two charges having the same charge magnitude experiencing an attracting force of 3.60N when the charges are 30cm apart.what is t
1 answer:
The charges have opposite sign and magnitude
Explanation:
The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:
where:
is the Coulomb's constant
are the two charges
r is the separation between the two charges
In this problem, we have:
F = 3.60 N is the force between the two charges
r = 30 cm = 0.30 m is their separation
The two charges have same magnitude, so
So we can rewrite the equation as
And solving for q:
Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so
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Answer:
∑ τ =0, L₀ =
Explanation:
In a circular turning movement, when the arms are extended and then contracted in two possibilities:
- They are lowered the force of gravity is what pulls them, the tension of the muscle becomes zero to allow this movement.
In this movement the force is vertical(gravity) and the movement of the center of mass of each arm is vertical, so that the work is the weight value of the arm by the distance traveled by the center of mass.
- Another possibility is that the arms have stuck to the body, in this case the person's muscles perform the force, this force is horizontal and the displacement is the horizontal of the center of mass of the arms from the extended position to the contracted
In these movements the torque of the external force is equal for each arm, but in the opposite direction, so they are canceled where a net torque of zero, this causes the angular momentum to be preserved, which changes is the moment of inertia of the system and therefore you must also change the angular velocity to keep your product constant
∑ τ =0
L₀ =
I₀ w₀ = I w