Answer:
5.843 m
Explanation:
suppose that the arrow leave the bow with a horizontal speed , towards he bull's eye.
lets consider that horizontal motion
distance = speed * time
time = 40/ 37 = 1.081 s
arrow doesnot have a initial vertical velocity component. but it has a vertical motion due to gravity , which may cause a miss of the target.
applying motion equation
(assume g = 10 m/s²)
Arrow misses the target by 5.843m ig the arrow us split horizontally
If a man pushes on a wall with some force then according to Newton's third law, wall will also apply force on man with same magnitude but opposite in direction.
Car A will have highest speed is 83.3m/s .
<h3>What is speed ? </h3>
The rate of change of position of an object in any direction.
The S.I unit is m/s . Speed is a scalar quantity it defines only magnitude not direction
.
speed = distance /time
In case of Car A ,
We have given distance 150Km in 3 min ,
First we have convert the distance km to m
150×1000m
then conversion of min to sec
38×60sec
speed = 15000/180
speed = 83.3m/sec
In case of Car B
we have given 800m in 150 min
lets convert the time into second
150×60
Speed = 800/150×60
speed = 0.88m/ s
In case of Car C
We have given here distance 250 Km and time in 8 hours
convert km to m
25000
and time into sec
88×60×60
speed = 0.86m/ s
Hence ,Car A has highest speed amongst them .
To learn more about speed click here
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Answer:
The maximum energy stored in the combination is 0.0466Joules
Explanation:
The question is incomplete. Here is the complete question.
Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Energy stored in a capacitor is expressed as E = 1/2CtV² where
Ct is the total effective capacitance
V is the supply voltage
Since the capacitors are connected in series.
1/Ct = 1/C1+1/C2+1/C3
Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF
1/Ct = 1/11.7 + 1/21.0 + 1/28.8
1/Ct = 0.0855+0.0476+0.0347
1/Ct = 0.1678
Ct = 1/0.1678
Ct = 5.96μF
Ct = 5.96×10^-6F
Since V = 125V
E = 1/2(5.96×10^-6)(125)²
E = 0.0466Joules
