The question is incomplete. The complete question is :
To measure the effective coefficient of friction in a bone joint, a healthy joint (and its immediate surroundings) can be removed from a fresh cadaver. The joint is inverted, and a weight is used to apply a downward force F⃗ d on the head of the femur into the hip socket. Then, a horizontal force F⃗ h is applied and increased in magnitude until the femur head rotates clockwise in the socket. The joint is mounted in such a way that F⃗ h will cause clockwise rotation, not straight-line motion to the right. The friction force will point in a direction to oppose this rotation.
Draw vectors indicating the normal force n⃗ (magnitude and direction) and the frictional force f⃗ f (direction only) acting on the femur head at point A.
Assume that the weight of the femur is negligible compared to the applied downward force.
Draw the vectors starting at the black dot. The location, orientation and relative length of the vectors will be graded
Solution :
The normal force represented by N is equal to the downward force, 
which is equal in magnitude but it is opposite in direction.
Also the frictional force acts always to oppose the motion because the bone starts moving in a clockwise direction. The frictional force that will be applied to the right direction so that the movement or the rotation at A is opposed.
I need help to I have a test and I need help to
The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m
Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:
Substitute numerical values:
The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
Learn more about Gaussian sphere here:
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PART A)
By Snell's law we know that
here we know that
now from above equation we have
so it will refract by angle 39.3 degree
PART B)
Here as we can see that image formed on the other side of lens
So it is a real and inverted image
Also we can see that size of image is lesser than the size of object here
Here we can use concave mirror to form same type of real and inverted image
PART C)
As per the mirror formula we know that
so image will form at 30 cm from mirror
it is virtual image and smaller in size
