Short answer, 10kg, it must be the same mass as Ball 1
Long calculation to prove the answer:
M1 = Mass of Ball 1
M2 = Mass of Ball 2
U1 = Initial speed of Ball 1
U2 = Initial speed of Ball 2
V1 = Final speed of Ball 1
V2 = Final speed of Ball 2
M1U1 + M2U2 = M1V1 + M2V2
10 x 20 + M2 x 20 = 10 x 40 + M2 x 0
200 + 20M2 = 400
20M2 = 200
M2 = 200 / 20
M2 = <u>10 kg</u>
Alternatively you could say that because the velocity of ball 1 increased by the exact velocity of ball 2, their masses must be equal, so both ball 1 and 2 are 10 kg.
Answer:![410.90\times 10^{-6} V](https://tex.z-dn.net/?f=410.90%5Ctimes%2010%5E%7B-6%7D%20V)
Explanation:
Given
![A=5.5 mm^2](https://tex.z-dn.net/?f=A%3D5.5%20mm%5E2)
![n=919 turns/cm\approx 85400 turns/m](https://tex.z-dn.net/?f=n%3D919%20turns%2Fcm%5Capprox%2085400%20turns%2Fm)
![\omega =226 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D226%20rad%2Fs)
According to the Faraday law of induction, induced emf is given by
![E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}](https://tex.z-dn.net/?f=E%3D-%5Cfrac%7B%5Cmathrm%7Bd%7D%20%5Cphi%20_B%7D%7B%5Cmathrm%7Bd%7D%20t%7D)
Magnetic field ![\phi _B=B\cdot A](https://tex.z-dn.net/?f=%5Cphi%20_B%3DB%5Ccdot%20A)
![B=\mu _0ni=\mu _0ni_osin\left ( \omega t\right )](https://tex.z-dn.net/?f=B%3D%5Cmu%20_0ni%3D%5Cmu%20_0ni_osin%5Cleft%20%28%20%5Comega%20t%5Cright%20%29)
![B=4\pi \times 10^{-7}\times 85400\times 3.08\times \sin (226t)](https://tex.z-dn.net/?f=B%3D4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%5Ctimes%2085400%5Ctimes%203.08%5Ctimes%20%5Csin%20%28226t%29)
![B=0.3305\sin (226t)](https://tex.z-dn.net/?f=B%3D0.3305%5Csin%20%28226t%29)
![\phi _{B}=0.3305\times \sin (226t)\times 5.5\times 10^{-6}](https://tex.z-dn.net/?f=%5Cphi%20_%7BB%7D%3D0.3305%5Ctimes%20%5Csin%20%28226t%29%5Ctimes%205.5%5Ctimes%2010%5E%7B-6%7D)
![\phi _{B}=1.818\times 10^{-6}\sin (226t)](https://tex.z-dn.net/?f=%5Cphi%20_%7BB%7D%3D1.818%5Ctimes%2010%5E%7B-6%7D%5Csin%20%28226t%29)
![E=-\frac{\mathrm{d} \phi _B}{\mathrm{d} t}](https://tex.z-dn.net/?f=E%3D-%5Cfrac%7B%5Cmathrm%7Bd%7D%20%5Cphi%20_B%7D%7B%5Cmathrm%7Bd%7D%20t%7D)
![E=-1.818\times 10^{-6}\times 226\cos (226t)](https://tex.z-dn.net/?f=E%3D-1.818%5Ctimes%2010%5E%7B-6%7D%5Ctimes%20226%5Ccos%20%28226t%29)
![E=-410.90\times 10^{-6}\cos (226t)](https://tex.z-dn.net/?f=E%3D-410.90%5Ctimes%2010%5E%7B-6%7D%5Ccos%20%28226t%29)
Amplitude of EMF induced![=410.90\times 10^{-6} V](https://tex.z-dn.net/?f=%3D410.90%5Ctimes%2010%5E%7B-6%7D%20V)
Answer:
The electric flux remains unchanged
Explanation:
From Gauss law the Electric flux is directly proportional to the number of electric field lines passing through a surface. The number of field lines passing through a surface become if the radius is doubled becomes 1/4th that is when radius of the Gaussian surface is doubled, but at the same time, the surface area has increased 4 times , so the electric flux remains unchanged
All you would do is for a, 10 times 2 is 20 so it would be 20-dB
For b, 10 times 4 is 40 so it would be 40-dB
For c, 10 times 8 is 80 so it would be 80-dB
Answer:
5.8 m/s
Explanation:
Let v = velocity of bike relative to Betty = -12.0 m/s (since the bike is moving away from betty).
u = velocity of ball relative to bike = + 17.8 m/s
and V = velocity of ball relative to Betty.
So, by Galilean relativity,
V = v + u
V = -12.0 m/s + 17.8 m/s
V = 5.8 m/s
So, the velocity of the ball as measured by Betty is 5.8 m/s