A 1kg cannon ball.is fired horizontally with an initial velocity of 5m/s. If the cannon was atop a wall 20m above the ground, wh
1 answer:
Answer:
Ek = 196.2 [J]
Explanation:
The question concerns the KE kinetic energy.
That is, we must find the kinetic energy at the moment the cannon is fired and the kinetic energy of when the ball hits the ground after having fallen 20 meters.
At the moment when the ball is fired it is 20 meters above ground level. If the ground level is taken as the reference level of potential energy, where it is equal to zero, in this way when the ball is at the highest (20 meters) you have the maximum potential energy.
In this way, the energy in the initial state is equal to the sum of the kinetic energy plus the potential energy. As the energy is conserved this same energy will be present when the ball hits the ground, where the potential energy is zero and will have only kinetic energy.
The kinetic energy in the initial state can be easily calculated by means of the following equation.
Therefore the change in KE
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Answer:
Acceleration of the plane, a = 5.4 g
Explanation:
It is given that,
Speed of the jet plane, v = 1890 km/h = 525 m/s
Radius of the arc, r = 5.20 km = 5200 m
The plane is moving in the circular path, the centripetal acceleration will act on it. It is given by :
We know that, the value of g is,
So, the acceleration of the plane is 5.4 g. Hence, this is the required solution.
Hi there!
We can approach this problem in many ways, but to show you how I arrive to the final conclusion, I will begin by solving the circuit with an assigned value for the power source.
Let's use a power source value of 6V (produces nice numbers).
Recall the following rules.
Capacitors in series:
- Voltage ADDS up.
- Charge is EQUAL across each.
- Total capacitance uses the reciprocal rule.
Capacitors in parallel:
- Voltage is EQUAL across each.
- Charge ADDS up.
- Total capacitance is simply the sum.
Solving for total capacitance:
Using rules for capacitors in series and parallel, the total capacitance of the circuit is 1.5 pF.
Thus, the total charge is:
9 pC will go through the parallel combination and the individual capacitor in series with the combination.
Since the voltage adds up, we can find the amount of voltage across the 3pF capacitor with the remaining going through the branches of the parallel combination.
Therefore, 3V goes through each branch since 6V - 3V = 3V.
Solving for the charge for each capacitor:
<u>Thus, the capacitor with the greatest charge is the 3 pF capacitor. </u>
To explain without all of the work above, the equivalent capacitance of the parallel combination (1 pF + 2pF = 3pF) is equivalent to the capacitance of the capacitor in series (3pF). Thus, the voltage across the parallel capacitors (since voltage is the same across branches in parallel) and the series capacitor is equal. However, since charge SUMS UP for capacitors in parallel, they would have less charge than the single capacitor in series.