<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
<span>principal quantum number (n) </span>represents the relative overall energy of each orbital
Hope this helps!
Answer:
Approximately (assuming that the acceleration due to gravity is .)
Explanation:
Let denote the first piston's contact area with the fluid. Let denote the second piston's contact area with the fluid.
Similarly, let and denote the size of the force on the two pistons. Since the person is placing all her weight on the first piston:
.
Since both pistons fit into cylinders, the two contact surfaces must be circles. Keep in mind that the area of a square is equal to times its radius, squared:
- .
- .
By Pascal's Law, the pressure on the two pistons should be the same. Pressure is the size of normal force per unit area:
.
For the pressures on the two pistons to match:
.
, , and have all been found. The question is asking for . Rearrange this equation to obtain:
.
Evaluate this expression to obtain the value of , which represents the force on the piston with the larger diameter:
.
V = u + a*t = 1100ft/s + (1000*10) ft/s = 11100 ft/s
Answer is <span>11,100 ft/s </span>
Answer:
There will be no change in the angular momentum of the system.
Explanation:
Total angular momentum of the system will remain unchanged . We can apply law of conservation of momentum because no external torque is acting on the system . There is increase in the momentum of inertia due to dropping of ball of putty . In order to conserve angular momentum , the system decreases its angular velocity . Hence the final angular momentum remains unchanged .