200 Hz = 200 cycles per sec
<span>1 cycle, the period = 1/200 = 0.005 seconds, or 5 milli seconds.</span>
At point C because it is at the lowest position.
Answer:
0.0109 m ≈ 10.9 mm
Explanation:
proton speed = 1 * 10^6 m/s
radius in which the proton moves = 20 m
<u>determine the radius of the circle in which an electron would move </u>
we will apply the formula for calculating the centripetal force for both proton and electron ( Lorentz force formula)
For proton :
Mp*V^2 / rp = qp *VB ∴ rp = Mp*V / qP*B ---------- ( 1 )
For electron:
re = Me*V/ qE * B -------- ( 2 )
Next: take the ratio of equations 1 and 2
re / rp = Me / Mp ( note: qE = qP = 1.6 * 10^-19 C )
∴ re ( radius of the electron orbit )
= ( Me / Mp ) rp
= ( 9.1 * 10^-31 / 1.67 * 10^-27 ) 20
= ( 5.45 * 10^-4 ) * 20
= 0.0109 m ≈ 10.9 mm
Answer:
Basic kinematics, negating drag and assuming ideal conditions, we use the equation:
d=vi*t+1/2*a*t^2
Since vi is 0 (we know this because you’re dropping it, not throwing it)…
…and the only acceleration acting on it is gravity, a=9.8 m/s^2…
…we get
d=1/2(9.8)(5)^2
Explanation:
Some quick mental math tells us that this is about 125 m.
Plugging it in, we find it to be 122.5 m.
The answer is TRUE, I'm pretty sure.
