The complete and comprehensive solution is attached.
The answer is B(t) = constants x I(t)
Please take precaution on the point that it is an independent field of its radial position, if the point is measured well in the solenoid. (also the radial position is the axis of its symmetry)
Answer:
μ = 0.692
Explanation:
In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.
Attached is an image with the respective forces:
A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.
Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.
The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.
The process of solving this problem can be seen in the attached image.
Answer: It would increase.
Explanation:
The equation for determining the force of the gravitational pull between any two objects is:

Where G is the universal gravitational constant, m1 is the mass of one body, m2 is the mass of the other body, and r^2 is the distance between the two objects' centers squared.
Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.
Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.
There are other forces at work here nevertheless we will imagine
it is just a conservation of momentum exercise. Also the given mass of the
astronaut is light astronaut.
The solution for this problem is using the formula: m1V1=m2V2 but
we need to get V1:
V1= (m2/m1) V2
V1= (10/63) 12 = 1.9 m/s will be the final speed of the astronaut after
throwing the tank.