First, we must find the vertical distance traveled upwards by the ball due to the throw. For this, we will use the formula:
2as = v² - u²
Because the final velocity v is 0 in such cases
s = -u²/2a; because both u and a are downwards, the negative sign cancels
s = 14.5² / 2*9.81
s = 10.72 meters
Next, to find the time taken to reach the ground, we need the height above the ground. This is:
45 + 10.72 = 55.72 m
We will use the formula
s = ut + 0.5at²
to find the time taken with the initial velocity u = 0.
55.72 = 0.5 * 9.81 * t²
t = 3.37 seconds
Answer:
Explanation:
Hello,
In this case, since the acceleration in terms of position is defined as its second derivative:
The purpose here is derive x(t) twice as follows:
Thus, the acceleration turns out 4.8 meters per squared seconds.
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Surface waves from an earthquake shake the ground back and forth and up and down. So basically in a circular motion. This wave is the most dangerous wave released during and earthquake and it comes after the p and s waves. The Surface wave is the last wave that comes after the other 2. So yea' these surface waves move in a circular motion.
Scientists review the work of other scientists to validate the results and check the evidence's reliability. The correct option among all the options that are given in the question is the fourth option or the last option or option "D". I hope that this is the answer that has come to your desired help.
The "meters" measurements refer to the diameter of the objective (the main big-moose lens or mirror).
But light collection depends on the AREA of the objective, and that means the SQUARE of the diameter.
So if you double the 'size' if the telescope, it gathers FOUR TIMES as much light as before.