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.
We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
f' = frequency observed by the police car after sound reflected from the vehicle and comes back to police car = 1250 Hz
f = frequency emitted by the police car = 1200 Hz
V = speed of sound = 340 m/s
v = speed of vehicle = ?
frequency observed by the police car is given as
f' = f (V + v)/(V - v)
inserting the values in the above equation
1250 = 1200 (340 + v)/(340 - v)
v = 6.9 m/s
Answer:
C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂
Explanation:
You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.
Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.
Stone 1
y₁ = v₀₁ t + ½ g t²
y₁ = 0 + ½ g t²
Rock2
It comes out a little later, let's say a second later, we can use the same stopwatch
t ’= (t-t₀)
y₂ = v₀₂ t ’+ ½ g t’²
y₂ = 0 + ½ g (t-t₀)²
y₂ = + ½ g (t-t₀)²
Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to
S = y₁ -y₂
S = ½ g t²– ½ g (t-t₀)²
S = ½ g [t² - (t²- 2 t to + to²)]
S = ½ g (2 t t₀ - t₀²)
S = ½ g t₀ (2 t -t₀)
This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.
For t <to. The rock y has not left and the distance increases
For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time
passes
Now we can analyze the different statements
A) false. The difference in height increases over time
B) False S increases
C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂
