If the planet earth has no land masses, the idealized zonal
precipitation pattern would likely have regions that are wet in the equator and
there will be more of mid-latitudes if the earth has no land masses at all and
it does not exist.
Answer:
a ) 11.1 *10^3 m/s = 39.96 Km/h
b) T_{o2} =1.58*10^5 K
Explanation:
a)
= 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h
b)
M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol
gas constant R = 8.31 j/mol.K

So, 
multiply each side by M_{o2}, so we have

solving for temperature T_{o2}

In the question given,

T_{o2} =1.58*10^5 K
Given that in a parallel circuit:
R1 = 12 ohms
R2= 15 ohms
I = 12 A
I2 = 4 A
V=?
R=?
R3 =?
P=?
Since,
V= IR
or,
V2 = I2 * R2
V2= 4* 15
V2 = 60V
Since in a parallel circuit voltage remain same in all component of the circuit and is equal to the source voltage.
Therefore,
V= V1 = V2 = V3 = 60V
Since,
V= IR
R= V/I
R= 60/12
R= 5 ohm
That is total resistance is equal to 5 ohms.
Since for parallel circuit,
1/R= 1/R1 + 1/R2 + 1/R3
1/5= 1/12+ 1/15 + 1/R3
or
1/R3= 1/5- 1/12- 1/15
1/R3= 1/20
or
R3= 20 ohms
Since,
V=IR
I= V/R
I1= V1/ R1
I1= 60/12
I1= 5 A
I3= V3/R3
I3= 60/20
I3= 3A
Since,
P=VI
P= 60*12
P= 720 watt
P1= V1* I1
P1= 60* 5
P1= 300 watt
P2= V2* I2
P2= 60* 4
P2= 240watt
P3= V3*I3
P3= 60*3
P3= 180 watt
Hence we have,
R1= 12 ohms , R2= 15 ohms, R3= 20 ohms, R= 5 ohms
I1= 5A, I2= 4A, I3= 3A, I= 12 A
V1= V2= V3= V= 60V
P1= 300 watt, P2= 240 watt, P3 = 180 watt, P= 720 watt