<span>e=ca{\displaystyle e={\frac {c}{a}}}.</span>
<span>a+b= ?
3i +3j + (3i -3j) = ?
3i + 3j + 3i -3j =?
= 6i + 0j</span>
Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
Answer:
D) equal to the flux of electric field through the Gaussian surface B.
Explanation:
Flux through S(A) = Flux through S (B ) = Charge inside/ ∈₀