Answer:
Explanation:
A track 100m
The sprinter passes 12m mark in 1.8seconds
Let this be the initial point
d1 = 12m
t1 = 1.8seconds
And Passes the 56m mark in 6.7 seconds
Then, the second point is
d2 = 56m
t2 = 6.7 seconds
The sprinter velocity between this two position
Velocity is the rate of change of displacement
V = displacement / time taken
V = ∆x / ∆t
V = (x2 - x1) / (t2 - t1)
V = (56 - 12) / (6.7 - 1.8)
V = 44 / 4.9
V = 8.98 m/s
Then, the sprinter velocity between this two position is 8.98 m/s
The boy's momentum is 160 kg*m/s north.
The formula of momentum is p = mv, where p is momentum.
p = 40 kg * 4m/s north
p =160 kg*m/s north<span>Thank you for posting your question. I hope you found what you were after. Please feel free to ask me more.</span>
The electric field points in the +x direction and the particle's charge is positive, therefore the electric force points in the +x direction.
The electric force acting on the particle is:
F = Eq
F is the electric force, E is the electric field strength, and q is the charge.
Given:
E = 1290N/C
q = 6.58×10⁻⁶C
F = 1290(6.58×10⁻⁶) = 8.49×10⁻³N
The net force points in the +x direction, yet its magnitude is smaller than the magnitude of the electric force, so the magnetic force must point in the -x direction with a magnitude equal to the difference between the electric force and net force:
8.49×10⁻³ - 6.17×10⁻³
= <em>2.32×10⁻³N</em>
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The particle's velocity lies in the x-y plane and the magnetic field points in the +z direction. The velocity is perpendicular to the magnetic field, therefore the magnitude of the magnetic force can be calculated by:
F = qvB
F is the magnetic force, q is the charge, v is the velocity, and B is the magnetic field strength.
Given values:
F = 2.32×10⁻³N (solved for previously)
q = 6.58×10⁻⁶C
B = 1.27T
Plug in the values and solve for v:
2.32×10⁻³ = 6.58×10⁻⁶v(1.27)
v = 278m/s
Answer:air flow and the movement from where it’s coming from
Explanation: