<span>The payload will continue in orbit along with the space shuttle and their relative velocities will be close to zero. However, depending up the relative orientation of the space shuttle's center of mass and the payload's center of mass, the orbits of both the shuttle and payload is likely to slowly diverge from each other. To illustrate how low the relative velocities would be, let's assume a near worse case situation happens and the center of mass of the payload and the space shuttle differs after the release by 50 meters, then the resulting orbital periods of the shuttle and the released payload would differ by about 0.06 seconds from each other.</span>
Answer:
Explanation:
A )
Th expression for time period for a spring -mass system is as follows
T = 2π
So if mass is doubled , Time period will be √2 or 1.414 times or 5.9 s
B )
If the mass is halved , time period becomes 1 / √2 times or .70 times or 3.0 s
C)
Time period does not depend upon the amplitude of oscillation . So in this case time period will be unchanged ie 4.2 s
D )
As per the formula above , if spring constant k is doubled , time period will be 1 / √2 times or .70 times or 3.0 s
The specific gravity is the ratio of the density of oil to the density of water. Since the density of water is 1g/ml, the density of a substance is numerically the same as the specific gravity.
However, specific gravity is a dimensionless number, while density has units mass/volume.
So, the density of oil is .901 g/ml, but the specific gravity is just .901
In Imperial units, the density of the oil is 56.2 lb/ft^3, but the specific gravity is still .901
The initial velocity is
v(0) = 16.5 ft/s
While in the water, the acceleration is
a(t) = 10 - 0.
![\frac{dv}{dt} =10-0.8v \\\\ \frac{dv}{10-0.8v}=dt \\\\ \int_{16.5}^{v} \, \frac{dv}{10-0.8v} = \int_{0}^{t} dt \\\\ - \frac{1}{0.8} [ln(10-0.8v)]_{16.5}^{v}=t \\\\ ln \frac{10-0.8v}{-3.2}=-0.8t \\\\ \frac{0.8v -10}{3.2} =e^{-0.8t} \\\\ 0.8v = 10 + 3.2e^{-0.8t} \\\\ v=12.5+4e^{-0.08t}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D10-0.8v%20%5C%5C%5C%5C%20%20%5Cfrac%7Bdv%7D%7B10-0.8v%7D%3Ddt%20%5C%5C%5C%5C%20%5Cint_%7B16.5%7D%5E%7Bv%7D%20%5C%2C%20%20%5Cfrac%7Bdv%7D%7B10-0.8v%7D%20%20%3D%20%5Cint_%7B0%7D%5E%7Bt%7D%20dt%20%5C%5C%5C%5C%20-%20%5Cfrac%7B1%7D%7B0.8%7D%20%5Bln%2810-0.8v%29%5D_%7B16.5%7D%5E%7Bv%7D%3Dt%20%5C%5C%5C%5C%20ln%20%5Cfrac%7B10-0.8v%7D%7B-3.2%7D%3D-0.8t%20%5C%5C%5C%5C%20%20%5Cfrac%7B0.8v%20-10%7D%7B3.2%7D%20%20%3De%5E%7B-0.8t%7D%20%5C%5C%5C%5C%200.8v%20%3D%2010%20%2B%203.2e%5E%7B-0.8t%7D%20%5C%5C%5C%5C%20v%3D12.5%2B4e%5E%7B-0.08t%7D)
The velocity function is

It satisfies the condition that v(0) = 16.5 ft/s.
When t = 5.7s, obtain

The depth of the lake is
![d=\int_{0}^{5.7} \, (12.5+4e^{-0.8t})dt \\\\ = 12.5(5.7)+ \frac{4}{(-0.8)}[e^{-0.8t}]_{0}^{5.7} \\\\ =71.25-5(0.0105-1) =76.198 \, ft](https://tex.z-dn.net/?f=d%3D%5Cint_%7B0%7D%5E%7B5.7%7D%20%5C%2C%20%2812.5%2B4e%5E%7B-0.8t%7D%29dt%20%5C%5C%5C%5C%20%3D%2012.5%285.7%29%2B%20%5Cfrac%7B4%7D%7B%28-0.8%29%7D%5Be%5E%7B-0.8t%7D%5D_%7B0%7D%5E%7B5.7%7D%20%5C%5C%5C%5C%20%3D71.25-5%280.0105-1%29%20%3D76.198%20%5C%2C%20ft)
Answer:
The velocity at the bottom of the lake is 12.5 ft/s
The depth of the lake is 76.2 ft