Let the distance between the towns be d and the speed of the air be s.
distance = speed * time
convert the minutes time into hours.
When flying into the wind, ground speed will be air speed MINUS wind speed, hence the against the wind trip is described by:
d
s−15
=
7
3
return trip is then :
d
s+15
=
7
5
Cross-multiplying both we get the two-variable system:
3d=7∗(s−15)5d=7∗(s+15)
3d=7s−1055d=7s+105
subtract first equation from second equation we get
2d=210d=105km
Substitute the value of d in the above equations for s.
5∗105=7s+1057s=420s=60km/hr
Answer:
The first graph is showing the constant acceleration (1 m/s)
Explanation:
The second graph showing the flexible velocity therefore a in the graph is different at t1, t2, t3, t4
The last graph is showing constant velocity therefore there is no acceleration (a = 0)
Answer:
Explanation:
Calories to be burnt = 3500 - 2500 = 1000 Cals .
Efficiency of conversion to mechanical work is 25 % .
Work needed to burn this much of Cals = 1000 x 100 / 25 = 4000 Cals.
4000 Cals = 4.2 x 4000 = 16800 J .
Work done in one jump = kinetic energy while jumping
= 1/2 m v²
= .5 x 70 x 3.3²
= 381.15 J .
Number of jumps required = 16800 / 381.15
= 44 .
Answer:
friction reduces the efficiency of machines, thus we must reduce the friction force that is acting upon it.
