An arrow is shot straight up in the air at an initial speed of 15 m/s. After how much time is the arrow heading downward at a speed of 8 m/s?

2.35 seconds

Explanation:

Remark

This is a question where direction matters.  Let us call down + and up minus. It won't matter. The answer will be the same.

Formula

a = (vf - vi)/t

Givens

a = 9.8 m/s^2

vi = - 15 m/s

vf = 8m/s

Solution

9.8 = (8 - - 15)/t                 Multiply both sides by t

t * 9.8 = 23                       Divide by 9.8

t = 23/9.8

t = 2.35 s

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.