Answer:
q/m = 2177.4 C/kg
Explanation:
We are given;
Initial speed v_o = 5 × 10³ m/s = 5000 m/s
Now, time of travel in electric field is given by;
t_1 = D_1/v_o
Also, deflection down is given by;
d_1 = ½at²
Now,we know that in electric field;
F = ma = qE
Thus, a = qE/m
So;
d_1 = ½ × (qE/m) × (D_1/v_o)²
Velocity gained is;
V_y = (a × t_1) = (qE/m) × (D_1/v_o)
Now, time of flight out of field is given by;
t_2 = D_2/v_o
The deflection due to this is;
d_2 = V_y × t_2
Thus, d_2 = (qE/m) × (D_1/v_o) × (D_2/v_o)
d_2 = (qE/m) × (D_1•D_2/(v_o)²)
Total deflection down is;
d = d_1 + d_2
d = [½ × (qE/m) × (D_1/v_o)²] + [(qE/m) × (D_1•D_2/(v_o)²)]
d = (qE/m•v_o²)[½(D_1)² + D_1•D_2]
Making q/m the subject, we have;
q/m = (d•v_o²)/[E((D_1²/2) + (D_1•D_2))]
We have;
E = 800 N/C
d = 1.25 cm = 0.0125 m
D_1 = 26.0 cm = 0.26 m
D_2 = 56 cm = 0.56 m
Thus;
q/m = (0.0125 × 5000²)/[800((0.26²/2) + (0.26 × 0.56))]
q/m = 312500/143.52
q/m = 2177.4 C/kg
Answer:
The average speed of an object is defined as the distance traveled divided by the time elapsed. Velocity is a vector quantity, and average velocity can be defined as the displacement divided by the time.
Metal as it is a conductor of electricity
Explanation:
y = y₀ + v₀ t + ½ at²
For the first ball:
0 = h + v₀ t − 4.9t²
For the second ball:
0 = h − 4.9 (t−1)²
a) If h = 20.0, find v₀.
0 = 20 − 4.9 (t−1)²
t = 3.02 s
0 = 20 + v₀ (3.02) − 4.9 (3.02)²
v₀ = 8.18 m/s
Graph:
desmos.com/calculator/uk1wzkxybt
If v₀ is given, find h.
First, find t in terms of v₀:
h + v₀ t − 4.9t² = h − 4.9 (t−1)²
v₀ t − 4.9t² = -4.9 (t−1)²
v₀ t − 4.9t² = -4.9 (t² − 2t + 1)
v₀ t − 4.9t² = -4.9t² + 9.8t − 4.9
v₀ t = 9.8t − 4.9
(9.8 − v₀) t = 4.9
t = 4.9 / (9.8 − v₀)
Therefore:
h = 4.9 (4.9 / (9.8 − v₀) − 1)²
bi) If v₀ = 6.0 m/s:
h = 4.9 (1 / (9.8 − 6.0) − 1)²
h = 2.66 m
bii) If v₀ = 9.5 m/s:
h = 4.9 (1 / (9.8 − 9.5) − 1)²
h = 26.7 m
c) As found in part a, the time it takes for the first ball to land is:
t = 4.9 / (9.8 − v₀)
If v₀ is greater than 9.8 m/s, the time becomes negative, which isn't possible. Therefore, vmax = 9.8 m/s. At this speed, the ball would reach its highest point after 1 second, the same time that the second ball is dropped. Two balls dropped at the same time from different heights cannot land at the same time.
d) If v₀ is less than 4.9 m/s, the time for the first ball to land becomes less than 1 second. Which means it will have already landed before the second ball is dropped. Therefore, vmin = 4.9 m/s.
Answer:
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