A charge is placed on a spherical conductor of radius r1. This sphere is then connected to a distant sphere of radius r2 (not eq
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Answer:
The excess charge is
Explanation:
According to the question, we are informed that
The diameter is
The potential of the surface is
The radius of the sphere is
by plugging in given values
Where k is Coulomb's constant with a value
Based on the question stating there are no other charges, Q represents the excess charge
Therefore
inserting the numerical values
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