Answer:
The planet that is farther from the star must have a time period greater.
Explanation:
We can determine the ratio of the period's planet with the radius of the circular orbit in the star's equatorial plane:
(1)
Where:
r: is the radius of the circular orbit of the planet and the star
T: is the period
G: is the gravitational constant
M: is the mass of the planet
From equation (1) we have:
(2)
Where k is a constant
From equation (2) we have that of the two planets, the planet that is farther from the star must have a time period greater.
I hope it helps you!
In solid and liquid the matter can occupy the 90 in³ and 157.1 in³ volume.
The matter in gaseous state can be expanded to occupy the volumes of the container.
<h3>
Volume of each of the container</h3>
The volume of each of the container is calculated as follows;
<h3>Volume of the rectangular container</h3>
V = 5 in x 6 in x 3 in
V = 90 in³
<h3>Volume of the cylindrical container</h3>
V = πr²h
V = (π)(2.5 in)²(8 in)
V = 157.1 in³
<h3>Volume of the matter</h3>
Vm = 3 in x 4 in x 5 in
Vm = 60 in³
<h3>Matter in solid and liquid state</h3>
Matter has fixed volume in solid and liquid state.
In solid and liquid the matter can occupy the 90 in³ and 157.1 in³ volume.
<h3>Matter in gaseous state</h3>
Matter has no definite volume in gaseous state.
The matter in gaseous state can be expanded to occupy the volumes of the container.
Learn more about states of matter here:
#SPJ1
Explanation:
The position vector r:
The velocity vector v:
The acceleration vector a:
Answer:
a)n= 3.125 x electrons.
b)J= 1.515 x A/m²
c) =1.114 x m/s
d) see explanation
Explanation:
Current 'I' = 5A =>5C/s
diameter 'd'= 2.05 x m
radius 'r' = d/2 => 1.025 x m
no. of electrons 'n'= 8.5 x
a) the amount of electrons pass through the light bulb each second can be determined by:
I= Q/t
Q= I x t => 5 x 1
Q= 5C
As we know that: Q= ne
where e is the charge of electron i.e 1.6 x C
n= Q/e => 5/ 1.6 x
n= 3.125 x electrons.
b) the current density 'J' in the wire is given by
J= I/A => I/πr²
J= 5 / (3.14 x (1.025x )²)
J= 1.515 x A/m²
c) The typical speed'' of an electron is given by:
=
=1.515 x / 8.5 x x |-1.6 x |
=1.114 x m/s
d) According to these equations,
J= I/A
= =
If you were to use wire of twice the diameter, the current density and drift speed will change
Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.
Also drift velocity will decrease as it is inversely proportional to the area