Answer:
The answer is the kinetic energy of sled B after it crash is 6439j.
Answer:
3/7 ω
Explanation:
Initial momentum = final momentum
I(-ω) + (2I)(3ω) + (4I)(-ω/2) = (I + 2I + 4I) ωnet
-Iω + 6Iω - 2Iω = 7I ωnet
3Iω = 7I ωnet
ωnet = 3/7 ω
The final angular velocity will be 3/7 ω counterclockwise.
The magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.
Speed of the proton = 5.02 × 10 ⁶ m /a
Angel of between the velocity and the magnetic force = 60 °
The magnitude of magnetic field B = 0.180 T
The magnitude of the magnetic force on the proton is,
Therefore, the magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.
To know more about magnetic force, refer to the below link:
brainly.com/question/23096032
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If the velocity of the train is v=s/t, where s is the distance and t is time, then v=400/5=80m/s. To get the vertical component of the velocity we need to multiply the velocity v with a sin(α): Vv=v*sin(α), where Vv is the vertical component of the velocity and α is the angle with the horizontal. So:
Vv=80*sin(10)=80*0.1736=13.888 m/s.
So the vertical component of the velocity of the train is Vv=13.888 m/s.
