Answer:
Final temperature of the aluminum = 41.8 °C
Explanation:
We have the equation for energy
E = mcΔT
Here m = 55 g = 0.055 kg
ΔT = T - 27.5
Specific heat capacity of aluminum = 921.096 J/kg.K
E = 725 J
Substituting
E = mcΔT
725 = 0.055 x 921.096 x (T - 27.5)
T - 27.5 = 14.31
T = 41.81 ° C = 41.8 °C
Final temperature of the aluminum = 41.8 °C
The kinetic energy would be 53,775J:)
Explanation:
Use Pythagorean theorem:
r² = x² + y²
r² = (-6.46 m)² + (-3.78 m)²
r = 7.48 m
Answer:
![\Delta t = 8 s](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%208%20s)
Explanation:
As we know that the angular acceleration of the wheel due to friction is constant
so we can use kinematics
![\theta = \omega_i t + \frac{1}{2}\alpha t^2](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Comega_i%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Calpha%20t%5E2)
so we have
![(65 \times 2\pi) = (2\pi \times 9)(10) + \frac{1}{2}(\alpha)(10^2)](https://tex.z-dn.net/?f=%2865%20%5Ctimes%202%5Cpi%29%20%3D%20%282%5Cpi%20%5Ctimes%209%29%2810%29%20%2B%20%5Cfrac%7B1%7D%7B2%7D%28%5Calpha%29%2810%5E2%29)
![130\pi = 180\pi + 50 \alpha](https://tex.z-dn.net/?f=130%5Cpi%20%3D%20180%5Cpi%20%2B%2050%20%5Calpha)
![\alpha = -\pi rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%20-%5Cpi%20rad%2Fs%5E2)
now time required to completely stop the wheel is given as
![\omega_f = \omega_i + \alpha t](https://tex.z-dn.net/?f=%5Comega_f%20%3D%20%5Comega_i%20%2B%20%5Calpha%20t)
![0 = (2\pi \times 9) + (-\pi) t](https://tex.z-dn.net/?f=0%20%3D%20%282%5Cpi%20%5Ctimes%209%29%20%2B%20%28-%5Cpi%29%20t)
![t = 18 s](https://tex.z-dn.net/?f=t%20%3D%2018%20s)
now time required to stop the wheel is given as
![\Delta t = 18 - 10](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%2018%20-%2010)
![\Delta t = 8 s](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%208%20s)
Answer:
The magnetic field inside the solenoid would decrease by a factor of 2.
Explanation:
The magnetic field, B, of a solenoid of length L, N windings, and radius b with a current, I, flowing through it is given as:
B = (N * r * I) / L
If the length of the solenoid is doubled, 2L,the magnetic field becomes:
B2 = (N * r * I) / 2L
B2 = ½ B
The magnetic field will decrease by a factor of 2.