Answer:
A) 21.2 kg.m/s at 39.5 degrees from the x-axis
Explanation:
Mass of the smaller piece = 200g = 200/1000 = 0.2 kg
Mass of the bigger piece = 300g = 300/1000 = 0.3 kg
Velocity of the small piece = 82 m/s
Velocity of the bigger piece = 45 m/s
Final momentum of smaller piece = 0.2 × 82 = 16.4 kg.m/s
Final momentum of bigger piece = 0.3 × 45 = 13.5 kg.m/s
since they acted at 90oc to each other (x and y axis) and also momentum is vector quantity; then we can use Pythagoras theorems
Resultant momentum² = 16.4² + 13.5² = 451.21
Resultant momentum = √451.21 = 21.2 kg.m/s at angle 39.5 degrees to the x-axis ( tan^-1 (13.5 / 16.4)
Answer:
The correct answer is - true.
Explanation:
Changes in the higher authorities or managers or coworkers can be a warning sign before a layoff, however, a company or organization needs to give an official notice before a particular time period before the layoff.
The layoff is a temporary suspension or permanent termination from the employment of an employee due to reasons related to business such as reducing manpower or staff and less work in the organization. These reasons can lead to a change in the behavior of the managers, coworkers, and subordinates.
Answer:
acceleration of the rocket is given as
![a = 7.45 m/s^2](https://tex.z-dn.net/?f=a%20%3D%207.45%20m%2Fs%5E2)
Explanation:
As we know that rocket starts from rest and then reach to final speed of 447 m/s after t = 1 min
so we have
![v_i = 0](https://tex.z-dn.net/?f=v_i%20%3D%200)
![v_f = 447 m/s](https://tex.z-dn.net/?f=v_f%20%3D%20447%20m%2Fs)
![t = 1 min = 60 s](https://tex.z-dn.net/?f=t%20%3D%201%20min%20%3D%2060%20s)
so we have
![a = \frac{v_f - v_i}{t}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bv_f%20-%20v_i%7D%7Bt%7D)
![a = \frac{447 - 0}{60}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B447%20-%200%7D%7B60%7D)
![a = 7.45 m/s^2](https://tex.z-dn.net/?f=a%20%3D%207.45%20m%2Fs%5E2)
Changes. :) I think... Whats your question
?
Answer:
a ) 11.1 *10^3 m/s = 39.96 Km/h
b) T_{o2} =1.58*10^5 K
Explanation:
a)
= 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h
b)
M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol
gas constant R = 8.31 j/mol.K
![v_{rms} = \sqrt{ \frac{3RT}{M}}](https://tex.z-dn.net/?f=v_%7Brms%7D%20%3D%20%5Csqrt%7B%20%5Cfrac%7B3RT%7D%7BM%7D%7D)
So, ![v_{rms,o2} =\sqrt{ \frac{3RT_{o2}}{M_{o2}}}](https://tex.z-dn.net/?f=v_%7Brms%2Co2%7D%20%3D%5Csqrt%7B%20%5Cfrac%7B3RT_%7Bo2%7D%7D%7BM_%7Bo2%7D%7D%7D)
multiply each side by M_{o2}, so we have
![v_{rms,o2}^2 *M_{o2} =3RT_{o2}](https://tex.z-dn.net/?f=v_%7Brms%2Co2%7D%5E2%20%2AM_%7Bo2%7D%20%3D3RT_%7Bo2%7D)
solving for temperature T_{o2}
![T_{o2} = \frac{v_{rms,o2}^2 *M_{o2}}{3R}](https://tex.z-dn.net/?f=T_%7Bo2%7D%20%3D%20%5Cfrac%7Bv_%7Brms%2Co2%7D%5E2%20%2AM_%7Bo2%7D%7D%7B3R%7D)
In the question given,![v_{rms} =v_{es}](https://tex.z-dn.net/?f=v_%7Brms%7D%20%3Dv_%7Bes%7D)
![T_{o2} = \frac{(11.1*10^3)^2 *32.0*10^{-3}}{3*8.31}](https://tex.z-dn.net/?f=T_%7Bo2%7D%20%3D%20%5Cfrac%7B%2811.1%2A10%5E3%29%5E2%20%2A32.0%2A10%5E%7B-3%7D%7D%7B3%2A8.31%7D)
T_{o2} =1.58*10^5 K