Answer:
5080.86m
Explanation:
We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:
We must consider that it's launched from the ground (
) and from rest (
), with an upwards acceleration 
that lasts a time t=9.7s.
We calculate then the height achieved in part 1:
And the velocity achieved in part 1:
We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (
) and its initial velocity is the one achieved in part 1 (
), now in free fall, which means with a downwards acceleration 
. For the data we have it's faster to use the formula 
, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:
Then, to get 
, we do:
And we substitute the values:
Is this a question? Please provide more information.
Answer:
W = 1,307 10⁶ J
Explanation:
Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)
F = G m₁ M / r²
W = ∫ F. dr
W = G m₁ M ∫ dr / r²
we integrate
W = G m₁ M (-1 / r)
We evaluate between the limits, lower r = R_ Moon and r = ∞
W = -G m₁ M (1 /∞ - 1 / R_moon)
W = G m1 M / r_moon
Body weight is
W = mg
m = W / g
The mass is constant, so we can find it with the initial data
For the capsule
m = 1000/32 = 165 / g_moon
g_moom = 165 32/1000
.g_moon = 5.28 ft / s²
I think it is easier to follow the exercise in SI system
W_capsule = 1000 pound (1 kg / 2.20 pounds)
W_capsule = 454 N
W = m_capsule g
m_capsule = W / g
m = 454 /9.8
m_capsule = 46,327 kg
Let's calculate
W = 6.67 10⁻¹¹ 46,327 7.36 10²² / 1.74 10⁶
W = 1,307 10⁶ J
Answer:
27%
Explanation:
15.999 divided by 58.32 = .27433128
Move the decimal place over 2 places.
27%
