A uniform metal rod, with a mass of 2.6 kg and a length of 1.5 m, is attached to a wall by a hinge at its base. A horizontal wir
1 answer:
Answer:
the horizontal force acting on rod is - 25.909 N
veritical force is 25.48 N
Explanation:
Given data:
mass of rod is m 2.6 kg
length of rod is 1.5 m
Hinge distance from the wire is 0.70 m
Apply torque law about hinge
Tension in the wire is
= -25.909 N
the horizontal force acting on rod is
Fx = T
FX = - 25.909 N
vertical force is
Fy = mg
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Explanation:
Formula to calculate electric field because of the plate is as follows.
E =
=
=
Now, we will consider that equilibrium of forces are present there. So,
ma = qE
a =
=
According to the third equation of motion,
or, d =
=
= 0.254 m
Thus, we can conclude that the proton will travel 0.254 m before reaching its turning point.
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
