The land form is formed by lava flows of low viscosity
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Your average speed is 20 mph, but ONLY IF YOU'RE IN THE CAR during those 2 hours.
We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:

Where

differs for each wavelenght, that means violet and red will have different refractive indices in the glass.
In the second figure provided details are given on which are the angles in question,

is the distance between both rays.


At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:

For red we have:

We finally have:
Power can be defined as the rate at which work is accomplished.
Option D is the correct answer.
<h3>
</h3><h3>
Power </h3>
The work done by an object in a given time interval is called the power of that object.
Suppose an external force F is applied to any object for the time interval T seconds. Due to this external force, the object will perform some amount of work for the time T seconds. This work W done by the object for the time interval T seconds is called the power of that object.
Power can be defined in mathematical term which is given below.

Thus the power can also be defined as the work done by the object per unit time interval.
Hence we can conclude that option D is the correct answer.
To know more about power, follow the link given below.
brainly.com/question/1618040.
The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as
- t=0.476v
- t=1.967v
- V2=4.323v
<h3>What is the potential across the capacitor?</h3>
Question Parameters:
A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.
at
- t = 1.0 seconds
- 5.0 seconds
- 20.0 seconds.
Generally, the equation for the Voltage is mathematically given as
v(t)=Vmax=(i-e^{-t/t})
Therefore
For t=1
V=5(i-e^{-1/10})
t=0.476v
For t=5s
V2=5(i-e^{-5/10})
t=1.967
For t=20s
V2=5(i-e^{-20/10})
V2=4.323v
Therefore, the values of voltages at the various times are
- t=0.476v
- t=1.967v
- V2=4.323v
Read more about Voltage
brainly.com/question/14883923
Complete Question
A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.
Determine the potential across the capacitor when t = 1.0 seconds, 5.0 seconds, 20.0 seconds.