Answer: 247.67 V
Explanation:
Given
Potential At A 
Potential at 
when particle starts from A it reaches with velocity
at Point while when it starts from C it reaches at point B with velocity 
Suppose m is the mass of Particle
Change in Kinetic Energy of particle moving under the Potential From A to B

Change in Kinetic Energy of particle moving under the Potential From C to B

Divide 1 and 2 we get

on solving we get


Answer:
Heyyy hope this helps
Convection currents describe the rising, spread, and sinking of gas, liquid, or molten material caused by the application of heat.
Answer:
The three types of thermal expansions are Linear expansion , Superficial expansion and Cubical expansion
Answer:
Explanation:
This problem can be solved easily if we represent velocity in the form of vector.
The velocity of 351 was towards easterly direction so
V₁ = 351 i
The velocity of 351 was towards south west making - 48° with east or + ve x direction.
V₂ = 351 Cos 48 i - 351 sin 48 j
V₂ = 234.86 i - 260.84 j
Change in velocity
= V₂ - V₁ = 234.86 i - 260.84 j - 351 i
= -116.14 i - 260.84 j
acceleration
= change in velocity / time
(-116.14 i - 260.84 j )/ 1
= -116.14 i - 260.84 j
magnitude = 285.53 ms⁻²
Direction
Tan θ = 260.84 / 116.14 = 2.246
θ = 66 degree south of west .
Answer:
I think this is how you do it
Explanation: