Answer:
sorry but I can understand the question
Answer:
19.3m/s
Explanation:
Use third equation of motion
where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity
insert values to get answer
![v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s](https://tex.z-dn.net/?f=v%5E2-0%5E2%3D2%289.81m%2Fs%5E2%29%2838%2F2%29%5C%5Cv%5E2%3D9.81m%2Fs%5E2%20%2A38%5C%5Cv%5E2%3D372.78%5C%5Cv%3D%5Csqrt%5B%5D%7B372.78%7D%20%5C%5Cv%3D19.3m%2Fs)
For any mass m:
a = F/m
v = √2*F/m*s = √2F/sm = k/√m
Momentum = mv = k√m
Energy = 1/ mv² = 1/2 m.k²/m = 1/2k²
SO
Both will have same energy
The larger mass will have greater momentum
25m north east hope this helps
Answer: F
Out of the page.
Explanation:
For an electron with a charge of -e, the magnitude of the force on it is F = BeV
Where
F = force on the electron
e = charge ( electrons )
V = velocity
B = magnetic field
F is the force acting on all the electrons in a wire which gives rise to the F = BIL
Where
I = current
L = length of the wire
The force F is always at the right angle to the particle's velocity and its direction can be found using the left hand rule.
When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed out of the page.
