3-m-high large tank is initially filled with water. The tank water surface is open to the atmosphere, and a sharp-edged 10-cm-di
ameter orifice at the bottom drains to the atmosphere through a horizontal 80-m-long pipe. If the total irreversible head loss of the system is determined to be 1.5 m, determine the initial velocity of the water from the tank. Disregard the effect of the kinetic energy correction factors.
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Answer:
Explanation:
An image of the bond resulting from the search is attached.
Consider the force directed towards thymine as negative.
For the O-H-N combination:
The resulting force from this combination is:
In the case of the N-H-N combination:
The total force acting from this combination is:
The force that thymine applies on adenine is:
When rounded to three significant figures, the net force is
b)
The negative value indicates that the force is attractive, as it is aimed towards thymi.
c)
The force acting on the electron due to the proton is:
Since the electron and proton carry opposite charges, the force on the electron points towards the proton.
d)
The ratio of the above forces is:
Therefore, the bonding strength of the electron in the hydrogen atom is 9.3 times greater than the bonding force between adenine and thymine molecules.
Answer:
The distance before stopping is 1.52 m,
velocity is 4.0 m/s on the y-axis
Explanation:
The particle’s motion is two-dimensional due to acceleration along both the x and y axes; each axis can be addressed independently for calculations.
a) At the moment the particle starts to reverse, its velocity should be zero (Vfx = 0)
Vfₓ = V₀ₓ + aₓ t
t = - V₀ₓ/aₓ
t = - 2.4/(-1.9)
t= 1.26 s
When the particle stops, we calculate its position
X1 = V₀ₓ t + ½ aₓ t²
X1= 2.4 1.26 + ½ (-1.9) 1.26²
X1= 1.52 m
At this point, the particle begins returning.
b) The velocity comprises both x and y components.
For the x section, Vₓ = 0 m/s indicates a halt, but the y component retains a velocity
Vfy= Voy + ay t
Vfy= 0 + 3.2 1.26
Vfy = 4.0 m/s
Thus, the velocity reads as
V = (0 x^ + 4.0 y^) m/s
c) To graph the motion, we create a table listing position x and y at given time intervals; let's begin the calculations for equations
X = V₀ₓ t+ ½ aₓ t²
Y = Voy t + ½ ay t²
X= 2.4 t + ½ (-1.9) t²
Y= 0 + ½ 3.2 t²
X= 2.4 t – 0.95 t²
Y= 1.6 t²
With these equations, we construct two graphs for position against time, one for the x-axis and another for the y-axis
Chart for graphing
Time (s) x(m) y(m)
0 0 0
0.5 0.960 0.4
1 1.45 1.6
1.50 1.46 3.6
2.00 1.00 6.4
Answer with Explanation:
Concepts and reasoning
The principle for addressing this question is that a capacitor in an RC circuit allows current to flow until fully charged. Once charged, it prevents any further current from moving through. Conversely, the situation is different with an inductor in an RL circuit. In accordance with Faraday's law, an inductor generates an electromagnetic force to counteract the applied voltage, but when no change in flux occurs, it behaves akin to a regular wire as if the inductor is absent.
In the accompanying diagram, a resistor is connected in series with a capacitor.
As we observe
the voltage across both the capacitor and the source.Voltage across a resistor in an RC circuit.
Voltage across a resistor in an RL circuit.
