Answer:
The car strikes the tree with a final speed of 4.165 m/s
The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m
Explanation:
First we need to calculate the initial speed 
Once we have the initial speed, we can isolate the final speed from following equation:
Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.
To do that, we replace 62 m in the first formula, as follows:
Answer:
C. Between North and West
Explanation:
Since all have equal masses and the red ball and green ball are moving in south and east direction, the blue ball would most likely be moving between the north and West direction.
Answer:R=1607556m
θ=180degrees
Explanation:
d1=74.8m
d2=160.7km=160.7km*1000
d2=160700m
d3=80m
d4=198.1m
Using analytical method :
Rx=-(160700+75*cos(41.8))= -160755.9m
Ry= -(74.8+75sin(41.8))-198.1=73m
Magnitude, R:
R=√Rx+Ry
R=√160755.9^2+20^2=160755.916
R=160756m
Direction,θ:
θ=arctan(Rx/Ry)
θ=arctan(-73/160755.9)
θ=-7.9256*10^-6
Note that θ is in the second quadrant, so add 180
θ=180-7.9256*10^6=180degrees
Answer:
Explanation:
Given:
Steam Mass rate, ms = 1.5 kg/min
= 1.5 kg/min × 1 min/60 sec
= 0.025 kg/s
Air Mass rate, ma = 100 kg/min
= 100 kg/min × 1 min/60 sec
= 1.67 kg/s
A.
Extracting the specific enthalpy and temperature values from property table of “Saturated water – Pressure table” which corresponds to temperature at 0.07 MPa.
xf, quality = 0.9.
Tsat = 89.9°C
hf = 376.57 kJ/kg
hfg = 2283.38 kJ/kg
Using the equation for specific enthalpy,
hi = hf + (hfg × xf)
= 376.57 + (2283.38 × 0.9)
= 2431.552 kJ/kg
The specific enthalpy of the outlet, h2 = hf
= 376.57 kJ/kg
B.
Rate of enthalpy (heat exchange), Q = mass rate, ms × change in specific enthalpy
= ms × (hi - h2)
= 0.025 × (2431.552 - 376.57)
= 0.025 × 2055.042
= 51.37455 kW
= 51.38 kW.
