Answer:
Electric field,
Explanation:
It is given that,
Radius of the circular loop, r = 13 cm = 0.13 m
Electric flux in the positive x direction,
Electric flux in the positive y direction,
The formula of the electric flux is given by :
In x- direction,
Where, is the electric field in x direction.
So, the x-component of the electric field is 470.87 N/C. Hence, this is the required solution.
Answer:
Explanation:
It is a case of adiabatic expansion .
T₁ , T₂ are initial and final temperature , V₁ and V₂ are initial and final volume.
Given ,
V₂ = 3 V₁ and T₁ = 292 . γ for air is 1.4 .
1.552 = 292 / T₂
T₂ = 188 K .
Range of a projectile motion is given by
R = v cos θ / g (v sin θ + sqrt(v^2 sin^2 θ + 2gy_0)); where R = 188m, θ = 41°, g = 9.8m/s^2, y_0 = 0.9
188 = v cos 41° / 9.8 (v sin 41° + sqrt(v^2 sin^2 41° + 2 x 9.8 x 0.9)) = 0.07701(0.6561v + sqrt(0.4304 v^2 + 17.64)) = 0.05053v + 0.07701sqrt(0.4304v^2 + 17.64)
0.07701sqrt(0.4304v^2 + 17.64) = 188 - 0.05053v
0.005931(0.4304v^2 + 17.64) = 35344 - 19v + 0.002553v^2
0.002553v^2 + 0.1046 = 35344 - 19v + 0.002553v^2
19v = 35344 - 0.1046 = 35343.8954
v = 35343.8954/19 = 1860 m/s
Answer:
c. As the thread is stretched, the coating thins and its resistance increases; as the thread is relaxed, the coating returns nearly to its original state.
Explanation:
This is in line with the result as shown in the data.
Stretching causes the coating to thin which in turn increases the resistance of the thread (conduction drops) releasing it backs returns the thread to nearly its initial values, but due to the strain already in the thread it does not return to its original values.
A compass has a magnitized needle that follows magnetic fields in the earth to lead you to the South I believe because of how strong the magnetic pull is.