Answer:
KE = 1/2 M v^2 kinetic energy of projectile
KE = 1/2 * 20 * (54 m/s)^2 = work done on projectile
W = 10 * 54^2 = work done on projectile
W = 29,160 J
D if the blue car started higher it would have more energy but since the red car is lower it is going faster because it’s going down a hill
Answer:
Explanation:
Given that:
If we let the piece of the close lose contact at ∠θ;
Then ; from force balance;
we have:
where;
Again:
Answer:
Explanation:
100 % increase makes an amount 2 times .
44 % increase = 42 MJ
100 % increase = 95.45 MJ
final kinetic energy = 2 x 95.45 MJ
= 190.9 x 10⁶ J
1/2 m v² = 190.9 x 10⁶
.5 x m x 9² = 190.9 x 10⁶
m = 4.71 x 10⁶ kg
= 4710 metric ton .
Answer:
α = 141.5° (counterclockwise)
Explanation:
If
q₁ = +q
q₂ = -q
q₃ < 0
b = 2*a
We apply Coulomb's Law as follows
F₁₃ = K*q₁*q₃ / d₁₃² = + K*q*q₃ / (2*a)² = + K*q*q₃ / (4*a²)
F₂₃ = K*q₂*q₃ / d₂₃² = - K*q*q₃ / (5*a²)
(d₂₃² = a² + (2a)² = 5*a²)
Then
∅ = tan⁻¹(2a/a) = tan⁻¹(2) = 63.435°
we apply
F₃x = - F₂₃*Cos ∅ = - (K*q*q₃ / (5*a²))* Cos 63.435°
⇒ F₃x = - 0.0894*K*q*q₃ / a²
F₃y = - F₂₃*Sin ∅ + F₁₃
⇒ F₃y = - (K*q*q₃ / (5*a²))* Sin 63.435° + (K*q*q₃ / (4*a²))
⇒ F₃y = 0.0711*K*q*q₃ / a²
Now, we use the formula
α = tan⁻¹(F₃y / F₃x)
⇒ α = tan⁻¹((0.0711*K*q*q₃ / a²) / (- 0.0894*K*q*q₃ / a²)) = - 38.5°
The real angle is
α = 180° - 38.5° = 141.5° (counterclockwise)
