A:a diver jumping Off a diving Board And falling to the swimming pool below
Answer:
The right solution is "
". A further explanation is given below.
Explanation:
The given values are:
Current,
I = ![2.6\times 10^5](https://tex.z-dn.net/?f=2.6%5Ctimes%2010%5E5)
Time,
t = 4.1 s
As we know,
⇒ ![Q=I\times t](https://tex.z-dn.net/?f=Q%3DI%5Ctimes%20t)
On substituting the given values, we get
⇒ ![=2.6\times 10^5\times 4.1](https://tex.z-dn.net/?f=%3D2.6%5Ctimes%2010%5E5%5Ctimes%204.1)
⇒ ![=1,066,000 \ C](https://tex.z-dn.net/?f=%3D1%2C066%2C000%20%5C%20C)
⇒ ![=10.66\times 10^5 \ C](https://tex.z-dn.net/?f=%3D10.66%5Ctimes%2010%5E5%20%5C%20C)
Answer:
The frictional torque is ![\tau = 0.2505 \ N \cdot m](https://tex.z-dn.net/?f=%5Ctau%20%20%3D%200.2505%20%5C%20N%20%5Ccdot%20m)
Explanation:
From the question we are told that
The mass attached to one end the string is ![m_1 = 0.341 \ kg](https://tex.z-dn.net/?f=m_1%20%3D%20%200.341%20%5C%20kg)
The mass attached to the other end of the string is ![m_2 = 0.625 \ kg](https://tex.z-dn.net/?f=m_2%20%3D%20%200.625%20%5C%20kg)
The radius of the disk is ![r = 9.00 \ cm = 0.09 \ m](https://tex.z-dn.net/?f=r%20%3D%209.00%20%5C%20cm%20%20%3D%200.09%20%5C%20m)
At equilibrium the tension on the string due to the first mass is mathematically represented as
![T_1 = m_1 * g](https://tex.z-dn.net/?f=T_1%20%3D%20%20m_1%20%2A%20%20g)
substituting values
![T_1 = 0.341 * 9.8](https://tex.z-dn.net/?f=T_1%20%3D%20%200.341%20%2A%209.8)
![T_1 = 3.342 \ N](https://tex.z-dn.net/?f=T_1%20%3D%20%203.342%20%5C%20N)
At equilibrium the tension on the string due to the mass is mathematically represented as
![T_2 = m_2 * g](https://tex.z-dn.net/?f=T_2%20%3D%20%20m_2%20%2A%20%20g)
![T_2 = 0.625 * 9.8](https://tex.z-dn.net/?f=T_2%20%3D%200.625%20%2A%209.8)
![T_2 = 6.125 \ N](https://tex.z-dn.net/?f=T_2%20%3D%206.125%20%5C%20N)
The frictional torque that must be exerted is mathematically represented as
![\tau = (T_2 * r ) - (T_1 * r )](https://tex.z-dn.net/?f=%5Ctau%20%20%3D%20%20%28T_2%20%2A%20r%20%29%20-%20%28T_1%20%2A%20r%20%29)
substituting values
![\tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )](https://tex.z-dn.net/?f=%5Ctau%20%20%3D%20%20%28%206.125%20%2A%200.09%20%29%20-%20%283.342%20%20%2A%200.09%20%29)
![\tau = 0.2505 \ N \cdot m](https://tex.z-dn.net/?f=%5Ctau%20%20%3D%200.2505%20%5C%20N%20%5Ccdot%20m)
Answer:
Angular acceleration = 5 rad /s ^2
Kinetic energy = 0.391 J
Work done = 0.391 J
P =6.25 W
Explanation:
The torque is given as moment of inertia × angular acceleration
angular acceleration = torque/ moment of inertia
= 10/2= 5 rad/ s^2
The kinetic energy is = 1/2 Iw^2
w = angular acceleration/time
=5/8= 0.625 rad /s
1/2 × 2× 0.625^2
=0.391 J
The work done is equal to the kinetic energy of the motor at this time
W= 0.391 J
The average power is = torque × angular speed
= 10× 0.625
P = 6.25 W
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