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: its a
Explanation: hope this help im sorry if its wrong
Answer:
86.6 lbs
Explanation:
Let the force is X.
Resultant force, R = 100 lbs
Other force is Y. Angle between resultant force and force X is 30°.
According to the diagram
X = 86.6 lbs
Other force Y
Y = 50 lbs
Answer: Released
Explanation: Energy is released in this reaction possibly in the form of heat thus it is an exergonic and or exothermic reaction.
Answer:
Acceleration = -3m/s²
Explanation:
Given the following data;
Initial velocity = 19m/s
Final velocity = 3m/s
Time = 2secs
To find the acceleration;
In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.
This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time. Mathematically, acceleration is given by the equation;
Substituting into the equation, we have;
Acceleration = -3m/s²
The value for her acceleration is negative because she is decelerating i.e her final velocity is lower than her initial velocity.
