Answer:
The x-component of the electric field at the origin = -11.74 N/C.
The y-component of the electric field at the origin = 97.41 N/C.
Explanation:
<u>Given:</u>
- Charge on first charged particle,

- Charge on the second charged particle,

- Position of the first charge =

- Position of the second charge =

The electric field at a point due to a charge
at a point
distance away is given by

where,
= Coulomb's constant, having value 
= position vector of the point where the electric field is to be found with respect to the position of the charge
.
= unit vector along
.
The electric field at the origin due to first charge is given by

is the position vector of the origin with respect to the position of the first charge.
Assuming,
are the units vectors along x and y axes respectively.

Using these values,

The electric field at the origin due to the second charge is given by

is the position vector of the origin with respect to the position of the second charge.

Using these values,

The net electric field at the origin due to both the charges is given by

Thus,
x-component of the electric field at the origin = -11.74 N/C.
y-component of the electric field at the origin = 97.41 N/C.
Folds and faults are difficult to identify because they occur in the interior of rocks and also due to the dense nature of the materials.
<h3>What are faults and folds?</h3>
Faults are lines of weakness are present in materials dues to uneven positioning of the particles of the material.
Folds occurs when infolds occur in materials.
Faults and folds usually occur in rocks.
Folds and faults are difficult to identify because they occur internally and also due to the dense nature of the materials.
Learn more about faults and folds at: brainly.com/question/14240712
Answer:
Give the child a lot of room to your side, which may mean moving closer to the oncoming vehicles.
Explanation:
I majored in Physics.
Here is the full question:
The rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by:

The radius k of the equivalent hoop is called the radius of gyration of the given body. Using this formula, find the radius of gyration of (a) a cylinder of radius 1.20 m, (b) a thin spherical shell of radius 1.20 m, and (c) a solid sphere of radius 1.20 m, all rotating about their central axes.
Answer:
a) 0.85 m
b) 0.98 m
c) 0.76 m
Explanation:
Given that: the radius of gyration
So, moment of rotational inertia (I) of a cylinder about it axis = 





k = 0.8455 m
k ≅ 0.85 m
For the spherical shell of radius
(I) = 




k = 0.9797 m
k ≅ 0.98 m
For the solid sphere of radius
(I) = 




k = 0.7560
k ≅ 0.76 m