The potential energy will be 1.46*10^-4J.
To find the answer, we have to know about the torque acting on a current loop in a uniform magnetic field.
<h3>How to find the potential energy of the loop?</h3>
- We have the expression for torque acting on a current loop in a uniform magnetic field as,
where; M is the magnetic dipole moment, B is the magnetic field , and theta is the angle between M and B.
- As we know that, the torque is equal to force times the perpendicular distance. Thus, it is equivalent to the work done. This work is stored as the potential energy in the loop.
- Thus, the potential energy will be,
Thus, we can conclude that, the potential energy will be 1.46*10^-4J.
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Time=50s
speed=25m/s
Distance = speed×time
=25×50
=1250m
DISTANCE TRAVELLED IS =1250m
D = 1/2 g t^2. It works out to 44.1 meters.
Linear expansivity, area expansivity and volume or cubic expansivity are
The orbiting velocity of the satellite is 4.2km/s.
To find the answer, we need to know about the orbital velocity of a satellite.
<h3>What's the expression of orbital velocity of a satellite?</h3>
- Mathematically, orbital velocity= √(GM/r)
- r = radius of the orbital, M = mass of earth
<h3>What's the orbital velocity of a satellite orbiting earth with a radius 3.57 times the earth radius?</h3>
- M= 5.98×10²⁴ kg, r= 3.57× 6.37×10³ km = 22.7×10⁶m
- Orbital velocity= √(6.67×10^(-11)×5.98×10²⁴/22.7×10⁶)
=4.2km/s
Thus, we can conclude that the orbiting velocity of the satellite is 4.2km/s.
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