Answer:
and
Explanation:
See attached figure.
E due to sphere
E due to particule
(1)
according to the law of gauss and superposition Law:
; electric field due to the small sphere with r1=R/4
then: 
(2)
on the other hand, for the particule:
⇒ 
(3)
We replace (2) y (3) in (1):
--------------------
if R<x<2R AND
remember that
then:
solving:
but: R<x<2R
so :
The answer is to this question D
Answer:
Explanation:
AVerage acceleration is the cjange in velocity with time
a = v-u/t
v is the final velocity = 48m/s
u is the initial velocity = 40m/s
t is the time = 6.5s
a = 48-40/6.5
a = 8/6.5
a = 1.23m/s²
Hence the magnitude of the car’s average acceleration during this period is 1.23m/s²
Answer:
The answer is 3.111111.
Explanation:
It runs 28 m in the first 9 s, and 28 divided by 9 equals 3.1 and the one goes on forever.
Answer:
Option B. 5 nC
Explanation:
From the question given above, the following data were obtained:
Capicitance (C) = 100 pF
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:
1 pF = 1×10¯¹² F
Therefore,
100 pF = 100 pF × 1×10¯¹² F / 1 pF
100 pF = 1×10¯¹⁰ F
Next, we shall determine the quantity of charge. This can be obtained as follow:
Capicitance (C) = 1×10¯¹⁰ F
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Q = CV
Q = 1×10¯¹⁰ × 50
Q = 5×10¯⁹ C
Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:
1 C = 1×10⁹ nC
Therefore,
5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C
5×10¯⁹ C = 5 nC
Thus, the quantity of charge is 5 nC
