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Elena-2011 [213]
3 years ago
14

If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s then the KINETIC ENERGY of the CANNON will be ("E4" means

"*10^4") *
1 point
4 J
40 J
400 J
None of the above
Physics
2 answers:
Alex3 years ago
7 0
40 j would be your answer
lilavasa [31]3 years ago
6 0

Answer:

M V = m v        conservation of momentum (Caps-cannon  Small-projectile)

V = m / M * V = 2 / 2000 * 200 m/s = .2 m/s    recoil velocity of cannon

KE = 1/2 M V^2 = 2000 / 2 kg * (.2 m/s)^2  = 40 kg m^2/s^2 = 40 J

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V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

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V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

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V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

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