The gravitational potential energy
gpe = mgh
![\tt gpe=75\times 9.8\times880=646,800~J](https://tex.z-dn.net/?f=%5Ctt%20gpe%3D75%5Ctimes%209.8%5Ctimes880%3D646%2C800~J)
Juan lives a hundred miles away from Bill then the average speed that he reaches Bill's Home in 50 seconds means that Juan lives 50 seconds away from bill because 50 + 50 seconds equals 100 seconds so Juansorry my bad Juan lives 50 miles away from Belle but he probably how I must have run to get to bill in 50 seconds .
Answer: D(t)= 50(4/5)^t
Explanation: If 1/5 of the temperature difference is lost each minute, that means 4/5 of the difference remains each minute. So each minute, the temperature difference is multiplied by a factor of 4/5 (or 0.8).
If we start with the initial temperature difference, 50° Celsius, and keep multiplying by 4/5, this function gives us the temperature difference t minutes after the cake was put in the cooler.
Explanation :
(1) Big bang : (1) The most accepted theory on the origin of the universe.
This theory shows the expanding of the universe from high density and high-temperature states.
(2) Steady state : (3) All is the same and will always stay the same.
Steady state means that the properties of any system remain the same always.
(3) Oscillating universe : (4) Agrees with the big bang theory but insists the universe expanded much quicker.
Oscillating universe theory is the result of big bang theory.
(4) Inflation Choices : (2) it's like an inflating and deflating balloon that never stops.
In cosmology, cosmic inflation or deflation is just the expanding and contraction of the universe.
So, the statements and the choices are related as:
(1)-(1)
(2)- (3)
(3)-(4)
(4)-(2)
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.