Upward and downward forces cancel out. Net force is 8 newtons to the right
Answer:
Explanation:
A proton of charge
q=+1.609×10^-19C
Orbit a radius of 12cm
r=0.12m
Magnetic Field of 0.31T
Angle between velocity and field is 90°
a. Because the magnetic force F supplies the centripetal force Fc.
The magnitude of the magnetic force F on a charge q moving at a speed v in a magnetic field of strength B is given by
F = qvB sin θ
And the centripetal force is given as
Fc=mv²/r
Where m is mass of proton
m=1.673×10^-27kg
Then, F=Fc
qvB sin θ=mv²/r
qBSin90=mv/r
rqB=mv
Then, v=rqB/m
v=0.12×1.609×10^-19×0.31/1.673×10^-23
v=3577692.78m/s
v=3.58×10^6m/s
b. Since,
F=qVBSin90
F=1.609×10^-19×3.58×10^6×0.31
F=1.785×10^-13 N.
Is the component perpendicular to the surface on contact of the contact force <span />
Answer:
<u><em>The truck was moving 16.5 m/s during the time it took to stop, which was 3 seconds. </em></u>
- <u><em>Initial velocity = 33 m/s</em></u>
- <u><em>Final velocity = 0 m/s</em></u>
- <u><em>Average velocity = (33 + 0) / 2 m/s = 16.5 m/s</em></u>
Explanation:
- <u><em>First, how long does it take the truck to come to a complete stop?</em></u>
- <u><em>( 33 m/s ) / ( 11 m / s^2 ) = 3 seconds</em></u>
- <u><em>Then we can look at the average velocity between when the truck started decelerating and when it came to a complete stop. Because the deceleration is constant (always 11m/s^2) we can use this trick.</em></u>
Hello!
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.
Data:
Hooke represented mathematically his theory with the equation:
F = K * Δx
On what:
F (elastic force) = 2 N
K (elastic constant) = 4 N/cm
Δx (deformation or elongation of the elastic medium or distance from a spring) = ?
Solving:




simplify by 2


Answer:
B.) 1/2 cm
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I Hope this helps, greetings ... Dexteright02! =)