Answer: the top one
Explanation: hope this helps
Answer:
Average force = 67 mn
Explanation:
Given:
Initial velocity u = 0 m/s
Final velocity v = 67 m/s
Time t = 1 ms = 0.001 sec.
Computation:
Using Momentum theory
Change in momentum = F × Δt
(v-u)/t = F × Δt
F × 0.001 = (67 - 0)/0.001
F= 67,000,000
Average force = 67 mn
Let's cut through the weeds and the trash
and get down to the real situation:
A stone is tossed straight up at 5.89 m/s .
Ignore air resistance.
Gravity slows down the speed of any rising object by 9.8 m/s every second.
So the stone (aka Billy-Bob-Joe) continues to rise for
(5.89 m/s / 9.8 m/s²) = 0.6 seconds.
At that timer, he has run out of upward gas. He is at the top
of his rise, he stops rising, and begins to fall.
His average speed on the way up is (1/2) (5.89 + 0) = 2.945 m/s .
Moving for 0.6 seconds at an average speed of 2.945 m/s,
he topped out at
(2.945 m/s) (0.6 s) = 1.767 meters above the trampoline.
With no other forces other than gravity acting on him, it takes him
the same time to come down from the peak as it took to rise to it.
(0.6 sec up) + (0.6 sec down) = 1.2 seconds until he hits rubber again.
B: heat is transferred as thermal energy by the interaction of moving particles
Answer:
Explanation:
A function f(x) is a Probability Density Function if it satisfies the following conditions:
Given the function:
(1)p(x) is greater than zero since the range of exponents of the Euler's number will lie in 
(2)
![\int_{0}^{\infty} p(x)=\int_{0}^{\infty} \dfrac{1}{r}e^{-x/r}\\=\dfrac{1}{r} \int_{0}^{\infty} e^{-x/r}\\=-\dfrac{r}{r}\left[e^{-x/r}\right]_{0}^{\infty}\\=-\left[e^{-\infty/r}-e^{-0/r}\right]\\=-e^{-\infty}+e^{-0}\\SInce \: e^{-\infty} \rightarrow 0\\e^{-0}=1\\\int_{0}^{\infty} p(x)=1](https://tex.z-dn.net/?f=%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20p%28x%29%3D%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20%5Cdfrac%7B1%7D%7Br%7De%5E%7B-x%2Fr%7D%5C%5C%3D%5Cdfrac%7B1%7D%7Br%7D%20%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20e%5E%7B-x%2Fr%7D%5C%5C%3D-%5Cdfrac%7Br%7D%7Br%7D%5Cleft%5Be%5E%7B-x%2Fr%7D%5Cright%5D_%7B0%7D%5E%7B%5Cinfty%7D%5C%5C%3D-%5Cleft%5Be%5E%7B-%5Cinfty%2Fr%7D-e%5E%7B-0%2Fr%7D%5Cright%5D%5C%5C%3D-e%5E%7B-%5Cinfty%7D%2Be%5E%7B-0%7D%5C%5CSInce%20%5C%3A%20e%5E%7B-%5Cinfty%7D%20%5Crightarrow%200%5C%5Ce%5E%7B-0%7D%3D1%5C%5C%5Cint_%7B0%7D%5E%7B%5Cinfty%7D%20p%28x%29%3D1)
The function p(x) satisfies the conditions for a probability density function.
