Speed of the car given initially
v = 18 m/s
deceleration of the car after applying brakes will be
a = 3.35 m/s^2
Reaction time of the driver = 0.200 s
Now when he see the red light distance covered by the till he start pressing the brakes
![d_1 = v* t](https://tex.z-dn.net/?f=d_1%20%3D%20v%2A%20t)
![d_1 = 18* 0.200 = 3.6 m](https://tex.z-dn.net/?f=d_1%20%3D%2018%2A%200.200%20%3D%203.6%20m)
Now after applying brakes the distance covered by the car before it stops is given by kinematics equation
![v_f^2 - v_i^2 = 2 a s](https://tex.z-dn.net/?f=v_f%5E2%20-%20v_i%5E2%20%3D%202%20a%20s)
here
vi = 18 m/s
vf = 0
a = - 3.35
so now we will have
![0^2 - 18^2 = 2*(-3.35)(s)](https://tex.z-dn.net/?f=0%5E2%20-%2018%5E2%20%3D%202%2A%28-3.35%29%28s%29)
![s = 48.35 m](https://tex.z-dn.net/?f=%20s%20%3D%2048.35%20m)
So total distance after which car will stop is
![d = d_1 + d_2](https://tex.z-dn.net/?f=%20d%20%3D%20d_1%20%2B%20d_2)
![d = 48.35 + 3.6 = 51.95 m](https://tex.z-dn.net/?f=%20d%20%3D%2048.35%20%2B%203.6%20%3D%2051.95%20m)
So car will not stop before the intersection as it is at distance 20 m
To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.
By definition the energy balance is simply given by the change between the two states:
![|\Delta E_{ij}| = |E_i-E_j|](https://tex.z-dn.net/?f=%7C%5CDelta%20E_%7Bij%7D%7C%20%3D%20%7CE_i-E_j%7C)
Our states are given by
![E_1 = -11.7eV](https://tex.z-dn.net/?f=E_1%20%3D%20-11.7eV)
![E_2 = -4.2eV](https://tex.z-dn.net/?f=E_2%20%3D%20-4.2eV)
![E_3 = -3.3eV](https://tex.z-dn.net/?f=E_3%20%3D%20-3.3eV)
In this way the energy balance for the states would be given by,
![|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\](https://tex.z-dn.net/?f=%7C%5CDelta%20E_%7B12%7D%7C%20%3D%20%7CE_1-E_2%7C%5C%5C%7C%5CDelta%20E_%7B12%7D%7C%20%3D%20%7C-11.7-%28-4.2%29%7C%5C%5C%7C%5CDelta%20E_%7B12%7D%7C%20%3D%207.5eV%5C%5C)
![|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV](https://tex.z-dn.net/?f=%7C%5CDelta%20E_%7B13%7D%7C%20%3D%20%7CE_1-E_3%7C%5C%5C%7C%5CDelta%20E_%7B13%7D%7C%20%3D%20%7C-11.7-%28-3.3%29%7C%5C%5C%7C%5CDelta%20E_%7B13%7D%7C%20%3D%208.4eV)
![|\Delta E_{23}| = |E_2-E_3|\\|\Delta E_{23}| = |-4.2-(-3.3)|\\|\Delta E_{23}| = 0.9eV](https://tex.z-dn.net/?f=%7C%5CDelta%20E_%7B23%7D%7C%20%3D%20%7CE_2-E_3%7C%5C%5C%7C%5CDelta%20E_%7B23%7D%7C%20%3D%20%7C-4.2-%28-3.3%29%7C%5C%5C%7C%5CDelta%20E_%7B23%7D%7C%20%3D%200.9eV)
Therefore the states of energy would be
Lowest : 0.9eV
Middle :7.5eV
Highest: 8.4eV
Answer:
776.6 w
1.04 hp
Explanation:
given:
Mass, m = 190kg
height change, h = 25m
time elapsed, t = 60 s
acceleration due to gravity, g = 9.81 m/s²
Potential energy required raising 190 kg of water to a height of 25m
= mgh
= 190 x 9.81 x 25
= 46,597.5 J
Power required in 60 s
= Energy required ÷ time elapsed
= 46,597.5 ÷ 60
= 776.6 Watts (Use conversion 1 W = 0.00134102 hp)
= 776.6 w x 0.00134102 hp/w
= 1.04 hp
<u>Answer:</u>
2N/cm
<u>Step-by-step explanation:</u>
According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:
![F=kx](https://tex.z-dn.net/?f=F%3Dkx)
where,
is the force which is stretching or compressing the spring,
is the spring constant; and
is the distance the spring is stretched.
Substituting the given values to find the elastic constant
to get:
![F=kx](https://tex.z-dn.net/?f=F%3Dkx)
![4=k(2)](https://tex.z-dn.net/?f=4%3Dk%282%29)
![k=\frac{4}{2}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B4%7D%7B2%7D)
![k=2](https://tex.z-dn.net/?f=k%3D2)
Therefore, the elastic constant is 2 Newton/cm.
To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.
The linear mass density is given as,
![\mu = \frac{m}{l}](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7Bm%7D%7Bl%7D)
![\mu = \frac{8.79*10^{-3}}{70*10^{-2}}](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7B8.79%2A10%5E%7B-3%7D%7D%7B70%2A10%5E%7B-2%7D%7D)
![\mu = 0.01255kg/m](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.01255kg%2Fm)
The expression for the wavelength of the standing wave for the second overtone is
![\lambda = \frac{2}{3} l](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7B2%7D%7B3%7D%20l)
Replacing we have
![\lambda = \frac{2}{3} (70*10^{-2})](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7B2%7D%7B3%7D%20%2870%2A10%5E%7B-2%7D%29)
![\lambda = 0.466m](https://tex.z-dn.net/?f=%5Clambda%20%3D%200.466m)
The frequency of the sound wave is
![f_s = \frac{v}{\lambda_s}](https://tex.z-dn.net/?f=f_s%20%3D%20%5Cfrac%7Bv%7D%7B%5Clambda_s%7D)
![f_s = \frac{344}{0.768}](https://tex.z-dn.net/?f=f_s%20%3D%20%5Cfrac%7B344%7D%7B0.768%7D)
![f_s = 448Hz](https://tex.z-dn.net/?f=f_s%20%3D%20448Hz)
Now the velocity of the wave would be
![v = f_s \lambda](https://tex.z-dn.net/?f=v%20%3D%20f_s%20%5Clambda)
![v = (448)(0.466)](https://tex.z-dn.net/?f=v%20%3D%20%28448%29%280.466%29)
![v = 208.768m/s](https://tex.z-dn.net/?f=v%20%3D%20208.768m%2Fs)
The expression that relates the velocity of the wave, tension on the string and linear mass density is
![v = \sqrt{\frac{T}{\mu}}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cmu%7D%7D)
![v^2 = \frac{T}{\mu}](https://tex.z-dn.net/?f=v%5E2%20%3D%20%5Cfrac%7BT%7D%7B%5Cmu%7D)
![T= \mu v^2](https://tex.z-dn.net/?f=T%3D%20%5Cmu%20v%5E2)
![T = (0.01255kg/m)(208.768m/s)^2](https://tex.z-dn.net/?f=T%20%3D%20%280.01255kg%2Fm%29%28208.768m%2Fs%29%5E2)
![T = 547N](https://tex.z-dn.net/?f=T%20%3D%20547N)
The tension in the string is 547N
PART B) The relation between the fundamental frequency and the
harmonic frequency is
![f_n = nf_1](https://tex.z-dn.net/?f=f_n%20%3D%20nf_1)
Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore
![n=3](https://tex.z-dn.net/?f=n%3D3)
Then,
![f_3 = 3f_1](https://tex.z-dn.net/?f=f_3%20%3D%203f_1)
Rearranging to find the fundamental frequency
![f_1 = \frac{f_3}{3}](https://tex.z-dn.net/?f=f_1%20%3D%20%5Cfrac%7Bf_3%7D%7B3%7D)
![f_1 = \frac{448Hz}{3}](https://tex.z-dn.net/?f=f_1%20%3D%20%5Cfrac%7B448Hz%7D%7B3%7D)
![f_1 = 149.9Hz](https://tex.z-dn.net/?f=f_1%20%3D%20149.9Hz)