All categories

3 years ago

3. What would the seasons be like if Earth's orbit had the eccentricity value of Mars?

2 answers:

4 0

According to Kepler's laws, the mass of a planet has almost no effect on its orbit; the mass of the sun is what controls things. Even though Earth is 10 times heavier than Mars, it would still trundle along Mars's old path. Both Mars and Earth are perpetually falling toward the sun, and all falling bodies fall at the same rate.

densk [106]3 years ago

3 0

Answer:

123456789

Explanation:

You might be interested in

The intensity of the sound from a certain source is measured at two points along a line from the source. The points are separate

Artemon [7]

Answer:

The source is at a distance of 4.56 m from the first point.

Solution:

As per the question:

Separation distance between the points, d = 11.0 m

Sound level at the first point, L = 66.40 dB

Sound level at the second point, L'= 55.74 dB

Now,

$L = 10log_{10}\frac{I}{I_{o}}$

$I = I_{o}10^{\frac{L}{10}} = I_{o}10^{0.1L} = 10^{- 12}\times 10^{0.1\times 66.40} = 10^{- 5.36}$

$L' = 10log_{10}\frac{I'}{I_{o}}$

$I' = I_{o}10^{\frac{L'}{10}} = 10^{- 12}\times 10^{0.1\times 55.74} = 10^{- 6.426}$

where

$I_{o} = 10^{- 12} W/m^{2}$

I = Intensity of sound

Now,

$I = \frac{P}{4\pi R^{2}}$

Similarly,

$I' = \frac{P}{4\pi (R + 11.0)^{2}}$

Now,

$\frac{I}{I'} = \frac{(R + 11.0)^{2}}{R^{2}}$

$\frac{10^{- 5.36}}{10^{- 6.426}} = \frac{(R + 11.0)^{2}}{R^{2}}$

$R ^{2} + 22R + 121 = 11.64R^{2}}$

$10.64R ^{2} - 22R - 121 = 0$

Solving the above quadratic eqn, we get:

R = 4.56 m

8 0

3 years ago

How does Newton's third law of motion describe forces?

svetlana [45]

I would say your answer is B, since Newton's 3rd law is, "For every action, there is an equal and opposite reaction."
It's talking about pairs of actions. Sorry if I'm wrong.

3 0

3 years ago

The effective nuclear charge for an atom is less than the actual nuclear charge due to

Dmitrij [34]

Answer:

Repulsion

Explanation, due to this repulsion with the proton and the electron, each electron experiences a nuclear charge that is somewhat less than the actual nuclear charge.

3 0

4 years ago

There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the differe

Usimov [2.4K]

An example that illustrates the difference is the circular motion

Explanation:

Let's start by reminding the definition of the two quantities:

- Speed is a scalar quantity that tells "how fast" an object is moving, regardless of its direction of motion.

Speed can be calculate as:

$speed = \frac{d}{t}$

where:

d is the distance travelled

t is the time taken

- Velocity is instead a vector quantity, given by:

$velocity = \frac{d}{t}$

where;

d is the displacement of the object (displacement is a a vector connecting the initial position to the final position of motion)

t is the time taken

Since it is a vector, velocity has both a magnitude and a direction, therefore it also takes into account the direction of motion of the object.

For an object in motion in a straight line, speed and velocity are the same. However, this is not always the case.

In fact, an example of motion in which the two quantities are different is the circular motion. Consider for example the object making one complete revolution along the circle. Therefore, its average speed is the ratio between the length of the perimeter (the distance) divided by the time taken:

$speed = \frac{2\pi r}{t}$