Answer:
F = 85.24 N
T = 30.84 N
Explanation:
The parameters given are
Mass of crates = m₁ = 3 kg and m₂ = 5 kg
Component of masses acting along the plane = mgsin(θ)
Which gives;
F - (3×9.81×sin(30) + 10 + 5×9.81×sin(30) + 17) = m×a
So that we have;
2 = (F - 66.24)(3 + 5)
F = 16 + 66.24 = 85.24 N
The tension T between the crates = F × m₁/(m₁+m₂) = 85.24 × 3/(3+5) = 30.84 N
Answer:
14 N and 2 N
Explanation:
We have two vectors:
a = 8 N
b = 6 N
When the two vectors are in the same direction, their resultant is simply given by the sum of the magnitudes of the two vectors. Therefore, we will have:
Vice-versa, when the two vectors are in opposite directions, we have to consider one of the two vectors as being negative: therefore, the resultant will be given by the difference between the magnitudes of the two vectors. Therefore, in this case, we will have:
Answer:
q = 1.96 10⁴ C
Explanation:
The elective force is given by
= q E
Where E is the electric field and q the charge.
Let's use Newton's law of equilibrium for the case of the suspended drop
F_{e} –W = 0
F_{e} = W
q E = m g
q = m g / E
Let's calculate
q = 2.0 10⁵ 9.8 / 100
q = 1.96 10⁴ C