Answer:
38 cm from q1(right)
Explanation:
Given, q1 = 3q2 , r = 60cm = 0.6 m
Let that point be situated at a distance of 'x' m from q1.
Electric field must be same from both sides to be in equilibrium(where EF is 0).
=> k q1/x² = k q2/(0.6 - x)²
=> q1(0.6 - x)² = q2(x)²
=> 3q2(0.6 - x)² = q2(x)²
=> 3(0.6 - x)² = x²
=> √3(0.6 - x) = ± x
=> 0.6√3 = x(1 + √3)
=> 1.03/2.73 = x
≈ 0.38 m = 38 cm = x
Answer:
The answer to your question is Ke = 72 J
Explanation:
Kinetic energy depends on the speed of and object and its mass.
Data
mass = m = 4 kg
speed = v = 6 m/s
distance = d = 8 m
Kinetic energy = ke = ?
Formula
Ke = (1/2) mv²
Substitution
Ke = (1/2) (4)(6)²
Simplification
Ke = (1/2)(4)(36)
Ke = (1/2)(144)
Ke = 72 Joules
Result
Ke = 72 J
Answer:
<u>ω = 1.7 rad/s</u>
Explanation:
Conservation of angular momentum
Assuming the rod is initially hanging vertically at rest.
Initial angular momentum is carried by the bullet only
L = Iω = (mR²)(v/R) = mvR = 0.020(200)(0.7) = 2.8 kg•m²/s
the same angular momentum exists after impact, only the moment of inertia has increased by that of the rod. I = ⅓mR²
2.8 = (⅓(10)(0.7²) + 0.020(0.7²))ω
2.8 = (1.64313333...)ω
ω = 1.70406134...
When potential energy <u>decreases</u>, kinetic energy increases.
