According to newton's first law, an object traveling in a circle does not have a force acting on it.

Light rays change direction when they hit a mirror. The phenomenon is known as reflection. Light rays travels in a straight light. They strike the surface of the mirror at a particular angle called incident angle. It is the angle between the ray and normal at the point of contact. The rays leaves the surface making the same angle with the normal called reflection angle but in different direction.

8 0

2 years ago

What is the moment of inertia of a 2.0 kg, 20-cm-diameter disk for rotation about an axis (a) through the center, and (b) throug

FinnZ [79.3K]

Answer:

(a) I=0.01 kg.m²

(b) I=0.03 kg.m²

Explanation:

Given data

Mass of disk M=2.0 kg

Diameter of disk d=20 cm=0.20 m

To Find

(a) Moment of inertia through the center of disk

(b) Moment of inertia through the edge of disk

Solution

For (a) Moment of inertia through the center of disk

Using the equation of moment of Inertia

$I=\frac{1}{2}MR^{2}\\ I=\frac{1}{2}(2.0kg)(0.20m/2)^{2}\\ I=0.01 kg m^{2}$

For (b) Moment of inertia through the edge of disk

We can apply parallel axis theorem for calculating moment of inertia

$I=(1/2)MR^{2}+MD\\ Here\\D=R\\I=(1/2)(2.0kg)(0.20m/2)^{2}+(2.0kg)(0.20m/2)^{2}\\ I=0.03kgm^{2}$

8 0

3 years ago

The linear charge density on the inner conductor is and the linear charge density on the outer conductor is

Lubov Fominskaja [6]

Complete Question

The complete question is shown on the first uploaded image (reference for Photobucket )

Answer:

The electric field is $E = -1.3 *10^{-4} \ N/C$

Explanation:

From the question we are told that

The linear charge density on the inner conductor is $\lambda _i = -26.8 nC/m = -26.8 *10^{-9} C/m$

The linear charge density on the outer conductor is

$\lambda_o = -60.0 nC/m = -60.0 *10^{-9} \ C/m$

The position of interest is r = 37.3 mm =0.0373 m

Now this position we are considering is within the outer conductor so the electric field at this point is due to the inner conductor (This is because the charges on the conductor a taken to be on the surface of the conductor according to Gauss Law )

Generally according to Gauss Law

$E (2 \pi r l) = \frac{ \lambda_i }{\epsilon_o}$

=> $E = \frac{\lambda _i }{2 \pi * \epsilon_o * r}$

substituting values

$E = \frac{ -26 *10^{-9} }{2 * 3.142 * 8.85 *10^{-12} * 0.0373}$

$E = -1.3 *10^{-4} \ N/C$

The negative sign tell us that the direction of the electric field is radially inwards

=> $|E| = 1.3 *10^{-4} \ N/C$

5 0

3 years ago