The energy is 3.06 electronvolts, E = 3.06eV
1eV = 1.6 * 10^-19 J
3.06 eV = 3.06* 1.6 * 10^-19 J = 4.896 * 10^-19 J
d=? v=2.5 u=0 and t=5 therefore the formula to be used to find the distance my brother covered is d=1/2(v-u)t
d=1/2(2.5-0)5
=6.15m
2NO2 means that there is 2 oxygen atoms and one nitrogen with two sets of that. So its the third one
Kepler’s three law is the answer. Kepler’s 3 is the amount
of time it takes to orbit the sun is related to size and distance. Kepler’s 3 is one of the planetary motion and
can be stated as all planets move in elliptical orbits, having the sun sits at
one of the foci.
Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.