(1 parsec) is the distance at which an object has a parallax of 1 arcsecond. The distance is about 3.26 light years.
Another way to understand it is: The distance from which the Earth's orbit appears 1 arcsecond across.
For a parallax angle of 1/2 arcsecond, the distance is <em>2 parsecs </em>(about 6.52 light years).
1 arcsecond is 1/3600 of a degree, 0.00028 degree.
<span>Density is a value for
mass, such as kg, divided by a value for volume, such as m3. Density is a
physical property of a substance that represents the mass of that substance per
unit volume. It is a property that can be used to describe a substance. We calculate as follows:
</span><span>Volume = 60.0 g ( 1 mL / 0.70 g ) = 85.71 mL
Therefore, the correct answer is option B.</span>
Answer:
A.B = -38
Explanation:
A = 2i + 9j and B = -i - 4j.
So, A.B = (2i + 9j).(-i - 4j)
= 2i.(-i) + 2i.(-4j) + 9j.(-i) + 9j.(-4j)
= -2i.i - 8i.j - 9j.i - 36j.j
since i.i = 1, j.j = 1, i.j = 0 and j.i = 0, we have
A.B = -2(1) - 8(0) - 9(0) - 36(1)
A.B = -2 - 0 - 0 - 36
A.B = -38
Answer:
a) x = ⅔ d
, b) the charge must be negative, c) Q
Explanation:
a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing
∑ F = 0
-F₁₂ + F₂₃ = 0
F₁₂ = F₂₃
let's replace the values
k Q Q / r₁₂² = k Q 4Q / r₂₃²
Q² / r₁₂² = 4 Q² / r₂₃²
suppose charge 3 is placed at point x
r₁₂ = x
r₂₃ = d-x
we substitute
1 / x² = 4 / (d-x) 2
1 / x = 2 / (d-x)
x = 2 (x-d)
x = 2x -2d
3x = 2d
x = ⅔ d
b) The sign of the charge must be negative, to have an attractive charge on the two initial charges
c) Q
Answer:
A) 1000 joules
Explanation:
In general work is given by the equation:
(1)
A) With 
the displacement and 
the force applied, because the force and the displacement are parallel (the crate is pushed horizontally) 
is simply 
, and because the path is a straight line and the force is constant work is:
(2),
B) The work-energy theorem says that the total work on a body is equal to the change on kinetic energy:
(3)
The total work on the crate is the work done by the push and plus the work of the friction 
(4) , as (A) because forces are parallel to the displacement 
(5) and 
(6), the due friction always has negative sign because is opposite to the displacement, using (6), (5) and (4) on (3):
(3)
C) The energy is lost by friction, so the amount of energy turned into heat is the work the friction does:
(3)
