Answer:

Explanation:
The index of refraction is equal to the speed of light c in vacuum divided by its speed v in a substance, or
. For our case we want to use
, which for our values is equal to:

Which we will express with 3 significant figures (since a product or quotient must contain the same number of significant figures as the measurement with the <em>least</em> number of significant figures):

Answer:
Explanation:
Given that,
Current in loops are
i1 = 12A
i2 = 20A
The loops are 3.4cm apart
The magnetic field at the center is found to be zero, so when want to find the radius of bigger loop
Magnetic Field is given as
B= μoi/2πr
Where,
μo is a constant = 4π×10^-7 Tm/A
r is the distance between the two wires
i is the current in the wires
B is the magnetic field
NOTE
Field due to large loop should be equal to the smaller loop.
B1 = B2
μo•i1 / 2π•r1 = μo•i2 / 2π•r2
Then, μo, 2π cancels out, so we have
i1 / r1 = i2 / r2
Make r2 subject of formula
i1•r2 = i2•r1
r2 = i2•r1 / i2
r2 = 20×3.4/12
r2 = 5.67cm
The radius of the bigger loop is 5.67cm.
Which excerpt are you talking about?
Answer:
Time = t = 6.62 s
Explanation:
Given data:
Height = h = 215 m
Initial velocity =
= 0 m/s
gravitational acceleration = g = 9.8 m/s²
Time = t = ?
According to second equation of motion

As initial velocity is zero, So the first term of right hand side of above equation equal to zero.

t² = 
t =
t = 
t = 6.62 s
(a) 392 N/m
Hook's law states that:
(1)
where
F is the force exerted on the spring
k is the spring constant
is the stretching/compression of the spring
In this problem:
- The force exerted on the spring is equal to the weight of the block attached to the spring:

- The stretching of the spring is

Solving eq.(1) for k, we find the spring constant:

(b) 17.5 cm
If a block of m = 3.0 kg is attached to the spring, the new force applied is

And so, the stretch of the spring is

And since the initial lenght of the spring is

The final length will be
