Answer:
The ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Explanation:
In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).
We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.
So, first we need to determine the components of the velocity of the ball, like this:
![V_{Ax}=V_{A}cos\theta](https://tex.z-dn.net/?f=V_%7BAx%7D%3DV_%7BA%7Dcos%5Ctheta)
![V_{Ax}=(21m/s)cos(-14^{o})](https://tex.z-dn.net/?f=V_%7BAx%7D%3D%2821m%2Fs%29cos%28-14%5E%7Bo%7D%29)
![V_{Ax}=20.38 m/s](https://tex.z-dn.net/?f=V_%7BAx%7D%3D20.38%20m%2Fs)
![V_{Ay}=V_{A}sin\theta](https://tex.z-dn.net/?f=V_%7BAy%7D%3DV_%7BA%7Dsin%5Ctheta)
![V_{Ay}=(21m/s)sin(-14^{o})](https://tex.z-dn.net/?f=V_%7BAy%7D%3D%2821m%2Fs%29sin%28-14%5E%7Bo%7D%29)
![V_{Ay}=-5.08 m/s](https://tex.z-dn.net/?f=V_%7BAy%7D%3D-5.08%20m%2Fs)
we pick the positive one, so it takes 0.317s for the ball to hit on point A.
so now we can find the distance from the net to point A with this time. We can find it like this:
![x_{A}=V_{Ax}t](https://tex.z-dn.net/?f=x_%7BA%7D%3DV_%7BAx%7Dt)
![x_{A}=(20.38m/s)(0.317s)](https://tex.z-dn.net/?f=x_%7BA%7D%3D%2820.38m%2Fs%29%280.317s%29)
![x_{A}=6.46m](https://tex.z-dn.net/?f=x_%7BA%7D%3D6.46m%20)
Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:
![V_{Bx}=20.88 m/s](https://tex.z-dn.net/?f=V_%7BBx%7D%3D20.88%20m%2Fs)
![V_{By}=-2.195 m/s](https://tex.z-dn.net/?f=V_%7BBy%7D%3D-2.195%20m%2Fs)
![y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}](https://tex.z-dn.net/?f=y_%7BBf%7D%3Dy_%7BB0%7D%2BV_%7B0%7Dt-%5Cfrac%7B1%7D%7B2%7Dat%5E%7B2%7D)
![0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}](https://tex.z-dn.net/?f=0%3D2.1m%2B%28-2.195m%2Fs%29t-%5Cfrac%7B1%7D%7B2%7D%28-9.8m%2Fs%5E%7B2%7D%29t%5E%7B2%7D)
![-4.9t^{2}-2.195t+2.1=0](https://tex.z-dn.net/?f=-4.9t%5E%7B2%7D-2.195t%2B2.1%3D0)
![t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-b%5Cpm%5Csqrt%7Bb%5E%7B2%7D-4ac%7D%7D%7B2a%7D)
![t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-%28-2.195%29%5Cpm%5Csqrt%7B%28-2.195%29%5E%7B2%7D-4%28-4.9%29%282.1%29%7D%7D%7B2%28-4.9%29%7D)
t= -0.9159s or t=0.468s
we pick the positive one, so it takes 0.468s for the ball to hit on point B.
so now we can find the distance from the net to point B with this time. We can find it like this:
![x_{B}=V_{Bx}t](https://tex.z-dn.net/?f=x_%7BB%7D%3DV_%7BBx%7Dt)
![x_{B}=(20.88m/s)(0.468s)](https://tex.z-dn.net/?f=x_%7BB%7D%3D%2820.88m%2Fs%29%280.468s%29)
![x_{B}=9.77m](https://tex.z-dn.net/?f=x_%7BB%7D%3D9.77m)
So once we got the two distances we can now find the difference between them:
![x_{B}-x_{A}=9.77m-6.46m=3.31m](https://tex.z-dn.net/?f=x_%7BB%7D-x_%7BA%7D%3D9.77m-6.46m%3D3.31m)
so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.
<h2>First stage of sleep Deprivation Subject </h2>
During the first stage of sleep deprivation, the subject is NREM which stands for non-rapid eye movement. In this condition, we are not sleeping in the depth. It can be said as dreamless sleep. On electro-encephalography recording, the brain waves are not fast so they have high voltage.
In this condition, the breathing, heart rate and blood pressure is low. The sleep is comparatively tranquil. NREM lasts for 90 minutes to 120 minutes. It accounts for about 75% of the normal sleep time. Rapid eye movements do not occur.
Answer:
a) this stone was first called Schorl stone
b)the first modern name for this mineral was Black tourmaline
Explanation:
hope this helps you!
:)
Answer: small cars can stop and go fast big trucks can not
Explanation: