Hello!
Using the equation for the electric field produced by a source charge:
E = Electric Field Strength ( 2.86 × 10⁵ N/C)
k = Coulomb's Constant ( 8.99 × 10⁹ Nm²/C²)
q = Charge of source charge (3 μC = 0.000003 C)
r = distance of test charge from source charge (m²)
We can rearrange the equation to solve for distance to make plugging in values easier. (Isolate for 'r').
Plug in the given values.
Answer:
he correct answer is 7 3 10⁻⁹ s
Explanation:
The speed of light is constant, so we can use the uniform motion ratios
v = x / t
t = x / v
let's calculate
t = 1 / 2.9972 10⁸8
t = 3.336 10⁻⁹ s
the correct answer is 7 3 10⁻⁹ s
Distance is a scalar quantity, as it has only one part.
Answer:
The refractive index of fluid 2 is 1.78
Explanation:
Refractive index , n = real depth/apparent depth
For the first fluid, n = 1.37 and apparent depth = 9.00 cm.
The real depth of the container is thus
real depth = n × apparent depth = 1.37 × 9.00 cm = 12.33 cm
To find the refractive index of fluid index of fluid 2, we use the relation
Refractive index , n = real depth/apparent depth.
Now,the real depth = 12.33 cm and the apparent depth = 6.86 cm.
So, n = 12.33 cm/6.86 cm = 1.78
So the refractive index of fluid 2 is 1.78