Answer
given,
mass of block (m)= 6.4 Kg
spring is stretched to distance, x = 0.28 m
initial velocity = 5.1 m/s
a) computing weight of spring
k x = m g


k = 224 N/m
b) 




c) 


d) 


e)


A = 0.682 m
Force =
=
F = 94.20 N
Answer:
31.25 m
25m/sec
Explanation:
Given :-
Time = 5sec
V = 0 (in going up)
U = 0 (in comming down)
Find :-
H and U by which it is thrown up
Since the total time is 5 sec ,therefore half time will be taken to go up and another half will be taken to go down .
We know that ,
V = U + gt
0 = U - 10*2.5
U = 25 m/sec
Also,
V² = U² +2gs
0 = 625 - 20s
s = 625/20 = 31.25 m
Answer:
Mình cũng không biết làm bạn ơi =)))
Explanation:
1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is

substituting the coordinates of the two charges, we get

2) Then, we can calculate the electrostatic force between the two charges

and

, which is given by

where

is the Coulomb's constant.
Substituting numbers, we get

and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.