Two large insulating parallel plates carry charge of equal magnitude, one positive and the other negative, that is distributed u
niformly over their inner surfaces. Rank the points 1 through 5 according to the magnitude of the electric field at the points, least to greatest.
A. 1, 2, 3, 4, 5
B. 2, then 1, 3, and 4 tied, then 5
C. 1, 4, and 5 tie, then 2 and 3 tie
D. 2 and 3 tie, then 1 and 4 tie, then 5
E. 2 and 3 tie, then 1, 4, and 5 tie
A. 1, 2, 3, 4, 5
B. 2, then 1, 3, and 4 tied, then 5
C. 1, 4, and 5 tie, then 2 and 3 tie
D. 2 and 3 tie, then 1 and 4 tie, then 5
E. 2 and 3 tie, then 1, 4, and 5 tie
1 answer:
Answer:
The correct choice is C: points 1, 4, and 5 are equal, followed by 2 and 3 being equal.
Explanation:
Here’s the breakdown:
The electric field from the positive sheets E₁ = б/2E₀
E₂ is from the negative sheet = -б/2E₀
At points 1, 4, and 5, the electric fields created by the sheets oppose each other.
At point 1, the total field is calculated as -E₁ + E₂ = 0.
Similarly, at point A, the total field results in -E₁ - E₂ = 0.
However, at any point in between the plates, the electric field is directed consistently in one way.
At points 2 and 3, the field is directed to the right.
Thus, we have:
E net = E₁ + E₂
= б/2E₀ + -б/2E₀
=б/E₀
Note: Please refer to the attached document for the full question accompanying this solution.
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Consider the diagram below.
m₁ = 8.5 x 10⁶ kg, the starship's mass
m₂ = 10⁴ kg, the shuttlecraft's mass
a₁ = acceleration of the starship
a₂ = the acceleration of the shuttle
F = 4 x 10⁴ N, the force exerted
Let y represent the distance covered by the starship
Let x denote the distance covered by the shuttlecraft
If t indicates the travel time, then
y = 0.5a₁t² (1)
x = 0.5a₂t² (2)
F = m₁a₁ = m₂a₂ (3)
Additionally,
x + y = 14000 m (4)
From (2), we derive
a₁ = (4 x 10⁴ N)/(8.5 x 10⁶ kg) = 4.706 x 10⁻³ m/s²
a₂ = (4 x 10⁴ N)/(10⁴ kg) = 4 m/s²
From (1), (2) and (4), we find
0.5*(t s)²*(4 + 4.706 x 10⁻³ m/s²) = 14000 m
2.002353t² = 14000
t² = 6991.774 s²
t = 83.617 s
Thus
x = 0.5*4*6991.774 = 13984 m = 13.984 km
y = 0.5*4.706 x 10⁻³*6991.774 = 16.452 m
The starship moves roughly 16.5 meters while towing the shuttlecraft by 13.98 kilometers.
Result: The starship shifts by 16.5 m (to the nearest tenth)
m₁ = 8.5 x 10⁶ kg, the starship's mass
m₂ = 10⁴ kg, the shuttlecraft's mass
a₁ = acceleration of the starship
a₂ = the acceleration of the shuttle
F = 4 x 10⁴ N, the force exerted
Let y represent the distance covered by the starship
Let x denote the distance covered by the shuttlecraft
If t indicates the travel time, then
y = 0.5a₁t² (1)
x = 0.5a₂t² (2)
F = m₁a₁ = m₂a₂ (3)
Additionally,
x + y = 14000 m (4)
From (2), we derive
a₁ = (4 x 10⁴ N)/(8.5 x 10⁶ kg) = 4.706 x 10⁻³ m/s²
a₂ = (4 x 10⁴ N)/(10⁴ kg) = 4 m/s²
From (1), (2) and (4), we find
0.5*(t s)²*(4 + 4.706 x 10⁻³ m/s²) = 14000 m
2.002353t² = 14000
t² = 6991.774 s²
t = 83.617 s
Thus
x = 0.5*4*6991.774 = 13984 m = 13.984 km
y = 0.5*4.706 x 10⁻³*6991.774 = 16.452 m
The starship moves roughly 16.5 meters while towing the shuttlecraft by 13.98 kilometers.
Result: The starship shifts by 16.5 m (to the nearest tenth)
Answer:
A) 328 m
B) 80.22 m/s
C) 8.18 sec
Explanation:
A)
The rocket's initial acceleration is 2.25 m/s²
Time until engine failure is 15.4 s
Initial velocity u during takeoff = 0 m/s
Distance covered while the engine is functional =?
We apply Newton's laws for this calculation
S = ut + a
Here, S represents the distance traveled under the rocket's thrust
S = (0 x 15.4) + (2.25 x
)
S = 0 + 266.81 m = 266.81 m
Before the engine fails, the final velocity can be found using:
v = u + at
v = 0 + (2.25 x 15.4) = 34.65 m/s
Once the engine fails, the rocket decelerates under gravitational pull at g = -9.81 m/s² (acting downwards)
The upward initial velocity when freefall begins is v = 34.65 m/s
The final velocity is reached at peak height, where the rocket halts, therefore:
u = 0 m/s
The distance covered during this freefall will be s =?
Utilizing the equation
=
+ 2gs
=
+ 2(-9.81 x s)
0 = 1200.6 - 19.62s
-1200.6 = -19.62s
s = -1200.6/-19.62 = 61.19 m
Thus,
Maximum Height = 266.81 m + 61.19 m = 328 m
B)
At maximum height, the rocket’s initial upward velocity drops to 0 m/s (the rocket completely stops)
The descent occurs freely under g = 9.81 m/s² (acting downwards)
The distance covered during the fall will be 328 m
Final velocity v just prior to impact =?
Applying =
+ 2gs
=
+ 2(9.81 x 328)
= 0 + 6435.36
v = = 80.22 m/s
The time taken before reaching the pad is found as follows
v = u + gt
80.22 = 0 + 9.81t
t = 80.22/9.81 = 8.18 sec