Answer:
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Answer: q2 = -0.05286
Explanation:
Given that
Charge q1 = - 0.00325C
Electric force F = 48900N
The electric field strength experienced by the charge will be force per unit charge. That is
E = F/q
Substitute F and q into the formula
E = 48900/0.00325
E = 15046153.85 N/C
The value of the repelled second charge will be achieved by using the formula
E = kq/d^2
Where the value of constant
k = 8.99×10^9Nm^2/C^2
d = 5.62m
Substitutes E, d and k into the formula
15046153.85 = 8.99×10^9q/5.62^2
15046153.85 = 284634186.5q
Make q the subject of formula
q2 = 15046153.85/ 28463416.5
q2 = 0.05286
Since they repelled each other, q2 will be negative. Therefore,
q2 = -0.05286
<h2>It will take 0.125 seconds to reach the net.</h2>
Explanation:
Initial speed, u = 34 ft/s = 10.36 m/s
Acceleration, a = -9.81 m/s²
Displacement, s = Final height - Initial height = 8 - 4 = 4 ft = 1.22 m
We have equation of motion, s = ut + 0.5 at²
Substituting
s = ut + 0.5 at²
1.22 = 10.36 x t + 0.5 x -9.81 x t²
4.905t² - 10.36 t + 1.22 = 0
t = 1.99 s or t = 0.125 seconds
Minimum time is 0.125 seconds.
It will take 0.125 seconds to reach the net.
<u>Answer:</u>
The height of ramp = 124.694 m
<u>Explanation:</u>
Using second equation of motion,
From the question,
u = 31 m/s; s = 156.3 m, a=0
substituting values
t = 
= 5.042 s
Similary, for the case of landing
t = 5.042 s; initial velocity, u =0
acceleration = acceleration due to gravity, g = 9.81 
Substituting in 
h = 124.694 m
So height of ramp = 124.694 m
