Position of B :
x = 4.66*cos 30 = 4.036
y = 3-4.66*sin 30 = 3-2.33 = 0.67
BC = √y^2+(x-1)^2 = √0.67^2+3.036^2 = 3.109
heading = arctan y/(x-1) = arctan 0.67/(3.036) = 12.44° south of west
hope this helps :)
Answer:
conserved
Explanation:
During this process the energy is conserved
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:
b because we apply Hooke's law
Explanation:
Hooke's law
