I'm not sure if this is correct but it's what I'll do
This is free-fall problem.
Stone A is thrown upward, at the point it falls down to the place where it was thrown, the velocity is -15m/s.
Now I choose the bridge is the origin. From the bridge, stone A and B fall the same distance which means Ya = Yb ( vertical distance )
Ya = Vo(t + 2) + 1/2a(t+2)^2
= -15(t + 2) + 1/2(9.8)(t^2 + 4t + 4)
= -15t - 30 + 4.5(t^2 + 4t + 4)
= -15t - 30 + 4.5t^2 + 18t + 18
= 4.5t^2 +3t - 12
Yb = Vo(t) + 1/2a(t)^2
= 0 + 4.5t^2
4.5t^2 = 4.5t^2 +3t - 12
0 = 3t - 12
4 = t
Time for Stone B is 4s
Time for Stone A is 6s
Answer:
el plomo será el más largo
Explanation:
Dado que;
longitud inicial (l1) = 4m
Longitud final l2
aumento de temperatura (θ) = 10 ° C
Coeficiente de expansión lineal α
Ahora para el hierro;
α = 11,7 x 10-6
Desde;
l2-l / l1θ = α
l2 = α l1θ + l1
l2 = l1 (αθ + 1)
l2 = 4 ((11,7 x 10-6 * 10) + 1)
l2 = 4.00044 m
Para el plomo
l2 = 4 ((27,3 x 10-6 * 10) + 1)
l2 = 4,00109 m
Para cobre
l2 = 4 ((16,7 x 10-6 * 10) + 1)
l2 = 4.000668 m
Por lo tanto, el plomo será el más largo
Answer:
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Explanation:
Let L represent Moe's height during the leap.
Moe's velocity v at any point in time during the leap is;
v = dL/dt = u - gt .......1
Where;
u = it's initial speed
g = acceleration due to gravity on Mars
t = time
The determine how far the cricket was off the ground when it became Moe’s lunch.
We need to integrate equation 1 with respect to t
L = ∫dL/dt = ∫( u - gt)
L = ut - 0.5gt^2 + L₀
Where;
L₀ = Moe's initial height = 0
u = 105m/s
t = 56 s
g = 3.75 m/s^2
Substituting the values, we have;
L = (105×56) -(0.5×3.75×56^2) + 0
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Answer:
Circled in Red...plus the scales, of course.
Explanation:
