Answer:
vB = 8.57[m/s]
Explanation:
Since there is no external force that generates momentum, the amount of momentum is conserved so you will have the following equation
![m_{A}*v_{A}=m_{B}*v_{B}\\Where:\\m_{A}= 90[kg]\\v_{A}=10[m/s]\\m_{B}=105[kg]\\\\Therefore:\\\\v_{B}=\frac{m_{A}*v_{A}}{m_{B}}\\ v_{B}=\frac{90*10}{105}\\v_{B}=8.57[m/s]](https://tex.z-dn.net/?f=m_%7BA%7D%2Av_%7BA%7D%3Dm_%7BB%7D%2Av_%7BB%7D%5C%5CWhere%3A%5C%5Cm_%7BA%7D%3D%2090%5Bkg%5D%5C%5Cv_%7BA%7D%3D10%5Bm%2Fs%5D%5C%5Cm_%7BB%7D%3D105%5Bkg%5D%5C%5C%5C%5CTherefore%3A%5C%5C%5C%5Cv_%7BB%7D%3D%5Cfrac%7Bm_%7BA%7D%2Av_%7BA%7D%7D%7Bm_%7BB%7D%7D%5C%5C%20v_%7BB%7D%3D%5Cfrac%7B90%2A10%7D%7B105%7D%5C%5Cv_%7BB%7D%3D8.57%5Bm%2Fs%5D)
The calculation of the centripetal acceleration of an object following a circular path is based on the equation,
a = v² / r
where a is the acceleration, v is the velocity, and r is the radius.
Substituting the known values from the given above,
4.4 m/s² = (15 m/s)² / r
The value of r from the equation is 51.14 m.
Answer: 51.14 m
Answer:
Explanation:
When the positively charged half shell is brought in contact with the electroscope, its needle deflects due to charge present on the shell.
When the negatively charged half shell is brought in contact with the positively charged shell , the positive and negative charge present on each shell neutralises each other .So both the shells lose their charges .The positive half shell also loses all its charges
When we separate the half shells , there will be no deflection in the electroscope because both the shell have already lost their charges and they have become neutral bodies . So they will not be able to produce any deflection in the electroscope.
Is there suppose to be an image?
Answer:
Explanation:
Given that :
mass of the SUV is = 2140 kg
moment of inertia about G , i.e 
= 875 kg.m²
We know from the conservation of angular momentum that:
![mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2](https://tex.z-dn.net/?f=mv_1%20%2A0.765%20%3D%20%5BI%2Bm%280.765%5E2%2B0.895%5E2%29%5D%20%5Comega_2)
![2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2](https://tex.z-dn.net/?f=2140v_1%2A0.765%20%3D%20%5B875%2B2140%280.765%5E2%2B0.895%5E2%29%5D%20%5Comega_2)
From the conservation of energy as well;we have :
^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7B2%7D%20%5B875%2B2140%280.765%5E2%2B0.895%5E2%29%5D%280.4262%20%5C%20v_1%29%5E2%20-2140%289.81%29%5B%5Csqrt%7B0.76%5E2%2B0.895%5E2%7D%20-0.765%5D%5D%20%3D0)
