Explanation:
a) Given in the y direction (taking down to be positive):
Δy = 50 m
v₀ = 0 m/s
a = 10 m/s²
Find: t
Δy = v₀ t + ½ at²
50 m = (0 m/s) t + ½ (10 m/s²) t²
t = 3.2 s
b) Given in the x direction:
v₀ = 12 m/s
a = 0 m/s²
t = 3.2 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (12 m/s) (3.2 s) + ½ (0 m/s²) (3.2 s)²
Δx = 38 m
The question is incomplete. The complete question is :
A plate of uniform areal density 
is bounded by the four curves:
where x and y are in meters. Point 
has coordinates 
and 
. What is the moment of inertia 
of the plate about the point ?
Solution :
Given :
and , 
, 
.
So,
, 
![$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$](https://tex.z-dn.net/?f=%24I%3D2%20%5Cint_1%5E2%20%5Cleft%28%20%5Cleft%5B%20%28x-1%29%5E2y%2B%5Cfrac%7B%28y%2B2%29%5E3%7D%7B3%7D%5Cright%5D_%7B-x%5E2%2B4x-5%7D%5E%7Bx%5E2%2B4x%2B6%7D%5Cright%29%20%5C%20dx%24)
So the moment of inertia is 
.
Answer:
D
Explanation:
If it is of very high intensity it will be 85-100%
Answer: 200 J
Explanation: In order to explain this we have consider that the work done in a electric field is given by:
Work= Q*ΔV=2*100=200J
