Answer:
v = 25 m / s
Explanation:
For this exercise we use the relations and kinematics
first part, the train accelerates from v₀ = 7.0 m/s to 13 m/s in a time t = 8.0 s
v = v₀ + a t
a =
a =
a = 0.75 m / s²
second part. Accelerate again for t = 16 s
v = v₁ + a t
for this interval the initial velocity is v₁ = 13 m / s
v = 13 + 0.75 16
v = 25 m / s
Answer:
+1.33 × C and -1.33 ×
C respectively.
Explanation:
Electric potential (V) is the work done in moving a unit positive charge from infinity to a reference point within an electric field. It is measured in volts.
V = ............. 1
where: k is a constant = 9 × N
, q is the charge and r is the distance between the charges.
From equation 1,
q = ............... 2
The charge on each ball can be determined as;
given that; V = 1.2 × , k = 9 ×
N
and r = 1.00 m.
From equation 2,
q =
= 1.33 × C
Thus, the charge on the first ball is +1.33 × C, while the charge on the second ball is -1.33 ×
C.