Answer: 6.36
Explanation:
Given
Radius of grindstone, r = 4 m
Initial angular speed of grindstone, w(i) = 8 rad/s
Final angular speed of the grindstone, w(f) = 12 rad/s
Time used, t = 4 s
Angular acceleration of the grinder,
α = Δw / t
α = w(f) - w(i) / t
α = (12 - 8) / 4
α = 4/4 = 1 rad/s²
Number of complete revolution in 4s =
Δθ = w(i).t + 1/2.α.t²
Δθ = 8 * 4 + 1/2 * 1 * 4²
Δθ = 32 + 1/2 * 16
Δθ = 32 + 8
Δθ = 40 rad/s
40 rad/s = 40/2π rpm = 6.36 rpm
Therefore, the grindstone does 6.36 revolutions during the 4 s interval
Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.
The correct answer would be yes