Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N
Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
Answer:
<em>The cyclist is traveling at 130 m/s</em>
Explanation:
<u>Constant Acceleration Motion
</u>
It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.
Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:
The cyclist initially travels at 10 /s and it's accelerating at a=6m/s^2. We need to know the new speed when t= 20 seconds have passed.
Apply the above equation:
The cyclist is traveling at 130 m/s
Answer:
η = 0.667 = 66.7%
Explanation:
The efficiency of the man can be given by the following formula:
η = output/input
where,
η = efficiency of man = ?
output = potential energy gain of the box = Wh
input = work done by man = Fd
Therefore,
where,
W = weight of box = 200 N
h = height gained by box = 1 m
F = force exerted by man = 60 N
d = length of ramp = 5 m
Therefore,
<u>η = 0.667 = 66.7%</u>
