Would presume you are asked to find the volume, since there is no second volume.
By General Gas Law:
P₁V₁/T₁ = P₂V₂/T₂
1.6 * 168 /255 = 1.3*V₂/285
V₂ = 1.6 * 168 * 285 / (1.3*255)
V₂ = 231.095
Final volume ≈ 231 cm³
Answer:
3100.05 N
Explanation:
= Force on piston = 45 N
= Radius of piston
= Radius of plunger
![\dfrac{r_2}{r_1}=8.3](https://tex.z-dn.net/?f=%5Cdfrac%7Br_2%7D%7Br_1%7D%3D8.3)
From Pascal's law we have
![\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}\\\Rightarrow F_2=\dfrac{F_1A_2}{A_1}\\\Rightarrow F_2=\dfrac{F_1\pi r_2^2}{\pi r_1^2}\\\Rightarrow F_2=\dfrac{F_1r_2^2}{r_1^2}\\\Rightarrow F_2=F_1\dfrac{r_2^2}{r_1^2}\\\Rightarrow F_2=45\times 8.3^2\\\Rightarrow F_2=3100.05\ N](https://tex.z-dn.net/?f=%5Cdfrac%7BF_1%7D%7BA_1%7D%3D%5Cdfrac%7BF_2%7D%7BA_2%7D%5C%5C%5CRightarrow%20F_2%3D%5Cdfrac%7BF_1A_2%7D%7BA_1%7D%5C%5C%5CRightarrow%20F_2%3D%5Cdfrac%7BF_1%5Cpi%20r_2%5E2%7D%7B%5Cpi%20r_1%5E2%7D%5C%5C%5CRightarrow%20F_2%3D%5Cdfrac%7BF_1r_2%5E2%7D%7Br_1%5E2%7D%5C%5C%5CRightarrow%20F_2%3DF_1%5Cdfrac%7Br_2%5E2%7D%7Br_1%5E2%7D%5C%5C%5CRightarrow%20F_2%3D45%5Ctimes%208.3%5E2%5C%5C%5CRightarrow%20F_2%3D3100.05%5C%20N)
The force is 3100.05 N
the answer is 5000, multiply 500 by 10.
Explanation:
Given that,
Initial speed of the car, u = 88 km/h = 24.44 m/s
Reaction time, t = 2 s
Distance covered during this time, ![d=24.44\times 2=48.88\ m](https://tex.z-dn.net/?f=d%3D24.44%5Ctimes%202%3D48.88%5C%20m)
(a) Acceleration, ![a=-4\ m/s^2](https://tex.z-dn.net/?f=a%3D-4%5C%20m%2Fs%5E2)
We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :
![s=\dfrac{v^2-u^2}{2a}](https://tex.z-dn.net/?f=s%3D%5Cdfrac%7Bv%5E2-u%5E2%7D%7B2a%7D)
![s=\dfrac{-(24.44)^2}{2\times -4}](https://tex.z-dn.net/?f=s%3D%5Cdfrac%7B-%2824.44%29%5E2%7D%7B2%5Ctimes%20-4%7D)
s = 74.66 meters
s = 74.66 + 48.88 = 123.54 meters
(b) Acceleration, ![a=-8\ m/s^2](https://tex.z-dn.net/?f=a%3D-8%5C%20m%2Fs%5E2)
![s=\dfrac{v^2-u^2}{2a}](https://tex.z-dn.net/?f=s%3D%5Cdfrac%7Bv%5E2-u%5E2%7D%7B2a%7D)
![s=\dfrac{-(24.44)^2}{2\times -8}](https://tex.z-dn.net/?f=s%3D%5Cdfrac%7B-%2824.44%29%5E2%7D%7B2%5Ctimes%20-8%7D)
s = 37.33 meters
s = 37.33 + 48.88 = 86.21 meters
Hence, this is the required solution.
Answer:
Part a)
![E = 3.66 eV](https://tex.z-dn.net/?f=E%20%3D%203.66%20eV)
Part b)
![\lambda = 508.5 nm](https://tex.z-dn.net/?f=%5Clambda%20%3D%20508.5%20nm)
Explanation:
Part a)
change in the energy due to decay of photon is given as
![E = h\nu](https://tex.z-dn.net/?f=E%20%3D%20h%5Cnu)
here we know that
![\nu = 8.88 \times 10^{14} Hz](https://tex.z-dn.net/?f=%5Cnu%20%3D%208.88%20%5Ctimes%2010%5E%7B14%7D%20Hz)
now we have
![E = (6.6 \times 10^{-34})(8.88 \times 10^{14})](https://tex.z-dn.net/?f=E%20%3D%20%286.6%20%5Ctimes%2010%5E%7B-34%7D%29%288.88%20%5Ctimes%2010%5E%7B14%7D%29)
![E = 5.86 \times 10^{-19} J](https://tex.z-dn.net/?f=E%20%3D%205.86%20%5Ctimes%2010%5E%7B-19%7D%20J)
![E = 3.66 eV](https://tex.z-dn.net/?f=E%20%3D%203.66%20eV)
Part b)
While electron return to its ground state it will emit a photon of energy 2/3rd of the total energy
so we have
![\Delta E = \frac{2}{3}(3.66 eV)](https://tex.z-dn.net/?f=%5CDelta%20E%20%3D%20%5Cfrac%7B2%7D%7B3%7D%283.66%20eV%29)
![\Delta E = 2.44 eV](https://tex.z-dn.net/?f=%5CDelta%20E%20%3D%202.44%20eV)
now to find the wavelength we have
![\Delta E = \frac{hc}{\lambda}](https://tex.z-dn.net/?f=%5CDelta%20E%20%3D%20%5Cfrac%7Bhc%7D%7B%5Clambda%7D)
![2.44 = \frac{1242}{\lambda}](https://tex.z-dn.net/?f=2.44%20%3D%20%5Cfrac%7B1242%7D%7B%5Clambda%7D)
![\lambda = 508.5 nm](https://tex.z-dn.net/?f=%5Clambda%20%3D%20508.5%20nm)