Answer:
1.17
Explanation:
Given that,
The refractive index of ice wrt air = 1.31
The refractive index of rock salt wrt air = 1.54
We need to find the refractive index of rock salt with respect to ice.
We know that,
refractive index = (speed of light in air or vaccum)/( speed of light in that medium)
So,
The speed of light in ice = c/(1.31)
The speed of light in rock salt = c/(1.54)
So, the refractive index of rock salt with respect to ice is :
So, the required refractive index of rock salt wrt ice is 1.17.
Answer:
Explanation:
m × 10n, where m is a number between 1 and 10 ( 1 ≤ |m| < 10 ) and the exponent n is a positive or negative integer.
Answer:
<h2>
t = 2.35s</h2>
Explanation:
Using one of the equation of motion expressed as v = u+at where;
v is the final velocity = 2.0m/s
u is the initial velocity = 25m/s
a is the acceleration = -a (since the acceleration is negative)
t is the time taken
Substituting the given parameters into the formula above;
2.0 = 25-9.81t
subtract 25 from both sides
2.0 - 25 = 25-9.81t - 25
-23 = -9.81t
Divide both sides by -9.81
-23/-9.81 = -9.81t/-9.81
t = 23.0/9.81
t = 2.35s
<em>Hence it took the car 2.35s to slow to a final velocity of 2.0 m/s</em>
The electric field strength of a uniform electric field is constant throughout the field. A perfectly uniform electric field has no variations in the entire field and is unattainable in the real world. However, two parallel plates can generate a field that resembles a perfectly uniform field with slight variations near the edge of the plates. <span>Electric fields are represented by drawing field lines that represent the direction of the field, as well as the strength of the field. More field lines represents a higher field strength. In a non-uniform electric field, the field lines tend to be curved and are more concentrated near the charges. In a uniform electric field, since the field strength does not vary, the field lines are parallel to each other and equally spaced. Uniform fields are created by setting up a potential difference between two conducting plates placed at a certain distance from one another. The field is considered to be uniform at the center of the plates, but varies close to the edge of the plates. The strength of the field depends on the potential difference applied to the plates and the distance by which they are separated. A higher potential difference or voltage results in a stronger electric field. The greater the distance between the plates, the weaker the field becomes. The electric field is therefore calculated as a ratio of the voltage between the plates to the distance they are separated by.</span>
