<span>(6.0x10^-22, -1.40x10^-21, 0) kg*m/s
Momentum is a conserved quantity. The total momentum of the system before and after the interactions will not change. So, let's look at the momentum before the interaction.
(3.2x10^-21, 0, 0) kg*m/s and (0,0,0) kg*m/s
After the interaction
(2.6x10^-21, 1.40x10^-21, 0) kg*m/s
and the other proton has to have a momentum that when added to this momentum equal the original value. Since the y and z vectors were initially 0, all we need for the y and x vector values of the result is to negate them. The x vector value will be
3.2x10^-21 - 2.6x10^-21 = 0.6x10^21 = 6.0x10^-22. So the other proton will have a momentum of
(6.0x10^-22, -1.40x10^-21, 0) kg*m/s</span>
Equation of velocity is given as
![V =[5.00m/s−(0.0180m/s^3)t^2]i^ + [2.00m/s+(0.550m/s^2)t]j^](https://tex.z-dn.net/?f=V%20%3D%5B5.00m%2Fs%E2%88%92%280.0180m%2Fs%5E3%29t%5E2%5Di%5E%20%2B%20%5B2.00m%2Fs%2B%280.550m%2Fs%5E2%29t%5Dj%5E)
at t = 7.93 s
so the magnitude of the velocity is given as
Part b)
the direction of the velocity is given as
part c)
for acceleration we know that
at t = 7.93 s
magnitude is given as
Part d)
for the direction of the motion
Answer:
Explanation:
The difference between the water level in the eudiometer tube and the water level in the beaker must be measured because we have to put into consideration, the pressure of the gases in the eudiometer tube. This said pressure of gas in the eudiometer must equal the atmospheric pressure. If or by chance, the water levels happens not to be at the same height, then this is not the case. And then, as a result, in order to account for the difference between both, while also being able to get accurate results, you have to find the difference or subtract the water levels and then go ahead in converting them to mmHg.
Momentum is transferred through the convert wich in this case is the golf club
Answer:
yes
Explanation:
As the formula is α= ΔL/L*ΔT where alpha (α) is the sign of coefficient of linear expansion
