Answer:

Explanation:
As we know that the formula of range is given as

now we know that
maximum value of the range of the projectile is given as

now we need to find such angles for which the range is half the maximum value
so we will have




Hooke's Law states that the extension is directly proportional to the force applied so:
F/x = constant
F₁/x₁ = F₂/x₂
2 / 0.02 = 1600 / x₂
x₂ = 16 m
Elastic work = 1/2 Fx
= 1/2 * 1600 * 16
= 12.8 kJ
<em>weight = (mass) x (gravity)</em>
Weight = (5.00 kg) x (9.81 m/s²)
weight = (5.00 x 9.81) (kg-m/s²)
<em>Weight = 49.05 Newton</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!!
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