Distance of lake a is 200 km at 20 degree north of east
distance between lake a and b is 230 km at 30 degree west of north
now the distance between base and lake b is given as

given that




now the total distance is


now the magnitude of the distance is given as


also the direction is given as


<em>so it is 277.4 km at 74.7 degree North of East</em>
Answer:
Q=185.84C
Explanation:
We have to take into account the integral

In this case we have a superficial density in coordinate system.
Hence, we have for R: x2 + y2 ≤ 4

but, for symmetry:
![Q=4\int_0^2\int_0^{\sqrt{4-x^2}}\rho dydx\\\\Q=4\int_0^2\int_0^{\sqrt{4-x^2}}(4x+4y+4x^2+4y^2) dydx\\\\Q=4\int_0^{2}[4x\sqrt{4-x^2}+2(4-x^2)+4x^2\sqrt{4-x^2}+\frac{4}{3}(4-x^2)^{3/2}]dx\\\\Q=4[46.46]=185.84C](https://tex.z-dn.net/?f=Q%3D4%5Cint_0%5E2%5Cint_0%5E%7B%5Csqrt%7B4-x%5E2%7D%7D%5Crho%20dydx%5C%5C%5C%5CQ%3D4%5Cint_0%5E2%5Cint_0%5E%7B%5Csqrt%7B4-x%5E2%7D%7D%284x%2B4y%2B4x%5E2%2B4y%5E2%29%20dydx%5C%5C%5C%5CQ%3D4%5Cint_0%5E%7B2%7D%5B4x%5Csqrt%7B4-x%5E2%7D%2B2%284-x%5E2%29%2B4x%5E2%5Csqrt%7B4-x%5E2%7D%2B%5Cfrac%7B4%7D%7B3%7D%284-x%5E2%29%5E%7B3%2F2%7D%5Ddx%5C%5C%5C%5CQ%3D4%5B46.46%5D%3D185.84C)
HOPE THIS HELPS!!
Answer:
Time according to earth clock (T0) = 0.22 years (Approx)
Explanation:
Given:
Time taken by light = 4.5 years
Time taken by ship = 5 years
Speed of light = c
Speed of ship (v) = 90% of c = 0.9c
Find:
Time according to earth clock (T0) = ?
Computation:
Time dilation is ,

Time according to earth clock (T0) = 0.22 years (Approx)