Will mark as BRAINLIEST.....
1 answer:
Answer:
Explanation:
Initially the packet was ascending up with the balloon.
Taking upward as positive direction;
initial velocity, u = 4.9 m/s
final velocity = v m/s
initial height, h₁ = 100 m
final height, h₂ = 0
a = -9.8 m/s²
time taken = t seconds
Considering the second equation of motion.
S = ut + 1/2at^2
s = ut + 0.5at²
(h₂-h₁) = ut + 0.5at²
0- 100 = 4.9t + 0.5×(-9.8)×t²
-100 = 4.9t - 4.9t²
4.9t² -4.9t - 100 =0
t² -t - 20.41 = 0
Using : b±√(b²-4ac))/(2a).... 1
Where a =1, b = -1 c =-20.408
Substituting the values into equation 1
Solving it, we get t = 5.045s
v = u + at = 4.9 -9.8×5.045 = 4.9 - 49.44 = -44.54 m/s
The final velocity is -44.54 m/s, and it took 5.045s to reach the ground.
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Answer:
the ship's energy is greater than this and the crew member does not meet the requirement
Explanation:
In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship
W =∫ F dx = ΔK
Let's replace
∫ (α x³ + β) dx = ΔK
α x⁴ / 4 + β x = ΔK
Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J
x (α x³ + β) = - K₀
= K₀ + x (α x³ + β)
Assuming that the low limit is x = 0, measured from the cargo hangar
Let's calculate
= 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)
Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)
Kf = 2.7 10¹¹ - 1.1475 10¹¹
Kf = 1.55 10¹¹ J
In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement
We evaluate the kinetic energy if the System is well calibrated
W = x F₀ = –K₀
= K₀ + x F₀
We calculate
= 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶
= (2.7 -2.625) 10¹¹
= 7.5 10⁹ J