Explanation:
SUPONIENDO QUE LA ACELERACIÓN DE LA GRAVEDAD ES 
USANDO LA SEGUNDA LEY DE NEWTON:
<em>m</em> = 80.0 N/
= 8.16 kg
Answer:

Explanation:
Given that
For straight wire
Charge density= λ
For hollow metal cylinder
Charge density=2 λ
We know that electric filed for wire given as


Now the electric filed due to hollow metal cylinder


Now by considering the Gaussian surface r<R then only electric fild due to wire will present.So
At r<R

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Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.

We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)
- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)
Both of these act in opposite directions. Let's use the equation for torque:

Doing the summation using their respective lever arms:


Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:

Now, let's solve for 'T'.

Plugging in the values:

Answer:

Explanation:
We are given that
Diameter of wire=d=4.12 mm
Radius of wire=r

Current=I=8 A
Drift velocity=
We have to find the density of free electrons in the metal
We know that
Density of electron=
Using the formula
Density of free electrons=
By using Area of wire=

Density of free electrons=