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3 years ago

How does the mass of and distance between the Sun and the Earth impact the number of days in a year?

Physics

1 answer:

Alexxandr [17]3 years ago

6 0

Answer:

Force of sun on earth can be written:

F = G M m / R^2

Centripetal force on earth can be written:

F= m w^2 R

Equating these:

G M / R^2 = w^2 R

Or w^2 = G (M / R^3)

Since no/ days per year is proportional to w (time for 1 revolution)

Increasing M (mass of sun) would increase w or decrease length of year

Increasing the mean radius R would decrease w or increase length of year

You might be interested in

Why might we expect venus and earth to be similar?

mina [271]

They have similar distances from the sun therefore receive similar amounts of solar energy but Venus has no atmosphere to protect it so greater variation in temp.
Both are rock planets unlike those further away from the sun which, because of their distance receive less solor energy are cold, are ice planets

5 0

3 years ago

If planet A is three times as far from planet C, then the period of its orbit will be __ times as long

liubo4ka [24]

I may be wrong, but I think you're trying to say that Planet-A is
<em>3 times as far from the sun</em> as Planet-C is.

If that's the real question, then the answer is that the period of Orbit-A
is about<em> 5.2</em> times as long as the period of Orbit-C .

Orbital period ≈ (proportional to) (the orbital distance) ^ 3/2 power.

This was empirically demonstrated about 350 years ago by Johannes
and his brilliant Kepple, and derived about 100 years later by Newton
from his formula for the forces of gravity.

6 0

3 years ago

Suppose that the resistance between the walls of a biological cell is 6.8 × 109 ω. (a) what is the current when the potential di

GenaCL600 [577]

(a) We can find the current flowing between the walls by using Ohm's law:
$I= \frac{\Delta V}{R}$
where $\Delta V=69 mV=0.069 V$ is the potential difference and $R=6.8\cdot 10^9 \Omega$ is the resistance. Substituting these values, we get
$I=1.01 \cdot 10^{-11} A$

(b) The total charge flowing between the walls is the product between the current and the time interval:
$Q=I \Delta t$
The problem says $\Delta t=0.86 s$ , so the total charge is
$Q=(1.01\cdot 10^{-11} A)(0.86 s)=8.73 \cdot 10^{-12} C$

The current consists of Na+ ions, each of them having a charge of $e=1.6 \cdot 10^{-19} C$ . To find the number of ions flowing, we can simply divide the total charge by the charge of a single ion:
$N= \frac{Q}{e} = \frac{8.73 \cdot 10^{-12}C}{1.6 \cdot 10^{-19}C} = 5.45 \cdot 10^7 ions$

4 0

3 years ago

You shake a bottle of soda and take off the cap. If the soda shoots out of the

kicyunya [14]

Answer:

A. 11.5 m

Explanation:

Given,

The initial velocity of the soda cap, u = 15 m/s

The soda bottle cap is projected vertically upwards,

Hence, the angle formed with the ground, Ф = 90°

The maximum height of the projectile is given by the formula,

$h_{max} =\frac{u^{2}sin^{2}\phi}{2g}$

Substituting the given values in the above equation

$h_{max} =\frac{15^{2}sin^{2}90}{2X9.8}$

= 11.5 m

Hence, the maximum height of the cap is h = 11.5 m

8 0

3 years ago

What is the acceleration of a 1000 kg car subject to a 550 N net force

Xelga [282]

Answer:

The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2

Explanation:

Given:

F = 550 N

m = 1000 kg

To Find:

a = ?

Solution:

So by the equation by Newton's 2nd Law of Motion,

F = m x a

550 N = 1000 kg x a

a = 550 N/ 1000 kg

a = 0.55 m/s^2

Therefore,

The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2

PLEASE MARK ME AS BRAINLIEST!!!

4 0

3 years ago