Answer:
B. In a vacuum, it is a constant value that does not depend on the observer.
Answer:
The distance is
Explanation:
From the question we are told that
The smallest shift is 
Generally a grid unit is
of an arcsec
This implies that 0.2 grid unit is 
The maximum distance at which a star can be located and still have a measurable parallax is mathematically represented as

substituting values


Note 
So
3.60 A = 3.60 coulombs of charge per second
(3.60 Coul/sec) x (15.3 sec) = 55.08 coulombs of charge
1 coulomb of charge is carried by 6.25 x 10^18 electrons
Number of electrons =
(55.08 Coul) x (6.25 x 10^18 e/coul) = <em>3.4425 x 10^20 electrons</em>
Answer:
a) 
b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).
c) The material is peat, possibly.
d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.
Explanation:
Given:
- mass of aluminium,

- mass of water,

- initial temperature of the system,

- mass of copper block,

- temperature of copper block,

- mass of the other block,

- temperature of the other block,

- final equilibrium temperature,

We have,
specific heat of aluminium, 
specific heat of copper, 
specific heat of water, 
Using the heat energy conservation equation.
The heat absorbed by the system of the calorie-meter to reach the final temperature.



The heat released by the blocks when dipped into water:

where
specific heat of the unknown material
For the conservation of energy : 
so,


b)
This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).
c)
The material is peat, possibly.
d)
The material cannot be ice because ice doesn't exists at a temperature of 100°C.
Answer:
The nucleus has an overall positive charge as it contains the protons. Every atom has no overall charge (neutral). This is because they contain equal numbers of positive protons and negative electrons. These opposite charges cancel each other out making the atom neutral.