Answer:
Gravitational potential energy to kinetic energy to gravitational potential energy to kinetic energy to gravitational potential energy.
Explanation:
Starting at its maximum displacement the pendulum will have only gravitational potential energy, its velocity being 0m/s. When released, it will lose height, losing then gravitational potential energy as it gains speed, or kinetic energy. When the pendulum is at its lowest the gravitational potential energy will be at its minimum and the kinetic energy at its maximum (and so its speed), with value equal to the original gravitational potential energy. Then it starts gaining height again, reverting this process, gaining gravitational potential energy and losing kinetic energy until the velocity is 0m/s again, thus returning to the state of maximum gravitational potential energy (same as originally) and null kinetic energy, but on the opposite side of the oscillation. Then the pendulum comes back repeating the exact same process just descibed, until it finishes one oscillation when reaching the original point.
Answer:
The total amount of energy, kinetic plus potential, remains the same.
Explanation:
B4 the tackle:
<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>
<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>
<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>
<span>The vector triangle is right angled: </span>
<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>
<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>
<span>v(f) = 5.6 m/s (to 2 sig figs) </span>
<span>direction of v(f) is the same as the direction of the final momentum </span>
<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>
<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>
<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
Equation of power in a electrical circuit is given as:
![\bf P = I^2R](https://tex.z-dn.net/?f=%20%5Cbf%20P%20%3D%20I%5E2R)
I → Current flowing through the circuit.
R → Resistance of the circuit
We need to calculate power when;
Current (I) = 0.02 A
Resistance (R) = 30 Ω
By substituting values in the equation, we get:
![\rm \longrightarrow P = (0.02)^2 \times 30 \\ \\ \rm \longrightarrow P = 0.0004 \times 30 \\ \\ \rm \longrightarrow P = 0.012 \: W](https://tex.z-dn.net/?f=%20%5Crm%20%5Clongrightarrow%20P%20%3D%20%280.02%29%5E2%20%5Ctimes%2030%20%5C%5C%20%20%5C%5C%20%5Crm%20%5Clongrightarrow%20P%20%3D%200.0004%20%5Ctimes%2030%20%5C%5C%20%20%5C%5C%20%5Crm%20%5Clongrightarrow%20P%20%3D%200.012%20%5C%3A%20W)
Power in the circuit = 0.012 W