The correct answer is magnetic field, electric field, and charges.
<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>
Answer:3H2O + light-c3h603+302
Explanation:
Answer:
At time 10.28 s after A is fired bullet B passes A.
Passing of B occurs at 4108.31 height.
Explanation:
Let h be the height at which this occurs and t be the time after second bullet fires.
Distance traveled by first bullet can be calculated using equation of motion
Here s = h,u = 450m/s a = -g and t = t+3
Substituting
Distance traveled by second bullet
Here s = h,u = 600m/s a = -g and t = t
Substituting
Solving both equations
So at time 10.28 s after A is fired bullet B passes A.
Height at t = 7.28 s
Passing of B occurs at 4108.31 height.
