶kg, so the different layers of clouds in Jupiter will be submitted to a greater gravitational pull because it has a bigger mass, as is established in the law of universal gravitation:

F = G\frac{m1m2}{r^{2}}   (1)

Where m1 and m2 are the masses of two objects, G is the gravitational constant and r is the distance between the two objects.

As it can be seen in equation 1, the gravitational force is directly proportional to the product of the masses of the objects, so if the mass increase the gravitational force will do it too.  

For the case of Saturn, it has a lower mass so its layers of clouds will suffer a weaker gravitational pull. That leads to the three clouds being more spacing that the ones of Jupiter.    

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Answer: C and D

The equipment would have stayed in the same exact location indefinitely until the very moment the astronaut applied force to it.

The equipment will continue moving in the same direction indefinitely unless another force is applied to stop it.

Explanation: According to Newton's first law of motion which state that; A body at rest will continue to be at rest, or in linear motion will continue to move in a straight line, unless an external force act on it.

The equipment would have stayed in the same exact location indefinitely until the very moment the astronaut applied force to it.

immediately the astronaut apply force to the object by pushing in, Newton's first law will be manifested in which the equipment will continue moving in the same direction indefinitely unless another force is applied to stop it.

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3 years ago
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Arm ab has a constant angular velocity of 16 rad/s counterclockwise. At the instant when theta = 60
geniusboy [140]

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second.

<h3>How to determine the angular velocity of a collar</h3>

In this question we have a system formed by three elements, the element AB experiments a <em>pure</em> rotation at <em>constant</em> velocity, the element BD has a <em>general plane</em> motion, which is a combination of rotation and traslation, and the ruff experiments a <em>pure</em> translation.

To determine the <em>linear</em> acceleration of the collar (a_{D}), in inches per square second, we need to determine first all <em>linear</em> and <em>angular</em> velocities (v_{D}, \omega_{BD}), in inches per second and radians per second, respectively, and later all <em>linear</em> and <em>angular</em> accelerations (a_{D}, \alpha_{BD}), the latter in radians per square second.

By definitions of <em>relative</em> velocity and <em>relative</em> acceleration we build the following two systems of <em>linear</em> equations:

<h3>Velocities</h3>

v_{D} + \omega_{BD}\cdot r_{BD}\cdot \sin \gamma = -\omega_{AB}\cdot r_{AB}\cdot \sin \theta   (1)

\omega_{BD}\cdot r_{BD}\cdot \cos \gamma = -\omega_{AB}\cdot r_{AB}\cdot \cos \theta   (2)

<h3>Accelerations</h3>

a_{D}+\alpha_{BD}\cdot \sin \gamma = -\omega_{AB}^{2}\cdot r_{AB}\cdot \cos \theta -\alpha_{AB}\cdot r_{AB}\cdot \sin \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \cos \gamma   (3)

-\alpha_{BD}\cdot r_{BD}\cdot \cos \gamma = - \omega_{AB}^{2}\cdot r_{AB}\cdot \sin \theta + \alpha_{AB}\cdot r_{AB}\cdot \cos \theta - \omega_{BD}^{2}\cdot r_{BD}\cdot \sin \gamma   (4)

If we know that \theta = 60^{\circ}, \gamma = 19.889^{\circ}, r_{BD} = 10\,in, \omega_{AB} = 16\,\frac{rad}{s}, r_{AB} = 3\,in and \alpha_{AB} = 0\,\frac{rad}{s^{2}}, then the solution of the systems of linear equations are, respectively:

<h3>Velocities</h3>

v_{D}+3.402\cdot \omega_{BD} = -41.569   (1)

9.404\cdot \omega_{BD} = -24   (2)

v_{D} = -32.887\,\frac{in}{s}, \omega_{BD} = -2.552\,\frac{rad}{s}

<h3>Accelerations</h3>

a_{D}+3.402\cdot \alpha_{BD} = -445.242   (3)

-9.404\cdot \alpha_{BD} = -687.264   (4)

a_{D} = -693.867\,\frac{in}{s^{2}}, \alpha_{BD} = 73.082\,\frac{rad}{s^{2}}

The <em>linear</em> acceleration of collar D when <em>θ = 60°</em> is - 693.867 inches per square second. \blacksquare

<h3>Remark</h3>

The statement is incomplete and figure is missing, complete form is introduced below:

<em>Arm AB has a constant angular velocity of 16 radians per second counterclockwise. At the instant when θ = 60°, determine the acceleration of collar D.</em>

To learn more on kinematics, we kindly invite to check this verified question: brainly.com/question/27126557

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2 years ago
A positive test charge of 5.00 E-5 C is places in an electric field. The force on it is 0.751 N. The magnitude of the electric f
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Explanation:

Charge=5.00 E-5

Force=0.751N

F=qE

0.751=5.00 E-5*E

E=1.502*10⁴

option a

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3 years ago
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If a farsighted person has a near point that is 0.600 mm from the eye, what is the focal length f2f2f_2 of the contact lenses th
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Answer:

0.22mm

Explanation:

A far sighted person is a person suffering from long sightedness i.e such individual can only see far distant object clearly but not near distant object. The defect is corrected using convex lens.

Since convex lens is used, the focal (f) length of the lens is positive and the image distance (v) is also positive.

Using the lens formula,

1/f = 1/u + 1/v

Where u is the object distance = 0.35mm

v = 0.6mm

1/f = 1/0.35+1/0.6

1/f = 2.86 + 1.67

1/f = 4.53

f = 1/4.53

f = 0.22mm

The focal length of the contact lenses will be 0.22mm

5 0
3 years ago
once you decide that it is safe to move out of traffic, what steps should you follow to ensure safe passage?
Solnce55 [7]

First put your turn signal on, next check for any ongoing traffic and wait until it is clear lastly start to drift into the lane you need to clear away from traffic

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