At a high school science fair, Connor won first place for his replica of the Golden Gate Bridge. Connor liked the project so muc
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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:
Hello some parts of your question is missing below is the missing part
Convection coefficient = 11 w/m^2. °c
answer : 44.83 watts
Explanation:
Given data :
surface emissivity ( ε )= 0.95
head ( sphere) diameter( D ) = 0.25 m
Temperature of sphere( T ) = 35° C
Temperature of surrounding ( T∞ ) = 25°C
Temperature of surrounding surface ( Ts ) = 15°C
б = ( 5.67 * 10^-8 )
Determine the total rate of heat loss
First we calculate the surface area of the sphere
As =
= = 0.2 m^2
next we calculate heat loss due to radiation
Qrad = ε * б * As( ) ---- ( 1 )
where ;
ε = 0.95
б = ( 5.67 * 10^-8 )
As = 0.2 m^2
T = 35 + 273 = 308 k
Ts = 15 + 273 = 288 k
input values into equation 1
Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )
= 22.83 watts
Qrad ( heat loss due to radiation ) = 22.83 watts
calculate the heat loss due to convection
Qconv = h* As ( ΔT )
= 11*0.2 ( 35 -25 ) = 22 watts
Hence total rate of heat loss
= 22 + 22.83
= 44.83 watts