You can use the impulse momentum theorem and just subtract the two momenta.
P1 - P2 = (16-1.2)(11.5e4)=1702000Ns
If you first worked out the force and integrated it over time the result is the same
Answer:0.061
Explanation:
Given
Temperature of soup 
heat capacity of soup 
Here Temperature of soup is constantly decreasing
suppose T is the temperature of soup at any instant
efficiency is given by
integrating From 
to 
![W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}](https://tex.z-dn.net/?f=W%3Dc_v%5Cleft%20%5B%20T-T_C%5Cln%20T%5Cright%20%5D_%7BT_H%7D%5E%7BT_C%7D)
![W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]](https://tex.z-dn.net/?f=W%3Dc_v%5Cleft%20%5B%20%5Cleft%20%28%20T_C-T_H%5Cright%20%29-T_C%5Cleft%20%28%20%5Cln%20%5Cfrac%7BT_C%7D%7BT_H%7D%5Cright%20%29%5Cright%20%5D)
Now heat lost by soup is given by
Fraction of the total heat that is lost by the soup can be turned is given by
![=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bc_v%5Cleft%20%5B%20%5Cleft%20%28%20T_C-T_H%5Cright%20%29-T_C%5Cleft%20%28%20%5Cln%20%5Cfrac%7BT_C%7D%7BT_H%7D%5Cright%20%29%5Cright%20%5D%7D%7Bc_v%28T_C-T_H%29%7D)
The banking angle of the curved part of the speedway is determined as 32⁰.
<h3>
Banking angle of the curved road</h3>
The banking angle of the curved part of the speedway is calculated as follows;
V(max) = √(rg tanθ)
where;
- r is radius of the path
- g is acceleration due to gravity
V² = rg tanθ
tanθ = V²/rg
tanθ = (34²)/(190 x 9.8)
tanθ = 0.62
θ = arc tan(0.62)
θ = 31.8
θ ≈ 32⁰
Learn more about banking angle here: brainly.com/question/8169892
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The data convincingly show that wave frequency does not affect wave speed. An increase in wave frequency caused a decrease in wavelength while the wave speed remained constant. The last three trials involved the same procedure with a different rope tension.
Answer:
x sin nx = x cos nx
same as
theta / theta x (xsin (nx)) = sin (nx) + (nx) cos (nx)
