Answer:
λ= 5.24 × 10 ⁻² nC/cm
Explanation:
Given:
distance r = 4.10 cm = 0.041 m
Electric field intensity E = 2300 N/C
K = 9 x 10 ⁹ Nm²/C
To find λ = linear charge density = ?
Sol:
we know that E= 2Kλ / r
⇒ λ = -E r/2K (-ve sign show the direction toward the wire)
λ = (- 2300 N/C × 0.041 m) / 2 × 9 x 10 ⁹ Nm²/C
λ = 5.24 × 10 ⁻⁹ C/m
λ = 5.24 nC/m = 5.24 nC/100 cm
λ= 5.24 × 10 ⁻² nC/cm
Calculate the minimum speed record at the point B in order for the real question to reach the top
Answer:
15.6m/s
Completed Question;
For a short period of time, the frictional driving force acting on the wheels of the 2.5-Mg van is N= 600t^2 , where t is in seconds. If the van has a speed of 20 km/h when t = 0, determine its speed when t = 5
Explanation:
Mass m = 2500kg
Speed v1 = 20km/h = 20/3.6 m/s = 5.556 m/s
To determine speed v2;
Using the principle of momentum and impulse;
mv1 + ∫₀⁵ F dt = mv2
V1: How do you want the answer expressed?
V2:(I. other words..) In what terms do you want the answer expressed?
Ex,
m/s
ft/s
etc,.
I believe the answer is the second option.