T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r
where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance.
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect,
Point 1:
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r
Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m
The distance between the two points then is equal to 7.07 m.
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Answer:
Explanation:
a ) Let let the frictional force needed be F
Work done by frictional force = kinetic energy of car
F x 107 = 1/2 x 1400 x 35²
F = 8014 N
b )
maximum possible static friction
= μ mg
where μ is coefficient of static friction
= .5 x 1400 x 9.8
= 6860 N
c )
work done by friction for μ = .4
= .4 x 1400 x 9.8 x 107
= 587216 J
Initial Kinetic energy
= .5 x 1400 x 35 x 35
= 857500 J
Kinetic energy at the at of collision
= 857500 - 587216
= 270284 J
So , if v be the velocity at the time of collision
1/2 mv² = 270284
v = 19.65 m /s
d ) centripetal force required
= mv₀² / d which will be provided by frictional force
= (1400 x 35 x 35) / 107
= 16028 N
Maximum frictional force possible
= μmg
= .5 x 1400 x 9.8
= 6860 N
So this is not possible.
-- The student's distance traveled is 200 meters.
-- The student's displacement is 141 meters Northeast.
Convection is the answer i
think
