Answer:
3.1216 m/s.
Explanation:
Given:
M1 = 0.153 kg
v1 = 0.7 m/s
M2 = 0.308 kg
v2 = -2.16 m/s
M1v1 + M2v2 = M1V1 + M2V2
0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2
= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2
0.153V1 + 0.308V2 = -0.55818. i
For the velocities,
v1 - v2 = -(V1 - V2)
0.7 - (-2.16) = -(V1 - V2)
-(V1 - V2) = 2.86
V2 - V1 = 2.86. ii
Solving equation i and ii simultaneously,
V1 = 3.1216 m/s
V2 = 0.2616 m/s
Answer: 330.88 J
Explanation:
Given
Linear velocity of the ball, v = 17.1 m/s
Distance from the joint, d = 0.47 m
Moment of inertia, I = 0.5 kgm²
The rotational kinetic energy, KE(rot) of an object is given by
KE(rot) = 1/2Iw²
Also, the angular velocity is given
w = v/r
Firstly, we calculate the angular velocity. Since it's needed in calculating the Kinetic Energy
w = v/r
w = 17.1 / 0.47
w = 36.38 rad/s
Now, substituting the value of w, with the already given value of I in the equation, we have
KE(rot) = 1/2Iw²
KE(rot) = 1/2 * 0.5 * 36.38²
KE(rot) = 0.25 * 1323.5
KE(rot) = 330.88 J
If there is no existence of capacitors in our world there would be no electrical or electronic engineering.
A capacitor is a device that stores electrical energy in an electric field. It has two terminals and is a passive electrical component. Capacitance refers to a capacitor's effect. A capacitor commonly referred to as a condenser is one of the fundamental parts needed to create electronic circuits. Without fundamental parts like resistors, inductors, diodes, transistors, etc., a circuit's design is incomplete or it won't work properly.
Energy storage is capacitors' most popular application. Power conditioning, signal coupling or decoupling, electronic noise filtering, and remote sensing are further applications. Capacitors are employed in a wide variety of industries and have integrated into daily life due to their numerous applications.
There are numerous significant uses for capacitors. They are employed in digital circuits, for instance, to prevent the loss of data saved in big computer memories during a brief loss of power. The electric energy held in such capacitors keeps the data from being lost during a brief power outage.
To know more about capacitors refer to: brainly.com/question/14126841
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