It is powered by the Earth's rotation and the moon gives a little boost.
Answer:
2081.65 m
Explanation:
We'll begin by calculating the time taken for the load to get to the target. This can be obtained as follow:
Height (h) = 3000 m
Acceleration due to gravity (g) = 10 m/s²
Time (t) =?
h = ½gt²
3000 = ½ × 10 × t²
3000 = 5 × t²
Divide both side by 5
t² = 3000 / 5
t² = 600
Take the square root of both side
t = √600
t = 24.49 s
Finally, we shall determine the distance from the target at which the load should be released. This can be obtained as follow:
Horizontal velocity (u) = 85 m/s
Time (t) = 24.49 s
Horizontal distance (s) =?
s = ut
s = 85 × 24.49
s = 2081.65 m
Thus, the load should be released from 2081.65 m.
Answer:
This is net charge on the surface is Q = σ₀ x (y + 2by²)
Explanation:
The surface charge density is defined as the amount of charge Q per unit area A
σ = dq / dA
dq = σ dA
Since the surface is a rectangular region we use an xy coordinate system so the area difference
dA = dxdy
dq = σ dx dy
We replace, evaluate the integral
∫ dq = ∫ σ₀ (1 + yb) dxdy
realizamos laintegral de dx
Q -0 =σ₀ ∫ (1 + yb) (x-0) dy
Where we evaluate We must recognize that the charge Q must be zero by the time X = 0 and Y = 0. At the starting point Q = 0 for x = 0
We perform the other integral (dy)
Q = σ₀ x (y + 2y² b)
Evaluated between Y = 0 and Y = y
Q = σ₀ x (y + 2by²)
This is net charge on the surface
The weight of an object is given by:
where
F is the weight
m is the mass of the object
is the gravitational acceleration
The piece of ham in this problem has a mass of
m=630 g = 0.63 kg
therefore its weight is
