Gains at least one nutrient
Answer:
the electric field strength on the second one is 2.67 N/C.
Explanation:
the electric fiel on the first one is:
E1 = k×q/(r^2)
r^2 = k×q/(E1)
= (9×10^9)×(q)/(24.0)
= 375000000q
then the electric field on the second one is:
E2 = k×q/(R^2)
we know that R = 3r
R^2 = 9×r^2
E2 = k×q/(9×r^2)
= k×q/(9×375000000q)
= k/(9×375000000)
= (9×10^9)/(9×375000000)
= 2.67 N/C
Therefore, the electric field strength on the second one is 2.67 N/C.
B. F<em>spring = k(triangle)</em> x
Answer: vf = 51 m/s
d = 112 m
Explanation: Solution attached:
To find vf we use acceleration equation:
a = vf - vi / t
Derive to find vf
vf = at + vi
Substitute the values
vf = 3.5 m/s² ( 8.0 s) + 23 m/s
= 51 m/s
To solve for distance we use
d = (∆v)² / 2a
= (51 m/s - 23 m/s )² / 2 ( 3.5 m/s²)
= (28 m/s)² / 7 m/s²
= 784 m/s / 7 m/s²
= 112 m