Answer:
Time= 6.12*10^4s
mass flow rate m=0.98kg/s
Explanation:
Given
Volume= 60m^3
diamter= 2.5cm= 0.025m
radius= 0.0125m
area A= πr^2
area A= 3.142*0.0125^2
area A= 4.9*10^-4m^2
the velocity of the flow 2m/s
<u>volume flow rate </u>
V=vA
V=2* 4.9*10^-4
V=9.82*10^-4 m^3/s
<u>Time taken to fill the pool</u>
time= volume/volume flow rate
time= 60/9.82*10^-4
time= 6.12*10^4s
<u>Mass flow rate </u>
m= density *volume flow rate
Assuming the density of water to be 997kg/m^3
m= 997*9.82*10^-4
m=0.98kg/s
1 volt = 1 joule per coulomb
50 volts = 50 joules per coulomb
50 joules/coulomb times 6 coulombs = 300 joules
With same braking power you will be stopping faster on the original weight therefore the answer to fill the blank is increase. The stopping distance will increase as there'll be higher energy to dissipate than lighter cars applied with the braking force similar with that of the lighter car. Also the skid and drag will add to the distance as well as the inertia of the moving heavier vehicle would be greater as well.
Answer:
t=0.0625s
Explanation:
F=number of swings/time taken
DATA
Frequency=4.0Hz
number of swings from Q to R
=1/4
time taken=?
Frequency=number of swings/time taken
make t the subject of the formula
t=n/f
substitute the given date
t=0.25/4.0
t=0.0625s
option A is collect
