Gases expand to fill all the space available. Their particles fly all over the place, and they take the shape and volume of whatever you put them in.
Answer:
When the net force on an object decreases, the object's acceleration decreases 3rd option
Explanation:
Answer:
α = 5.75°
Explanation:
In this case, the problem states that both springs have identical lenghts and we also have theri constant. We want to know the angle of the rod with the horizontal. This can be found with the following expression:
sinα = Δx/L
α = sin⁻¹ (Δx/L) (1)
However, we do not have Δx. This can be found when half of the weight of the rod is balanced. In this way:
F₁ = k₁*x₁ ----> x₁ = F₁ / k₁ (2)
And the force is the weight in half so: F₁ = mg/2
Replacing in (2) we have:
x₁ = (1.3 * 9.8) / (2 * 58) = 0.1098 m
Doing the same thing with the other spring, we have:
x₂ = (1.3 * 9.8) / (2 * 36) = 0.1769 m
Now the difference will be Δx:
Δx = 0.1769 - 0.1098 = 0.0671 m
Finally, we can calculate the angle α, from (1):
α = sin⁻¹(0.0671 / 0.67)
<h2>
α = 5.75 °</h2>
Hope this helps
Answer:
-10 m/s^2
Explanation:
Acceleration = Change of velocity over time
25/2.5 = 10 m/s^2
Since we are deccelerating here (Going slower) then the number is written as negative
Answer:
d) 2Fr
Explanation:
We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².
We now evaluate the integral from r = +r to -r
W = ∫Fdr
= ∫(-e²/4πε₀r²)dr
= -∫e²dr/4πε₀r²
= -e²/4πε₀∫dr/r²
= -e²/4πε₀ × -[1/r] from r = +r to -r
W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.
Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.
So W = -2e²/4πε₀r = 2Fr.
So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr