Answer:
The total displacement is 102 km 
north of east.
Explanation:
We can treat this problem as a trigonometric one, so we need to calculate the total displacement on the north and east.
and
The total displacement is given by:
with an agle of:
Answer:
Pressure,P=6×10^3Pa
Explanation:
The gas has an ideal gas behaviour and ideal gas equation
PV=NKT
T= V/N p/K ...eq1
Average transitional kinetic energy Ktr=1.8×10-23J
Ktr=3/2KT
T=2/3Ktr/K....eq2
Equating eq1 and 2
V/N p/K = 2/3Ktr/K
Cancelling K on both sides
P= 2/3N/V( Ktr)
Substituting the value of N/V and dividing by 10^-6 to convert cm^3 to m^3
P = 2/3 (5.0×10^20)/10^-6 × 1.8×10^-23
P= 6 ×10^3Pa
The normal force is always perpendicular to the surface. So it would be straight to the left of the wall
<span>LOCATION Z, because it is only 2 away from the coast.
The rest are farther inland
hope this helps</span>
Answer:
The power dissipated in the 3 Ω resistor is P= 5.3watts.
Explanation:
After combine the 3 and 6 Ω resistor in parallel, we have an 2 Ω and a 4 Ω resistor in series.
The resultating resistor is of Req=6Ω.
I= V/Req
I= 2A
the parallel resistors have a potential drop of Vparallel=4 volts.
I(3Ω) = Vparallel/R(3Ω)
I(3Ω)= 1.33A
P= I(3Ω)² * R(3Ω)
P= 5.3 Watts
