It can be described as a constant variation
Answer:
Electric potential = 0.00054 V
Explanation:
We are given;
Charge; q = 3 pC = 3 × 10^(-12) C
Radius; r = 2 cm = 0.02 m
Formula for the electric potential of this surface will be;
V = kqr
Where;
K is a constant = 9 × 10^(9) N⋅m²/C².
Thus;
V = 9 × 10^(9) × 3 × 10^(-12) × 0.02
V = 0.00054 V
Answer:
it will take 4 seconds to travel 2 meters.
Answer:
a. k = (1/k₁ + 1/k₂)⁻¹ b. k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹
Explanation:
Since only one force F acts, the force on spring with spring constant k₁ is F = k₁x₁ where x₁ is its extension
the force on spring with spring constant k₂ is F = k₂x₂ where x₁ is its extension
Let F = kx be the force on the equivalent spring with spring constant k and extension x.
The total extension , x = x₁ + x₂
x = F/k = F/k₁ + F/k₂
1/k = 1/k₁ + 1/k₂
k = (1/k₁ + 1/k₂)⁻¹
B
The force on spring with spring constant k₃ is F = k₃x₃ where x₃ is its extension
Let F = kx be the force on the equivalent spring with spring constant k and extension x.
The total extension , x = x₁ + x₂ + x₃
x = F/k = F/k₁ + F/k₂ + F/k₃
1/k = 1/k₁ + 1/k₂ + 1/k₃
k = (1/k₁ + 1/k₂ + 1/k₃)⁻¹
Answer:
Noise level will be -18.2 watt
So option (b) will be correct answer
Explanation:
We have given sound intensity 
And threshold intensity
( in question it is given as
but its standard value is
)
Now noise level 
So the noise level will be -18.2
So option (b) will be correct answer