The head of the hammer is weighted
Answer:
Continuously changing magnetic field of the Sun
Explanation:
The Sun is made up of plasma and is not solid like our planet. When it rotates the whole of the Sun doesn't rotate with same speed. The equatorial part completes the rotation in just 25 days whereas the poles do it in 35 days. Due to this the magnetic field lines entangle and reorganize them regularly.
The places where the field line exit and enter the surface of the Sun, temperature drops by around 1000 K thus these spots appear black in color and are known as Sun spots.
The magnetic field is not permanent as it will keep changing due to differential rotation. This will result in the change in the no. of location of Sun spots.
If we track the no. of sunspots visible with respect to years we will notice that they follow a cycle of 10.6 years. This is known as Solar cycle in which there comes a solar minima when we see very few sunspots. When it is solar maxima we can see more than 100 sunspots.
coulomb's law. inverse square law of e static attraction and/or repulsion
Answer:
If the speed of the merry-go-round doubles, the force you will need to exert to hang on is 400 N.
Explanation:
Given;
initial force exerted to hang on, F₁ = 100 N
The force exerted on the merry-go-round in order to hang on must be an inward force known as centripetal force.
Centripetal force is given by;
![F_c = \frac{mv^2}{r} \\\\keeping \ "m" \ and \ "r" \ constant, we \ will \ have \ the \ following \ equation;\\\\\frac{F_c_1}{v_1^2} = \frac{F_c_2}{v_2^2} \\\\F_c_2 = \frac{F_c_1*v_2^2}{v_1^2}\\\\when \ the \ speed\ doubles \ i.e, v_2 = 2v_1\\\\ F_c_2 = \frac{F_c_1*(2v_1)^2}{v_1^2}\\\\ F_c_2 = \frac{F_c_1*4v_1^2}{v_1^2}\\\\F_c_2 = F_c_1 *4\\\\F_c_2 = 4(F_c_1)\\\\F_c_2 = 4 (100 \ N)\\\\F_c_2 = 400 \ N](https://tex.z-dn.net/?f=F_c%20%3D%20%5Cfrac%7Bmv%5E2%7D%7Br%7D%20%5C%5C%5C%5Ckeeping%20%5C%20%22m%22%20%5C%20and%20%5C%20%22r%22%20%5C%20constant%2C%20we%20%5C%20will%20%5C%20have%20%5C%20the%20%5C%20following%20%5C%20equation%3B%5C%5C%5C%5C%5Cfrac%7BF_c_1%7D%7Bv_1%5E2%7D%20%3D%20%5Cfrac%7BF_c_2%7D%7Bv_2%5E2%7D%20%5C%5C%5C%5CF_c_2%20%3D%20%5Cfrac%7BF_c_1%2Av_2%5E2%7D%7Bv_1%5E2%7D%5C%5C%5C%5Cwhen%20%5C%20the%20%5C%20speed%5C%20doubles%20%5C%20i.e%2C%20v_2%20%3D%202v_1%5C%5C%5C%5C%20F_c_2%20%3D%20%5Cfrac%7BF_c_1%2A%282v_1%29%5E2%7D%7Bv_1%5E2%7D%5C%5C%5C%5C%20F_c_2%20%3D%20%5Cfrac%7BF_c_1%2A4v_1%5E2%7D%7Bv_1%5E2%7D%5C%5C%5C%5CF_c_2%20%3D%20F_c_1%20%2A4%5C%5C%5C%5CF_c_2%20%3D%204%28F_c_1%29%5C%5C%5C%5CF_c_2%20%3D%204%20%28100%20%5C%20N%29%5C%5C%5C%5CF_c_2%20%3D%20400%20%5C%20N)
Therefore, If the speed of the merry-go-round doubles, the force you will need to exert to hang on is 400 N.