Answer: 15.8
Explanation:
You are given that the
Object distance U = 32 cm
Focal length F = 30.1 cm
First calculate the image distance V by using the formula
1/F = 1/U + 1/V
Substitute F and V into the formula
1/30.1 = 1/32 + 1/V
1/V = 1/30.1 - 1/32
1/V = 0.00197259
Reciprocate both sides
V = 506.94 cm
Magnification M is the ratio of image distance to object distance.
M = V/U
substitute the values of V and U into the formula
M = 506.94/32
M = 15.8
Therefore, the magnification of the image is 15.8 or approximately 16.
Answer:

Explanation:
The rotated angle is given by:

Since this is a quadratic equation it can be solved using:

Rewriting our equation:


Since
we discard the negative solution.

Answer:
When the starting and ending points are the same, the total work is zero.
Explanation:
option ( D )correct
A force is said to be conservative when the work done by the force in moving a particle from a point A to a point B is independent of the path followed between A and B and is the same for all the paths. The work done depends only on the particles initial and final positions. And when the initial and final position in conservative field are same the work done is said to be zero.
Answer:
A.)1.52cm
B.)1.18cm
Explanation:
angular speed of 120 rev/min.
cross sectional area=0.14cm²
mass=12kg
F=120±12ω²r
=120±12(120×2π/60)^2 ×0.50
=828N or 1068N
To calculate the elongation of the wire for lowest and highest point
δ=F/A
= 1068/0.5
δ=2136MPa
'E' which is the modulus of elasticity for alluminium is 70000MPa
δ=ξl=φl/E =2136×50/70000=1.52cm
δ=F/A=828/0.5
=1656MPa
δ=ξl=φl/E
=1656×50/70000=1.18cm

The answer is 45 degrees.
According to the Kinematics of projectile motion, if the purpose is to maximize range, optimum angle of landing is always 45 degrees.If the purpose is to maximize range & projection height is zero, the optimum angle of projection (and landing) is 45 degrees.