Answer:
Option E
Explanation:
In the presence of two point charges at the two vertices of an equilateral triangle, the resultant electric field at the third vertex due to these charges can not be zero whether the charges are identical or not.
The reason being that only of the x or y component of the field can be cancelled out in either case still the total field can't be reduced to zero.
This can only be achieved if another charge is present.
Answer:
The combined velocity is 8.61 m/s.
Explanation:
Given that,
The mass of a truck, m = 2800 kg
Initial speed of truck, u = 12 m/s
The mass of a car, m' = 1100 kg
Initial speed of the car, u' = 0
We need to find the combined velocity the moment they stick together. Let it is V. Using the conservation of momentum.

So, the combined velocity is 8.61 m/s.
Answer:
magnitude of A − B = 15.81 km
Explanation:
Vector A points in the negative y-direction and has a magnitude of 5 km. Vector B points in the positive x-direction and has a magnitude of 15 km.
According to Cartesian coordinate system, the resultant will start either from tail of A and ends at head of B and vice-versa.
A(0,-5)
B(15,0)
A - B = (-15 i - 5 j )
Magnitude of the vector is given by
|A - B| = 
|A - B| = 
|A - B| = 15.81 km
Answer:
the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Explanation:
This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period
v² = v₀² + 2 a₁ x
indicate that the initial velocity is zero
v² = 2 a₁ x
let's calculate
v =
v = 143.666 m / s
now for the second interval let's find the distance it takes to stop
v₂² = v² - 2 a₂ x₂
in this part the final velocity is zero (v₂ = 0)
0 = v² - 2 a₂ x₂
x₂ = v² / 2a₂
let's calculate
x₂ =
x₂ = 573 m
as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake