(a) Two point charges totaling 8.00 μC exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the

Answer:

d = 2021.6 km

Explanation:

This distance problem can be solved using vector analysis; it's best to find each plane's position components before applying the Pythagorean theorem to calculate the separation between them.

For Airplane 1:

Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

          z₁ = 18 103 sin 25 = 7,607 103 m = 7607 m

For Plane 2:

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 sin 25 = 8.452 103 m = 8452 m

To determine the distance between the planes using the Pythagorean theorem:

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Now, we perform the calculations:

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 + 9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

        d = 2021.6 km

Given:
a rod with a circular cross section is experiencing uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]

From the details provided, the cross-section area = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=F/A
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)

(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183 
= 0.0318 (to three significant figures)

(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm  (to three significant figures)

Answer:

Consequently, we find that:

Thus, the most suitable answer would be:

a. vA = vB/4

Explanation:

In this situation, we can establish the following conditions:

both wires share the same length

both wires carry an identical current

Both wires are constructed of the same material, indicating that the electron density (n) remains constant across both wires

We also know that where r signifies the radius.

Given that wires are cylindrical in shape, we can determine the area for each case:

Thus, we conclude that

Now we are aware that the drift velocity of an electron in a wire can be described by:

Where I denotes the current, n is the electron density, e represents the electron charge, and A signifies the area.

By substituting, we arrive at:

So we observe that:

Thus, the most fitting answer is:

a. vA = vB/4

The answer is 30 seconds.

Answer:

Acceleration(a) = 0.75 m/s²

Explanation:

Given:

Force(F) = 3 N

Mass of object(m) = 4 kg

Find:

Acceleration(a)

Computation:

Force(F) = ma

3 = (4)(a)

Acceleration(a) = 3/4

Acceleration(a) = 0.75 m/s²