Answer:
E =17,280 J
Explanation:
Since the voltage drops linearly from 6V to 4V, the average is 5V. The energy delivered is .
V is the average of initial and fianl voltage
A is the current passing
t is the time in which current is passed
E=V*A*t = 5*0.015*(64*3600) = 17,280 J
Answer:
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The specific heat capacity of different substances vary for the same reason that different substances have different melting and boiling points to one another. If the bonds between atoms are stronger, it will require more energy to heat up the substance.Sep 8, 2018
The net force acting on the airplane is 25N.
Forces acting on the paper airplane when it is in the air:
- The forward force generated by the engine, propeller, or rotor is called thrust. It resists or defeats the drag force. It operates generally perpendicular to the longitudinal axis. However, as will be discussed later, this is not always the case.
- Drag is an airflow disruption generated by the wing, rotor, fuselage, and other projecting surfaces that causes a backward, decelerating force. Drag acts backward and perpendicular to the relative wind, opposing thrust.
- Weight is the total load carried by airplane, including the weight of the crew, fuel, and any cargo or baggage. Due to the influence of gravity, weight pulls the airplane downward.
- Lift—acts perpendicular to the flight path through the center of lift and opposes the weight's downward force. It is produced by the air's dynamic influence on the airfoil.
Given.
Weight of the paper airplane, F1 = 16N
The force of air resistance, F2 = 9N
Net force = F1 + F2
Net force = 25N
Thus, the net force acting on the airplane is 25N.
Learn more about the net force here:
brainly.com/question/18109210
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Answer:
The water level rises at 11.76 
let be h: height and s: side of an equilateral triangle
Explanation:
The picture shows a diagram of the situation, first we have to determine the height of the trough as follows:
With the Pytagorean Theorem we can find out that:
Then, the area of an equilateral triangle, as any triangle, is a half of its base times its height:
Replacing values, we have:
That is the total area of the trough, but the problem specifies that it has been filled until 0.5 ft. Therefore, we have to find the cross section area of the water flow by substracting the total area minus the unfilled area of the trough:
Then, the cross section area of water flow is
Finally, to determine the speed of water flow at this point we solve for v, the flow formula:
