Answer:
eccentrcity of orbit is 0.22
Explanation:
GIVEN DATA:
Initial velocity of satellite = 8333.3 m/s
distance from the sun is 600 km
radius of earth is 6378 km
as satellite is acting parallel to the earth therefore
and radial component of given velocity is zero
we have
h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s
we know that
so
solvingt for 
therefore eccentrcity of orbit is 0.22
Answer:
24.72 kwh
Explanation:
Electric energy=potential energy=mgz where m is mass, g is acceleration due to gravity and z is the elevation.
Substituting the given values while taking g as 9.81 and dividing by 3600 to convert to per hour we obtain
PE=(108*9.81*84)/3600=24.72 kWh
Answer:
0
Explanation:
output =transfer function H(s) ×input U(s)
here H(s)=
U(s)=
for unit step function
output =H(s)×U(s)
=×
=
taking inverse laplace of output
output=t×
at t=0 putting the value of t=0 in output
output =0
Answer:
38 kJ
Explanation:
The solution is obtained using the energy balance:
ΔE=E_in-E_out
U_2-U_1=Q_in+W_in-Q_out
U_2=U_1+Q_in+W_in-Q_out
=38 kJ
