By Snell's law:
η = sini / sinr. i = 25, η = 1.33
1.33 = sin25° / sinr
sinr = sin25° / 1.33 = 0.4226/1.33 = 0.3177 Use a calculator.
r = sin⁻¹(0.3177)
r ≈ 18.52°
Option A.
God's grace.
If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula
![A_x = |A|cos(\alpha)\\\\A_y = |A|sin(\alpha)](https://tex.z-dn.net/?f=A_x%20%3D%20%7CA%7Ccos%28%5Calpha%29%5C%5C%5C%5CA_y%20%3D%20%7CA%7Csin%28%5Calpha%29)
Where | A | is the magnitude of the vector and
is the angle that it forms with the x axis in the opposite direction to the hands of the clock.
In this problem we know the value of Ax and Ay and we need the angle
.
Vector A is in the 4th quadrant
So:
![A_x = 6\\\\A_y = -6.5](https://tex.z-dn.net/?f=A_x%20%3D%206%5C%5C%5C%5CA_y%20%3D%20-6.5)
So:
![|A| = \sqrt{6^2 + (-6.5)^2}\\\\|A| = 8.846](https://tex.z-dn.net/?f=%7CA%7C%20%3D%20%5Csqrt%7B6%5E2%20%2B%20%28-6.5%29%5E2%7D%5C%5C%5C%5C%7CA%7C%20%3D%208.846)
So:
![Ay = -6.5 = 8.846cos(\alpha)\\\\sin(\alpha) = \frac{-6.5}{8.846}\\\\sin(\alpha) = -0.7348\\\\\alpha = sin^{- 1}(- 0.7348)](https://tex.z-dn.net/?f=Ay%20%3D%20-6.5%20%3D%208.846cos%28%5Calpha%29%5C%5C%5C%5Csin%28%5Calpha%29%20%3D%20%5Cfrac%7B-6.5%7D%7B8.846%7D%5C%5C%5C%5Csin%28%5Calpha%29%20%3D%20-0.7348%5C%5C%5C%5C%5Calpha%20%3D%20sin%5E%7B-%201%7D%28-%200.7348%29)
= -47.28 ° +360° = 313 °
= 313 °
Option 4.
The centripetal force acting on the ball will be 23.26 N.The direction of the centripetal force is always in the path of the center of the course.
<h3>What is centripetal force?</h3>
The force needed to move a body in a curved way is understood as centripetal force. This is a force that can be sensed from both the fixed frame and the spinning body's frame of concern.
The given data in the problem is;
m is the mass of A ball = 0.25 kg
r is the radius of circle= 1.6 m rope
v is the tangential speed = 12.2 m/s
is the centripetal force acting on the ball
The centripetal force is found as;
![\rm F_C = \frac{mv^2}{r} \\\\ F_C = \frac{0.25 \times (12.2)^2}{1.6} \\\\ F_C=23.26\ N](https://tex.z-dn.net/?f=%5Crm%20F_C%20%3D%20%5Cfrac%7Bmv%5E2%7D%7Br%7D%20%20%5C%5C%5C%5C%20F_C%20%3D%20%5Cfrac%7B0.25%20%5Ctimes%20%2812.2%29%5E2%7D%7B1.6%7D%20%20%5C%5C%5C%5C%20F_C%3D23.26%5C%20N)
Hence the centripetal force acting on the ball will be 23.26 N.
To learn more about the centripetal force refer to the link;
brainly.com/question/10596517
Explanation:
Calculate position vectors in a multidimensional displacement problem. Solve for the displacement in two or three dimensions. Calculate the velocity vector
Answer:
The drill's angular displacement during that time interval is 24.17 rad.
Explanation:
Given;
initial angular velocity of the electric drill,
= 5.21 rad/s
angular acceleration of the electric drill, α = 0.311 rad/s²
time of motion of the electric drill, t = 4.13 s
The angular displacement of the electric drill at the given time interval is calculated as;
![\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Comega%20_i%20t%20%5C%20%2B%20%5C%20%5Cfrac%7B1%7D%7B2%7D%5Calpha%20t%5E2%5C%5C%5C%5C%5Ctheta%20%3D%20%285.21%20%5C%20%5Ctimes%20%5C%204.13%29%20%5C%20%2B%20%5C%20%5Cfrac%7B1%7D%7B2%7D%280.311%29%284.13%29%5E2%5C%5C%5C%5C%5Ctheta%20%3D%20%2821.5173%20%29%20%5C%20%2B%20%5C%20%282.6524%29%5C%5C%5C%5C%5Ctheta%20%3D24.17%20%5C%20rad)
Therefore, the drill's angular displacement during that time interval is 24.17 rad.