Answer:
with teamwork
Explanation:
you need to use team work so the right answer is C
<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>
I’m pretty sure 14 is mutations
Answer:
a) 6.4 kJ
b) 43.4 kJ
Explanation:
a)
= Heat absorbed = 37 kJ
= Coefficient of performance = 5.8
= Work done
Heat absorbed is given as
=
37 = (5.8)
= 6.4 kJ
b)
= work per cycle required
= +
= 37 + 6.4
= 43.4 kJ
Illluminance is the measurement of photo-metric power. That means, illuminance is the rate of photo-metric flux that is received by a surface per area. It is usually expressed as a unit of W/m^2. Thus, from the choices, the answer we're looking for is illuminance.
