Answer:
Acceleration = 2.35 m/
Speed = 8.67 m/s
Explanation:
The coefficient of friction , u =0.3
The angle of incline = 30°
The two forces acting on block are weight and friction.
weight along the incline = mg cos60° =
= 0.5 mg
Friction along incline = umg cos30° = mg 
Friction along incline = 0.26 mg
Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg
Acceleration =
= 0.24 g = 2.35 m/
The height of incline = 8 m
Length of the inclined edge = 16 m


v= 8.67 m/s
<span>When reading a buret, the initial reading should be taken from the top of the glassware and the final volume should still taken at the top. If the buret is completely, the initial volume for most buret would be zero. though, there are some where their initial starts at 50 decreasing to zero.</span>
Answer:
Part a)

Part b)

Part c)
d = 0.072 m
Explanation:
Part a)
As we know that the radius of the charge particle in constant magnetic field is given by

now for single ionized we have


Part b)
Similarly for doubly ionized ion we will have the same equation



Part c)
The distance between the two particles are half of the loop will be given as


