Answer:
False
Explanation:
It is located in the little dipper whose stars are more faint.
When the heat source is removed from a fluid, convection currents in the fluid will eventually distribute heat uniformly throughout the fluid. When all of the fluid is at the same temperature, convection currents will stop.
The normal force is always perpendicular to the surface. So it would be straight to the left of the wall
Answer:
Part a)

Part b)

Explanation:
Part a)
Electric field due to large sheet is given as


now the electric field is given as


Now acceleration of an electron due to this electric field is given as



Now work done on the electron due to this electric field



So work done is given as



Part b)
Now we know that work done by all forces = change in kinetic energy of the electron
so we will have


