An electron with a speed of 1.7 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts
on it. The mass of the electron is 9.11 × 10-31 kg. Determine the vertical distance the electron is deflected during the time it has moved 29 mm horizontally.
2 answers:
Answer:
y = 77.74 10⁻⁵ m
Explanation:
For this exercise we can use Newton's second law
F = m a
a = F / m
a = 4.9 10⁻¹⁶ / 9.1 10⁻³¹
a = 0.538 10¹⁵ m / s
This is the vertical acceleration of the electron.
Now let's use kinematics to find the time it takes to move the
x= 29 mm = 29 10⁻³ m
On the x axis
v = x / t
t = x / v
t = 29 10⁻³ / 1.7 10⁷
t = 17 10⁻¹⁰ s
Now we can look for vertical distance at this time.
y = t + ½ a t²
y = 0 + ½ 0.538 10¹⁵ (17 10⁻¹⁰)²
y = 77.74 10⁻⁵ m
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Answer:
2(maximum), -2(minimum), -2(maximum).
Explanation:
V(t)= 2πcos πt--------------------------------------------------------------------------------(1).
Therefore, there is a need to integrate v(t) to get S(t).
S(t)= 2×sinπt + C ------------------------------------------------------------------------------(2).
Applying the condition given, we have s(0)= 0.
S(0)= 2sin ×π(0) + C.
Which means that; 0+C= 0. That is; C=0.
S(t)= 2 sin πt.
The mass moves to its highest positions at time,t=half(1/2=.5) and time,t=2.5.
Take note that; sin(π/2) = sin(5π/2) = 1 .
Also, the mass moves to its lowest position at time,t=(3/2); also, sin(3π/2) = -1.
Therefore, we have that 2 maximum; -2 minimum and -2 maximum.
