Answer:

Explanation:
Given:
- diameter of hose pipe,

- diameter of nozzle,

- volume flow rate,

<u>Now, flow velocity in hose:</u>



<u>Now, flow velocity in nozzle:</u>



We know the Bernoulli's equation:

when the two points are at same height then the eq. becomes




Answer:
Whether the force exerted by the locomotive on the wall was larger
Than the force the locomotive could exert on the wall.
Explanation:
The Newton's third law of motion States that every force have it's equal and opposite reaction force, whose magnitude is the same as the applied force. Therefore the magnitude of these opposite forces will be equal.
So we have;
F12=-F21
F12 is the force in a direction
-F21 is the force in the opposite direction.
Therefore we see that the magnitude of the force the locomotive exerts on the wall is equal to the one the wall exerts on the locomotive. Both magnitudes are equal but in opposite directions.
Answer:
25.13 cm
Explanation:
Mrs. Lynn's art class is making pencil holders out of soup cans
These soup cans are definitely in cylindrical forms.
Each student is first covering a can with felt.
A felt is a type of material that is used for covering other objects. Here a felt is used to cover soup cans.
So if the diameter of the bottom of the can is 8 cm (i.e d = 8cm)
What will the circumference of the piece of felt need to be in order to cover the bottom of the can?
The felt will also take the circumference of a cylinder; and which is given by the expression
C = 2 πr
since diameter (d) = 8 cm
radius (r) = 
r = 
r = 4 cm
C = 2 × π × 4 cm
C = 2 × 3.142 × 4 cm
C = 25.13 cm
∴ the circumference of the piece of felt need to be 25.13 cm in order to cover the bottom of the can.
By conducting exhaustive, high-intensity online research for about 15 seconds, I found a source that says the speed of sound in copper is 4600 m/s. (You could easily have completed the same research project in about 1/3 of the time it took you to type and post the question here.)
Time it takes = (distance) / (speed)
Time = (25,000 meters) / (4600 m/s)
Time = (25 / 4.6) km-sec/km
<em>Time = 5.43 seconds </em>
Answer:
7.6 g
Explanation:
"Well lagged" means insulated, so there's no heat transfer between the calorimeter and the surroundings.
The heat gained by the copper, water, and ice = the heat lost by the steam
Heat gained by the copper:
q = mCΔT
q = (120 g) (0.40 J/g/K) (40°C − 0°C)
q = 1920 J
Heat gained by the water:
q = mCΔT
q = (70 g) (4.2 J/g/K) (40°C − 0°C)
q = 11760 J
Heat gained by the ice:
q = mL + mCΔT
q = (10 g) (320 J/g) + (10 g) (4.2 J/g/K) (40°C − 0°C)
q = 4880 J
Heat lost by the steam:
q = mL + mCΔT
q = m (2200 J/g) + m (4.2 J/g/K) (100°C − 40°C)
q = 2452 J/g m
Plugging the values into the equation:
1920 J + 11760 J + 4880 J = 2452 J/g m
18560 J = 2452 J/g m
m = 7.6 g