Answer:
(a) 0.17 m
(b) 5.003 m
(c) 6.38 × 
N
(d) 7.37 ×
N
Explanation:
(a) The minimum value of 
will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.
(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.
<em>Hence, the maximum distance is 5.002 m</em>
(c) For minimum magnitude we use the minimum distance calculated in (a)
Minimum Distance = 0.17 m
For electrostatic force= 
×
(d) For maximum magnitude, we use the maximum distance calculated in (b)
Maximum Distance = 5.002 m
Using the formula for electrostatic force again:
F = 
F= 7.37×
N
Answer:
SDFK fbsdfasdgasdfgasdfg⊃⊃⊃⊃⊃⊃×∈⇔⇔⇔
Explanation:
Answer:
speed and time are Vf = 4.43 m/s and t = 0.45 s
Explanation:
This is a problem of free fall, we have the equations of kinematics
Vf² = Vo² + 2g x
As the object is released the initial velocity is zero, let's look at the final velocity with the equation
Vf = √( 2 g X)
Vf = √(2 9.8 1)
Vf = 4.43 m/s
This is the speed with which it reaches the ground
Having the final speed we can find the time
Vf = Vo + g t
t = Vf / g
t = 4.43 / 9.8
t = 0.45 s
This is the time of fall of the body to touch the ground
The answer is B artificial selection
