The correct answer is
C. Light can pass through Object B faster than it can pass through Object A.
In fact, the index of refraction of a material is defined as:

where c is the speed of light in vacuum and v is the speed of light in the material. Rearranging the equation, we can write the speed of light in the material as:

So we that, the smaller the refractive index n, the greater the speed of light in the material, v. In this problem, object B has lower refractive index than object A, so light travels faster in object B.
Answer:
The time constant is
Explanation:
From the question we are told that
The spring constant is 
The mass of the ball is 
The amplitude of the oscillation t the beginning is 
The amplitude after time t is 
The number of oscillation is 
Generally the time taken to attain the second amplitude is mathematically represented as
Here T is the period of oscillation

=> 
=> 
Generally the amplitude at time t is mathematically represented as

Here a is the damping constant so
at
, 
So

=> 
taking natural log of both sides
=>
=> 
Generally the time constant is mathematically represented as
=>
=>
I think that the answer is friction
If an object is thrown in an upward direction from the top of a building 1.60 x 102 ft. high at an initial velocity of 21.82 mi/h, what is its final velocity when it hits the ground? (Disregard wind resistance. Round answer to nearest whole number and do not reflect negative direction in your answer.)
this question is troubling me i guessed 96 ft/s
can someone help me out and explain it thanks so much!!!!!!
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.