The answer is d I took the test
Answer:B
Explanation:Grote reber was the first scientist to map the milky way galaxy using radio waves.
The energy in ordinary light is greater than the energy in ordinary sound
Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc. Additionally, there is heat and work, which is energy being transferred from one body to another.
Energy is a physical system's ability to perform labor. The capital letter E is a typical sign for energy. The joule, denoted by the letter J, is the common unit. The energy produced by one newton's (1 N) worth of force acting over one meter's (1 m) worth of displacement is measured in joules (1 J). Because it is a fundamental human requirement, energy plays a significant role in our daily lives.
To learn more about Energy please visit-
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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.
Wrap around a metal with wire instead of using wire alone.