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3 years ago

How would gravity cause planets to move if they did not have inertia?

Physics

2 answers:

steposvetlana [31]3 years ago

7 0

Answer: C.

Explanation: A-pe-x

daser333 [38]3 years ago

4 0

I’d think the answer would be C. i’m just kinda guessing but my thought process is this (as simply as i can put it because physics is confusing):

so for example say you throw a ball across a flat surface. inertia is what keeps the ball rolling straight in a line, so unless you were to maybe put your hand in front of the ball or something, it would just go straight forever.

this is what happens with the planets. they go in a straight line, but since there’s gravity, the planets are also being pulled towards the sun. so gravity and inertia are why the planets orbit in the circle pattern they do. so when we remove inertia, we’re removing the state in which the planets keep going straight while being pulled towards a center point (the sun). this causes gravity to be the only factor in the planets orbiting. so that being said, the planets would just be pulled towards the sun. :)

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Based on the simple blackbody radiation model described in class, answer the following question. The planets Mars and Venus have

Lera25 [3.4K]

Answer:

The extent of greenhouse effect on mars is $G_m = 87 K$

Explanation:

From the question we are told that

The albedo value of Mars is $A_1 = 0.15$

The albedo value of Mars is $A_2 = 0.15$

The surface temperature of Mars is $T_1 = 220 K$

The surface temperature of Venus is $T_2 = 700 K$

The distance of Mars from the sun is $d_m = 2.28*10^8 \ km = 2.28*10^8* 1000 = 2.28*10^{11} \ m$

The distance of Venus from sun is $d_v = 1.08 *10^{8} \ km = 1.08 *10^{8} * 1000 = 1.08 *10^{11} \ m$

The radius of the sun is $R = 7*10^{8} \ m$

The energy flux is $E = 6.28 * 10^{7} W/m^2$

The solar constant for Mars is mathematically represented as

$T = [\frac{E R^2 (1- A_1)}{\sigma d_m} ]$

Where $\sigma$ is the Stefan's constant with a value $\sigma = 5.6*10^{-8} \ Wm^{-2} K^{-4}$

So substituting values

$T = \frac{6.28 *10^{7} * (7*10^8)^2 * (1-0.15)}{(5.67 *10^{-8}) * (2.28 *10^{11})^2)}$

$T = 307K$

So the greenhouse effect on Mars is

$G_m = T - T_1$

$G_m = 307 - 220$

$G_m = 87 K$

3 0

4 years ago

A rotating object has an angular acceleration of α = 0 rad/s2. Which one or more of the following three statements is consistent

Murrr4er [49]

Answer:

A,B and C

Explanation:

Statement A

At all times, angular velocity is $\omega = 0\,{\rm{rad/s}$

Angular acceleration is the rate of change in angular velocity with respect to time.

Angular velocity and angular acceleration are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

Which when re-arranged becomes

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

There’s no change in angular velocity anytime when the angular velocity is $\omega = 0\,{\rm{rad/s}}$

The equation can be modified as follows:

$\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero hence statement A is valid.

Statement B

Angular acceleration is the rate of change in angular velocity with respect to time.

Angular velocity and angular acceleration are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

Which when re-arranged becomes

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

There’s no change in angular velocity anytime when the angular velocity is $\omega = 10\,{\rm{rad/s}}$ .The final and initial velocities remain the same.

The equation can be modified as follows:

$\begin{array}{c}\\\alpha = \frac{{10\,{\rm{rad/s}} - 10\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero and statement B is valid

Statement C

Angular velocity is defined as the change in the angular position with respect to time.

Angular velocity and angular displacement are related by

$\theta = \omega t$

Which can also be modified as:

${\theta _{\rm{f}}} - {\theta _{\rm{i}}}$

Note that the final position is ${\theta _{\rm{f}}}$ and initial position is ${\theta _{\rm{i}}}$

Modifying the equation to find the angular velocity we obtain

$\omega = \frac{{{\theta _{\rm{f}}} - {\theta _{\rm{i}}}}}{t}$

When the angular displacement has the same value at all times, the equation becomes

$\begin{array}{c}\\\omega = \frac{{{\theta _{\rm{i}}} - {\theta _{\rm{i}}}}}{t}\\\\ = 0\\\end{array}$

The angular velocity becomes zero.

Angular acceleration and angular velocity are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

The expression above can be rearranged as follows:

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

At all times, the angular velocity is $\omega = 0\,{\rm{rad/s}}$ hence initial and final velocities remain the same

We obtain

$\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero and statement C is valid.

Therefore, statements A,B and C are consistent .

4 0

4 years ago

What happens to the temperature and thermal energy of water when it is heated in a pot on the stove?

Gekata [30.6K]

Answer:

Both its temperature and its thermal energy will increase.

Explanation:

Objects are made up of particles. For example, water is made up of water molecules. The kinetic energy of a particle of mass $m$ and velocity $v$ is equal to $\displaystyle \frac{1}{2}\, m \cdot v^2$ ,

The thermal energy of an object measures the total kinetic energy of all its particles.
On the other hand, the temperature of that object measures the average kinetic energy of all these particles.

Water in that pot gains energy when the pot is heated. That would increase the total kinetic energy of these water molecules.

What about temperature? Assume that the number of water particles in the pot stays the same- In other words, assume that water in the pot does not evaporate. As the total kinetic energy of these water molecules increase, their average kinetic energy would also increase. As a result, it would appear that the temperature of water in that pot has increased.

7 0

4 years ago

Elle is pushing a box of mass 7.0 kilograms with the force of 25 newtons. If the force of friction is 2.6 newtons, what is the v

BigorU [14]

In the given question, there are several information's of immense importance. Using those given information's, it is easy to get to the required result. It is given that the mass of the box Ellie is pushing is 7 kg with a force of 25 newtons. the force of friction effecting the acceleration is 2.6 Newton.
We already know that
Force = mass * Acceleration
We also know that
Force = Force applied - Force of friction
   = (25 - 2.6) Newton
   = 22.4 Newton.
Now putting the value of force in the equation, we get
Force = Mass * Acceleration
22.4 = 7 * Acceleration.
Acceleration = 22.4/7
   = 3.2 m/s^2
So the value of acceleration is 3.2 m/s^2

3 0

3 years ago

What does waining mean?

Blizzard [7]

Answer:

(of the moon) have a progressively smaller part of its visible surface illuminated, so that it appears to decrease in size.

Explanation:

5 0

3 years ago

Read 2 more answers