How far apart would you have to place the poles of a 1. 5 v battery to achieve the same electric field?
1 answer:
To place the poles of a 1. 5 v battery to achieve the same electric field is 1.5×10−2 m
The potential difference is related to the electric field by:
∆V=Ed
where,
∆V is the potential difference
E is the electric field
d is the distance
what is potential difference?
The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.
We want to know the distance the detectors have to be placed in order to achieve an electric field of
E=1v/cm=100v/cm
when connected to a battery with potential difference
∆v=1.5v
Solving the equation,we find
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To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.
From the perspective of angular movement, we find the relationship with the tangential movement of velocity through
Where,
Angular velocity
v = Lineal Velocity
R = Radius
At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is
Where
Angular acceleration
Angular velocity
t = Time
Our values are
Replacing at the previous equation we have that the angular velocity is
Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s
At the same time the angular acceleration would be
Therefore the angular acceleration of a point on the outer edge of the tires is
Answer:
0.30581
0.24464
Explanation:
= Coefficient of static friction
= Coefficient of kinetic friction
= 75 N
= 60 N
Normal force
Frictional force
The coefficient of static friction is 0.30581
Kinetic force
The coefficient of kinetic friction is 0.24464