- Initial velocity (u) = 10 m/s
- Final velocity (v) = 22 m/s
- Time (t) = 12 s
- Mass (m) = 200 Kg
- Let the acceleration be a.
- By using the equation of motion,
v = u + at, we have
- 22 m/s = 10 m/s + 12 s × a
- or, 22m/s - 10 m/s = 12 s × a
- or, 12 m/s = 12 s × a
- or, a = 1 m/s^2
- Let the force be F.
- We know, F = ma
- Therefore, the force on the accelerated object (F)
- = ma
- = (200 × 1) N
- = 200 N
<u>Answer</u><u>:</u>
<u>b)</u><u> </u><u>2</u><u>0</u><u>0</u><u> </u><u>N</u>
Hope you could understand.
If you have any query, feel free to ask.
Answer: not according with truth or fact/ incorrect.
Explanation:
Answer:
X₃₁ = 0.58 m and X₃₂ = -1.38 m
Explanation:
For this exercise we use Newton's second law where the force is the Coulomb force
F₁₃ - F₂₃ = 0
F₁₃ = F₂₃
Since all charges are of the same sign, forces are repulsive
F₁₃ = k q₁ q₃ / r₁₃²
F₂₃ = k q₂ q₃ / r₂₃²
Let's find the distances
r₁₃ = x₃- 0
r₂₃ = 2 –x₃
We substitute
k q q / x₃² = k 4q q / (2-x₃)²
q² (2 - x₃)² = 4 q² x₃²
4- 4x₃ + x₃² = 4 x₃²
5x₃² + 4 x₃ - 4 = 0
We solve the quadratic equation
x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2 5
x₃ = [-4 ± 9.80] 10
X₃₁ = 0.58 m
X₃₂ = -1.38 m
For this two distance it is given that the two forces are equal
Answer:
Assessment zone
Explanation:
It is the assessment zone in various security zones where active and passive security measures are employed to identify, detect, classify and analyze possible threats inside the assessment zones.
