(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.
Answer:
<h3>The answer is 5.4 kg</h3>
Explanation:
The mass of the object can be found by using the formula
f is the force
a is the acceleration
From the question we have
We have the final answer as
<h3>5.4 kg</h3>
Hope this helps you
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It slows down. The molecules/particles aren't as close together, resulting in slower movement.
Since this is a projectile motion problem, break down each of the five kinematic quantities into x and y components. To find the range, we need to identify the x component of the displacement of the ball.
Let's break them down into components.
X Y
v₁ 32 cos50 m/s 32 sin50 m/s
v₂ 32 cos50 m/s ?
Δd ? 0
Δt ? ?
a 0 -9.8 m/s²
Let's use the following equation of uniform motion for the Y components to solve for time, which we can then use for the X components to find the range.
Δdy = v₁yΔt + 0.5ay(Δt)²
0 = v₁yΔt + 0.5ay(Δt)²
0 = Δt(v₁ + 0.5ayΔt), Δt ≠ 0
0 = v₁ + 0.5ayΔt
0 = 32sin50m/s + 0.5(-9.8m/s²)Δt
0 = 2<u>4</u>.513 m/s - 4.9m/s²Δt
-2<u>4</u>.513m/s = -4.9m/s²Δt
-2<u>4</u>.513m/s ÷ 4.9m/s² = Δt
<u>5</u>.00s = Δt
Now lets put our known values into the same kinematic equation, but this time for the x components to solve for range.
Δdₓ = v₁ₓΔt + 0.5(a)(Δt)²
Δdₓ = 32cos50m/s(<u>5</u>.00s) + 0.5(0)(<u>5</u>.00)²
Δdₓ = 32cos50m/s(<u>5</u>.00s)
Δdₓ = 10<u>2</u>.846
Therefore, the answer is A, 102.9m. According to significant digit rules, neither would be correct, but 103m is the closest to 102.9m so I guess that is what it is.