During a solar eclipse, the moon is positioned directly between earth and the sun. find the magnitude of the net gravitational force acting on the moon during the solar eclipse due to both earth and the sun. the masses of the sun, earth, and the moon are 1.99×10301.99×1030 kg, 5.98×10245.98×1024 kg, and 7.36×10227.36×1022 kg, respectively. the moon's mean distance from earth is 3.84×1083.84×108 m, and earth's mean distance from the sun is 1.50×10111.50×1011 m. the gravitational constant is =6.67×10−11 nm2/kg2g=6.67×10−11 nm2/kg2 .

Protons have a positive charge and, collectively, give an element its atomic number.

The figure is missing but we can answer anyway.the correct answer is "releasing the string". in fact, when the person is holding the bow and the arrow, the kinetic energy of the arrow is still zero (because the arrow is not moving). the moment the arrow is released, it acquires a velocity v, so its kinetic energy will become different from zero: where m is the mass of the arrow and v its velocity.  all the other options, instead, do not increase the kinetic energy of the system: -  pulling farther back on the string : this will just increase the elastic potential energy of the system-  holding the arrow higher: this will increase the gravitational potential energy of the system (which is given by u=mgh, where h is the height above the ground)-  decreasing the mass of the arrow: this will decrease the gravitational potential energy of the system (which is given by u=mgh, where m is the mass)