According to snells law
<span>n1 sin theta1 = n2 sin theta2
</span>n1 = 1.333 (water)
<span>n2 = 2.42 (diamond)</span>
it is given that theta =30 degrees so
by putting the values we have
<span>1.333 sin theta = 2.42 sin 30 </span>
<span>sin theta = (2.42/1.333) *0.5 =65.2 degree
</span>so our conclusion is
<span>the ray's angle of incidence θ1 on the diamond</span> = 65.2 degree.
hope this helps
Answer:


Explanation:
r = Radius = 2.7 cm
F = Force = 
A = Area = 
= Stress = 
E = Young's modulus = 
= Strain
= Original length = 67 cm
= Change in length
Young's modulus is given by

Strain is 
Strain is given by

The cylinder height decreases by 
Answer:
4. All of the above I think, not to sure about 1. but the rest are right so im like 90.99999 percent sure good luck
Answer:
i'm not sure if you are asking as a personal question or a book question so i'm taking it personal.
Explanation:
I was doing a simple task that was handed to me to test my responsibility and I agreed (knowing i am responsible :3). my first thought was "man , this is easy!" but then i started seeing the other kids slaking off and quiting their tasks. I thought that was against the rules, but then i saw my bff doing it too and i thought "this should be ok then!" so i did the same. other kids where still doing it. the teacher came, saw the ones still working and smiled... but when the teacher looked at the ones slaking off omg... his face was like * im gonna kill yall* we took one big gulp and whined. the teacher awarded the ones who completed the task... the others , we had to do our original task but doubled... for 3 weeks!!! it was awful!!!
I WOULD NEVER DO THAT AGAIN!!!
Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.