The answer is C.89,866. I multiplied 262x343.
<h3><u>Answer</u>;</h3>
= 0.6
<h3><u>Explanation</u>;</h3>
Using Pythagoras theorrem
Base² + height ² = Hypotenuse²
Thus;
Base² = 15² - 12²
= 81
Base = √81 = 9
But; cosine = adjacent/hypotenuse
Hence; cos θ = 9/15
<u>= 0.6 </u>
Answer:
94.13 ft/s
Explanation:
<u>Given:</u>
= time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
= distance to be moved by the rock long the horizontal = 98 yards
= displacement to be moved by the rock during the time of flight along the vertical = 0 yard
<u>Assume:</u>
= magnitude of initial velocity of the rock
= angle of the initial velocity with the horizontal.
For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.
Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.
On dividing equation (1) by (2), we have
Now, putting this value in equation (2), we have
Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.
1. In a single atom, no more than 2 electrons can occupy a single orbital? A. True
2. The maximum number of electrons allowed in a p sublevel of the 3rd principal level is?
B.6
3. A neutral atom has a ground state electronic configuration of 1s^2 2s^2. Which of the following statements concerning this atom is/are correct?
B. All of the above.
