Answer:
v = (78.0 i ^ - 70.27 j ^) m/s, v = 105 m / s
, θ = 318º
Explanation:
We have a projectile launch problem, let's start by calculating the time it takes to get through the canyon
y =
t - ½ gt2
As the motorcyclist comes out horizontally, the speed he has is the horizontal speed (vox) and the initial vertical speed is zero (I go = 0)
y = 0 - ½ g t2
t = √ 2y / g
t = √ (2 252 /9.8)
t = 7.17 s
Let's calculate the vertical speed for this time
=
- gt
= 0 - gt
= - 9.8 7.17
= - 70.27 m / s
We can give the result in two ways
First:
v = (78.0 i ^ - 70.27 j ^) m / s
Second:
using the Pythagorean theorem and trigonometry
v² = vₓ² +
²
v = √ [(78.0)² + (-70.42)²] = √ (11042.98)
v = 105 m / s
tan θ₁ =
y / vₓ
tan θ₁ = -70.42 / 78.0
θ₁ = 42º
If we measure this angle from the positive direction of the x-axis counterclockwise
θ = 360 - θ₁
θ = 360 - 42
θ = 318º
Answer:
The period of the harmonic vibration is inversely proportional to its frequency
Explanation:
The period of the simple harmonic vibration is actually the inverse of its frequency.
This means that, to get the period, all you have to do is divide "1" by the frequency, in other words:
period = 1/frequency
From this formula, we can note that as the frequency increases, the period decreases and vice versa.
Based on this, the relation between the period and the frequency is an inversely proportional relation.
Hope this helps :)
Answer:
R' = R/4
Explanation:
The resistance of a metal rod is R. It is given by the relation as follows :
![R=\rho\dfrac{l}{A}](https://tex.z-dn.net/?f=R%3D%5Crho%5Cdfrac%7Bl%7D%7BA%7D)
Where
l is the length and A is the area of cross-section
![A=\pi r^2=\pi (\dfrac{d}{2})^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%3D%5Cpi%20%28%5Cdfrac%7Bd%7D%7B2%7D%29%5E2)
If both its length and its diameter are quadrupled, it means,
l' = 4l
and d'= 4d
It means,
![A'=\pi (\dfrac{4d}{2})^2](https://tex.z-dn.net/?f=A%27%3D%5Cpi%20%28%5Cdfrac%7B4d%7D%7B2%7D%29%5E2)
Let new resistance be R'. So,
![R'=\rho\dfrac{l'}{A'}\\\\R'=\rho\dfrac{4l}{\pi (\dfrac{4d}{2})^2}\\\\=\rho \dfrac{4l}{\pi \dfrac{16d^2}{2}}\\\\=\dfrac{4}{16}\times \dfrac{\rho l}{\pi \dfrac{d^2}{2}}\\\\=\dfrac{1}{4}\times \dfrac{\rho l}{\pi \dfrac{d^2}{2}}\\\\R'=\dfrac{R}{4}](https://tex.z-dn.net/?f=R%27%3D%5Crho%5Cdfrac%7Bl%27%7D%7BA%27%7D%5C%5C%5C%5CR%27%3D%5Crho%5Cdfrac%7B4l%7D%7B%5Cpi%20%28%5Cdfrac%7B4d%7D%7B2%7D%29%5E2%7D%5C%5C%5C%5C%3D%5Crho%20%5Cdfrac%7B4l%7D%7B%5Cpi%20%5Cdfrac%7B16d%5E2%7D%7B2%7D%7D%5C%5C%5C%5C%3D%5Cdfrac%7B4%7D%7B16%7D%5Ctimes%20%5Cdfrac%7B%5Crho%20l%7D%7B%5Cpi%20%5Cdfrac%7Bd%5E2%7D%7B2%7D%7D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cdfrac%7B%5Crho%20l%7D%7B%5Cpi%20%5Cdfrac%7Bd%5E2%7D%7B2%7D%7D%5C%5C%5C%5CR%27%3D%5Cdfrac%7BR%7D%7B4%7D)
So, the correct option is (B) "R/4".
The answer for this would be A. since power is Joules/seconds and energy is rated in Joules