I am going to need a picture for this question
Answer:
The acceleration of the ball is 4.18 [m/s^2]
Explanation:
By Newton's second law we can find the acceleration of the ball
![F = m*a\\where:\\F = force applied [N] or [kg*m/s^2]\\m = mass of the ball [kg]\\a = acceleration [m/s^s]](https://tex.z-dn.net/?f=F%20%3D%20m%2Aa%5C%5Cwhere%3A%5C%5CF%20%3D%20force%20applied%20%5BN%5D%20or%20%5Bkg%2Am%2Fs%5E2%5D%5C%5Cm%20%3D%20mass%20of%20the%20ball%20%5Bkg%5D%5C%5Ca%20%3D%20acceleration%20%5Bm%2Fs%5Es%5D)
Now we have:
![a = F/m\\a = \frac{1.8 [kg*m/s^s]}{0.43[kg]} \\a = 4.18 [kg]](https://tex.z-dn.net/?f=a%20%3D%20F%2Fm%5C%5Ca%20%3D%20%5Cfrac%7B1.8%20%5Bkg%2Am%2Fs%5Es%5D%7D%7B0.43%5Bkg%5D%7D%20%5C%5Ca%20%3D%204.18%20%5Bkg%5D)
Answer:
Magnetic field, B = 0.199 T
Explanation:
It is given that,
Radius of circular loop, r = 11.7 cm = 0.117 m
Magnetic flux through the loop, 
The magnetic flux linked through the loop is :


Here, 

or


B = 0.199 T
So, the strength of the magnetic field is 0.199 T. Hence, this is the required solution.
In this problem, we are given that a ball is thrown upward and that it has an initial velocity of 5 m/s. Its path both on Earth and on the moon will be trajectory in nature. This is because while its horizontal velocity is maintained, its vertical velocity increases.