A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At
1 answer:
The electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.
To determine σ:
σ = Q/A
Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:
σ = Q/d²
Make this substitution in the equation for E:
E = Q/(2ε₀d²)
We see that E is inversely proportional to the square of d:
E ∝ 1/d²
The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude:
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Answer:
Explanation:
r = Radius of disk = 7.9 cm
N = Number of revolution per minute = 1190 rev/minute
Angular speed is given by
The angular speed is
r = 2.98 cm
Tangential speed is given by
Tangential speed at the required point is
Radial acceleration is given by
The radial acceleration is .
t = Time = 2.06 s
Distance traveled is given by
The total distance a point on the rim moves in the required time is .
Answer:
b. 0.20 m/s.
Explanation:
Given;
initial mass, m = 0.2 kg
maximum speed, v = 0.3 m/s
The total energy of the spring at the given maximum speed is calculated as;
K.E = ¹/₂mv²
K.E = 0.5 x 0.2 x 0.3²
K.E = 0.009 J
If the mass is changed to 0.4 kg
¹/₂mv² = K.E
mv² = 2K.E
Therefore, the maximum speed is 0.20 m/s