Answer:
only thing I think of when I see that is 'Just Wondering'
Explanation:
The sun’s gravitational attraction and the planet’s inertia keeps planets moving is circular orbits.
Explanation:
The planets in the Solar System move around the Sun in a circular orbit. This motion can be explained as a combination of two effects:
1) The gravitational attraction of the Sun. The Sun exerts a force of gravitational attraction on every planet. This force is directed towards the Sun, and its magnitude is

where
G is the gravitational constant
M is the mass of the Sun
m is the mass of the planet
r is the distance between the Sun and the planet
This force acts as centripetal force, continuously "pulling" the planet towards the centre of its circular orbit.
2) The inertia of the planet. In fact, according to Newton's first law, an object in motion at constant velocity will continue moving at its velocity, unless acted upon an external unbalanced force. Therefore, the planet tends to continue its motion in a straight line (tangential to the circular orbit), however it turns in a circle due to the presence of the gravitational attraction of the Sun.
Learn more about gravity:
brainly.com/question/1724648
brainly.com/question/12785992
#LearnwithBrainly
Explanation:
It is given that,
Magnetic field, B = 0.15 T
Charge on a proton, 
Mass of a proton, 
The cyclotron frequency is given by :


f = 2286785.40 Hz
or


Hence, this is the required solution.
Answer:a. Magnetic dipole moment is 0.3412Am²
b. Torque is zero(0)N.m
Explanation: The magnetic dipole moment U is given as the product of the number of turns n times the current I times the area A
That is,
U = n*I*A
But Area A is given as pi*radius² since it is a circular coil
Radius given is 5cm converting to meter we divide by 100 so we have our radius to be 0.05m. So area A is
A = 3.142*(0.05)² =7.86*EXP {-3} m²
Current I is 2 A
Number of turns is 20
So magnetic dipole moment U is
U = 20*2*7.86*EXP {-3}=0.3142A.m²
b. Torque is given as the cross product of the magnetic field B and magnetic dipole moment U
Torque = B x U =B*U*Sine(theta)
But since the magnetic field is directed parallel to the plane of the coil from the question, it means that the angle between them is zero and sine zero is equals 0(zero) if you substitute that into the formula for torque you will find out that your torque would equals zero(0)N.m