Answer:
q2 = -1.61*10^-5 C.
Explanation:
It was given that,
F = 0.985N
q1 = +8.40 X10-6 C
q2 = ?
r = 1.11 m
k = 9 x 10^9 (standard)
It generally follows that, if force is attractive, charge will be negative.
force, F = kq1q2/r^2
0.985 = 9*10^9*8.40*10^-6*q2/1.11^2
75600q2 = 0.985*1.11^2
75600q2 = 1.2136
q2 = 1.2136/75600 = 1.60529
q2 = -1.61*10^-5 C.
Answer:
Elastic potential energy, E = 3.26 J
Explanation:
It is given that,
Force constant of the spring, k = 5.2 N/m
Relaxed length of the spring, X = 2.45 m
When the mass is attached to the end of the spring, the vertical length of the spring is, x' = 3.57 m
To find,
The elastic potential energy stored in the spring.
Solution,
The extension in the length of the spring is given by :


x = 1.12 m
The elastic potential energy of the spring is given by :


E = 3.26 J
So, the elastic potential energy stored in the spring is 3.26 joules.
Answer:
6.5 m above the floor and 5 m above Christine's hand when it reaches the maximum height.
Explanation:
Let g = 10 m/s2 be the gravitational deceleration that affects the ball vertical motion so it comes to the maximum height at 0 speed. We can use the following equation of motion to find out the distance traveled by the ball from where it's thrown:

where v = 0 m/s is the final velocity of the ball when it reaches maximum level,
= 10m/s is the initial velocity of the ball when it starts, g = -10 m/s2 is the deceleration, and
is the distance traveled, which we care looking for:


So the ball is 5 m above Christine' hands when it reaches maximum height, and since the hand is 1.5 m above the floor, the ball is 5 + 1.5 = 6.5 m above the floor when it reaches maximum height.