Density=mass/volume therefore volume=mass/density; 55g/11.4g/cm^3= 4.82cm^3
Answer:
Explanation:
From the question;
We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.
We are to calculate the following task, i.e. to determine the electric field at the distances:
a) at 4.75 cm
b) at 20.5 cm
c) at 125.0 cm
Given that:
the charge (q) = 33.3 nC/m
= 33.3 × 10⁻⁹ c/m
radius of rod = 5.75 cm
a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.
Then, the electric field will be zero.
b) The electric field formula 

E = 1461.95 N/C
c) The electric field E is calculated as:

E = 239.76 N/C
Answer:
Una Mezcla Homogénea es aquella mezcla en la que las sustancias que la forman poseen una combinación uniforme.Son ejemplos de Mezclas Homogéneas: Compuesta
Explanation:
Aire (es una mezcla de gases homogénea formada principalmente por de nitrógeno, oxígeno, vapor de agua, dióxido de carbono...)
Leche (mezcla de agua, carbohidratos, proteínas...)
Bebida alcohólica (mezcla de agua y alcohol etílico)
Acero (mezcla de elementos aleados como el hierro, el carbono y otras sustancias)
Petróleo (mezcla de hidrocarburos)
Agua de mar (mezcla de agua, cloruro sódico y otras sustancias)
Mezcla de agua y sal disuelta
Agua azucarada (mezcla de agua y azúcar)
Aleación metálica (las aleaciones metálicas son mezclas en las que se combinan diferentes metales de una manera homogénea y definida)
Perfume (mezcla de agua y otras sustancias olorosas cuya composición es uniforme)
Answer:
C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
- So, we can write the following equation:

- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:

- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

- As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
The mass of the box
HOPE THIS HELPS❤️