Answer:
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Explanation:
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Answer:
<h3>1.01 s</h3>
Explanation:
Using the equation of motion S = ut+1/2gt² to solve the problem where;
u is the initial velocity of the chocolate = 0m/s
t is the time taken
g is the acceleration due to gravity = 9.81m/s²
S is the height of fall = 5.0m
Substituting the given parameter into the formula to get the time t we have;
5 = 0(t)+1/2(9.81)t²
5 = 4.905t²
t² = 5/4.905
t² = 1.019
t = √1.019
t = 1.009 secs
<em>Hence it will take 1.01 secs for me to catch the chocolate bar</em>
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Answer:
Let I and j be the unit vector along x and y axis respectively.
Electric field at origin is given by
E= kq1/r1^2 i + kq2/r2^2j
= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)
= (2.88i + 1.44j)*10^-3 N/C
Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3
F= (1.382 i + 0.691 j) *10^-21
Goodluck
Explanation:
It would be 1.5 meters im sure form that distance to me is that nswe
