Momentum = mass • velocity
v= 17.5/2.5
= 7 m/s
Hello, love! The answer is True, or T, on Edge2020.
Hope this helped!
~ V.
Answer:
The force of friction that acts on him is
![F_k=567N](https://tex.z-dn.net/?f=F_k%3D567N)
Explanation:
The firefighter with an acceleration of 3m/s^2 take the gravity acceleration as 10m/s^2 isn't necessary to know the coefficient of friction just to know the force of friction:
![F=m*a](https://tex.z-dn.net/?f=F%3Dm%2Aa)
![F=F_w-F_k](https://tex.z-dn.net/?f=F%3DF_w-F_k)
![m*a=F_w-F_k](https://tex.z-dn.net/?f=m%2Aa%3DF_w-F_k)
![F_w=81kg*10m/s^2=810N](https://tex.z-dn.net/?f=F_w%3D81kg%2A10m%2Fs%5E2%3D810N)
Sole to Fk
![81kg*3m/s^2=810N-F_k](https://tex.z-dn.net/?f=81kg%2A3m%2Fs%5E2%3D810N-F_k)
![F_k=810N-243N](https://tex.z-dn.net/?f=F_k%3D810N-243N)
![F_k=567N](https://tex.z-dn.net/?f=F_k%3D567N)
The answer is A) 1000 J
According to the law of conservation of energy, the total energy before an action is always equal to the total energy after the action.
So that is, the total energy is 8000J found as potential energy, 7000J has transformed into kinetic energy, then the thermal energy should be the remaining 1000J.
Hope this helps.
Answer:
1.034m/s
Explanation:
We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,
![m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s](https://tex.z-dn.net/?f=m_1%20%3D%2065000kg%5C%5Cv_1%20%3D%200.8m%2Fs%5C%5Cm_2%20%3D%2092000kg%5C%5Cv_2%20%3D%201.2m%2Fs)
<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,
![V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}](https://tex.z-dn.net/?f=V_%7Bcm%7D%20%3D%20%5Cfrac%7Bm_1v_1%2Bm_2v_2%7D%7Bm_1%2Bm_2%7D)
Substituting,
![V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}](https://tex.z-dn.net/?f=V_%7Bcm%7D%20%3D%20%5Cfrac%7B%2865000%2A0.8%29%2B%2892000%2A1.2%29%7D%7B92000%2B65000%7D)
![V_{cm} = 1.034m/s](https://tex.z-dn.net/?f=V_%7Bcm%7D%20%3D%201.034m%2Fs)
Part B)
For the Part B we need to apply conserving momentum equation, this formula is given by,
![m_1v_1+m_2v_2 = (m_1+m_2)v_f](https://tex.z-dn.net/?f=m_1v_1%2Bm_2v_2%20%3D%20%28m_1%2Bm_2%29v_f)
Where here
is the velocity after the collision.
![v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}](https://tex.z-dn.net/?f=v_f%20%3D%20%5Cfrac%7Bm_1v_1%2Bm_2v_2%7D%7Bm_1%2Bm_2%7D)
![v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}](https://tex.z-dn.net/?f=v_f%20%3D%20%5Cfrac%7B%2865000%2A0.8%29%2B%2892000%2A1.2%29%7D%7B92000%2B65000%7D)
![v_f = 1.034m/s](https://tex.z-dn.net/?f=v_f%20%3D%201.034m%2Fs)