Answer
given,
discharge rate from pipe = 1000 gallons/minutes
now,
flow rate in cubic meters per second
1 gallon = 0.00378541 m³
1 min = 60 s
Q = 
Q = 0.063 m³/s
flow rate in liters per minute
1 gallon = 3.78541 L
Q = 
Q = 3785.41 m³/min
flow rate in cubic feet per second
1 gallon = 0.133681 ft³
1 min = 60 s
Q = 
Q = 2.23 ft³/s
E=energy=5.09x10^5J = 509KJ
<span>M=mass=2250g=2.25Kg </span>
<span>C=specific heat capacity of water= 4.18KJ/Kg </span>
<span>ΔT= change in temp= ? </span>
<span>E=mcΔT </span>
<span>509=(2.25)x(4.18)xΔT </span>
<span>509=9.405ΔT </span>
<span>ΔT=509/9.405=54.1degrees </span>
<span>Initial temp = 100-54 = 46 degrees </span>
<span>Hope this helps :)</span>
Answer: Both cannonballs will hit the ground at the same time.
Explanation:
Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.
then the acceleration equation is only on the vertical axis, and can be written as:
a(t) = -(9.8 m/s^2)
Now, to get the vertical velocity equation, we need to integrate over time.
v(t) = -(9.8 m/s^2)*t + v0
Where v0 is the initial velocity of the object in the vertical axis.
if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s
and:
v(t) = -(9.8 m/s^2)*t
Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)
And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.
You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)
Tectonic plates is the unscrambled phrase in that jumble.
