The first ariplanr was made December 17, 1903
Answer:
Explanation:
a ) Between r = 0 and r = r₁
Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .
b ) From r = r₁ to r = r₂
At distance r , charge contained in the sphere of radius r
volume charge density x 4/3 π r³
q = Q x r³ / R³
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q x r³ / ε₀R³
E= Q x r / (4πε₀R³)
E ∝ r .
c )
Outside of r = r₂
charge contained in the sphere of radius r = Q
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q / ε₀
E = Q / 4πε₀r²
E ∝ 1 / r² .
Answer:
v₃ = 5 [m/s]
Explanation:
To solve this problem we must use the definition of linear momentum, which tells us that momentum is equal to the product of mass by Velocity.
P = m*v
where:
P = linear momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
We must also clarify that the momentum is preserved i.e. it is equal before the collision and after the collision
Pbeforecollision = Paftercollision
(m₁*v₁) + (m₂*v₂) = (m₁*v₃) + (m₂*v₄)
where:
m₁ = mass of the truck = 3000 [kg]
v₁ = velocity of the truck = 10 [m/s]
m₂ = mass of the car = 1000 [kg]
v₂ = velocity of the car before the collision = 0 (the car is parked)
v₃ = velocity of the truck after the collision [m/s]
v₄ = velocity of the car after the collision = 15 [m/s]
(3000*10) + (1000*0) = (3000*v₃) + (1000*15)
30000 = 3000*v₃ + 15000
3000*v₃ = 30000 - 15000
3000*v₃ = 15000
v₃ = 5 [m/s]
Answer: B. store electric charges.
Explanation: I JUST TOOK THE PF EXAM AND I GOT IT CORRECT!!!!
