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butalik [34]
3 years ago
6

Suppose a mad scientist went back in time to the beginning of the solar system and caused the newly formed Mars and Venus to swi

tch places. So now there is a Venus-sized planet where Mars was (let's call it Venus 2), and a Mars-sized planet where Venus was (let's call it Mars 2). How would these hypothetical new planets, Mars 2 and Venus 2, turn out in this alternate time line
Physics
1 answer:
Black_prince [1.1K]3 years ago
8 0

Answer:

Mars2   illuminated side of the planet is very  heat, dark side very cool

Venus 2 a warm planet with a constant temperature across the entire surface

Explanation:

For this hypothetical case, when changing the planets they are changed with their current characteristics.

Case of Mars2

In this case, there is a planet with a very thin atmosphere, so the solar radiation reaches the ground without damping it, causing a lot of noise, so the illuminated side of the planet is very heat and when the dark side turns due to the little atmosphere it loses everything the heat for which it is very cold.

This thermal stress between the two sides of the planet continues constantly creating possible fruit trees in its rocky systems.

Case of Venus 2

The planet has a high atmospheric density, but it is very far from the sun, so the amount of radiation that arrives slightly warms the planet, but due to the thin atmosphere the losses for the dark period are very small, so the entire planet it is heated until it reaches an almost uniform temperature over its entire surface.

In this case we have a warm planet with a constant temperature across the entire surface, regardless of which side is lit.

You might be interested in
An imbalance of _____________ causes an object to be charged.
dezoksy [38]
An imbalance of electrons causes an object to be charged. A charged substance is one that has a net charge that is not zero. If the number of protons and that of electrons is equal in an object, the object is not neutral, but it has zero charge. Loss or gain of electrons determine the charge of an object, when an object loses electrons it becomes positively charged, but when it gains electrons it becomes negatively charged.
5 0
3 years ago
a stone is thrown horizonttaly from a cliff of a hill with an initial velocity of 30m/s it hits the ground at a horizontal dista
ELEN [110]

Answer:

a) Time = 2.67 s

b) Height = 35.0 m

Explanation:

a) The time of flight can be found using the following equation:

x_{f} = x_{0} + v_{0_{x}}t + \frac{1}{2}at^{2}   (1)

Where:

x_{f}: is the final position in the horizontal direction = 80 m

x_{0}: is the initial position in the horizontal direction = 0

v_{0_{x}}: is the initial velocity in the horizontal direction = 30 m/s

a: is the acceleration in the horizontal direction = 0 (the stone is only accelerated by gravity)

t: is the time =?  

By entering the above values into equation (1) and solving for "t", we can find the time of flight of the stone:  

t = \frac{x_{f}}{v_{0}} = \frac{80 m}{30 m/s} = 2.67 s

b) The height of the hill is given by:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final position in the vertical direction = 0

y_{0}: is the initial position in the vertical direction =?

v_{0_{y}}: is the initial velocity in the vertical direction =0 (the stone is thrown horizontally)            

g: is the acceleration due to gravity = 9.81 m/s²

Hence, the height of the hill is:

y_{0} = \frac{1}{2}gt^{2} = \frac{1}{2}9.81 m/s^{2}*(2.67 s)^{2} = 35.0 m  

I hope it helps you!

5 0
3 years ago
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

5 0
2 years ago
the goat weighs 900 N and is 1 meter from the fulcrum. The strongman pulls down on the lever 3 meters from the fulcrum. What is
garik1379 [7]

Answer: The smallest effort = 300N

Explanation:

Using one of the condition for the attainment of equilibrium:

Clockwise moment = anticlockwise moments

900 × 1 = 3 × M

Where M = the weight of the strong man

3M = 900

M = 900/3 = 300N

Therefore, 300N is the smallest effort that the strongman can use to lift the goat

6 0
3 years ago
Two marbles, one twice as heavy as the other, are dropped to the ground from the roof of a building. Just before hitting the gro
Andreyy89

Answer:

 B. twice as much kinetic energy

Explanation:

Lets take the mass of the first marble =2 m

the mass of the second marble = m

We know that velocity of particle does not depends on their mass that is the velocity of both mass will be same after dropping from the roof.

We know that kinetic energy of a mass is given as

KE=\dfrac{1}{2}Mv^2

Kinetic energy for heavier mass

KE=\dfrac{1}{2}\times 2m\times V^2

Kinetic energy for light mass

KE'=\dfrac{1}{2}\times m\times V^2

KE=2 KE '

Form above two equation we can say that ,the kinetic energy for the heavier mass is twice the lighter mass.

Therefore the answer will be B.

7 0
3 years ago
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