Answer:
The tension in the cable when the craft was being lowered to the seafloor is 4700 N.
Explanation:
Given that,
When the craft was stationary, the tension in the cable was 6500 N.
When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N.
The drag force of 1800 N will act in the upward direction. As it was lowered or raised at a steady rate, so its acceleration is 0. As a result, net force is 0. So,
T + F = W
Here, T is tension
F = 1800 N
W = 6500 N
Tension becomes :
So, the tension in the cable when the craft was being lowered to the seafloor is 4700 N.
Answer:
d = 329.81m
Explanation:
V_f = V_0+a*t
V_f = Velocity final
V_0 = Velocity initial
a = acceleration
t = time
V_f = (0m/s)+(9.81m/s²)*(8.2s)
V_f = 80.442m/s
d = ((V_f-V_0)/2)*t
d = distance
d = ((80.442m/s-0m/s)/2)*(8.2s)
d = 329.81m
Answer:
D
Explanation:
F = G m1 m2 / r^2 now double r
F = G m1m1/ (2r)^2
F = 1/4 G m1m2/r^2 <===== this is 1/4 of the original
I’ve answered this problem before and there were 2 parts in
this problem.
The solution would be like this for this specific problem:
<span>A.
</span><span>Vf = Vi +
Vex*ln(Mi / Mf) </span><span>
<span>0.002 * 3e8m/s = 0 + 2000m/s * ln(Mi / Mf) </span>
<span>300 = ln(Mi / Mf) </span>
<span>1.9e130 = Mi / Mf </span></span>
<span>B.
</span><span>4000m/s =
2000m/s * ln(Mi / Mf) </span><span>
<span>2 = ln(Mi / Mf) </span>
<span>7.389 = Mi / Mf </span>
<span>Mf = Mi / 7.389 = 0.135*Mi<span> </span></span></span>
The answer is “Impulse acting on it” according to the impulse-momentum theorem.
