A potential difference of 71 mV is developed across the ends of a 12.0-cm-long wire as it moves through a 0.27 T uniform magneti
c field at a speed of 6.0 m/s. The magnetic field is perpendicular to the axis of the wire.
Required:
What is the angle between the magnetic field and the wire's velocity?
Required:
What is the angle between the magnetic field and the wire's velocity?
1 answer:
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Answer:
Explanation:
We are given that
Surface area of membrane=
Thickness of membrane=
Assume that membrane behave like a parallel plate capacitor.
Dielectric constant=5.9
Potential difference between surfaces=85.9 mV
We have to find the charge resides on the outer surface of membrane.
Capacitance between parallel plate capacitor is given by
Substitute the values then we get
Capacitance between parallel plate capacitor=
V=
Hence, the charge resides on the outer surface=
Answer:
The orbital speed of this second satellite is 5195.16 m/s.
Explanation:
Given that,
Orbital radius of first satellite
Orbital radius of second satellite
Mass of first satellite
Mass of second satellite
Orbital speed of first satellite = 4800 m/s
We need to calculate the orbital speed of this second satellite
Using formula of orbital speed
From this relation,
Now,
Put the value into the formula
Hence, The orbital speed of this second satellite is 5195.16 m/s.