Of the charge Q initially on a tiny sphere, a portion q is to be transferred to a second, nearby sphere. Both spheres can be tre
1 answer:
Answer:
2. Smaller value of q = 1/4Q
3. Larger value of q = 3/4Q
Explanation:
1. Please check the attachment for the solution of part 1. Due to time factor I couldn't type out the solution properly.
2. The smaller value is
F = 75% = 75/100
= 3/4 * Q²/16piEod² ---3
When we equate 1 and 3
We get 3Q²/16 = Qq - q² from here we cross multiply to get
3Q² = 16(Qq-q²)
3Q² = 16Qq-16q²
16q² - 16Qq + 3Q² = 0
When solve this out using the quadratic equation formula, we have:
Q/2 +- 2Q/8
We can get the smaller value of q as
Q/2-2Q/8
Solve using LCM we get
2Q/8 = 1/4Q = Q/4
C. The larger value of q
Q/2+2Q/8
= 6Q/8
= 3Q/4
Please use the attachment it will guide you to understand this better.
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Answer:
Explanation:
F = Magnetic force = 4.11 N
= Net current
= Current in one of the wires = 7.68 A
B = Magnetic field = 0.59 T
= Angle between current and magnetic field =
= Length of wires = 2.64 m
= Current in the other wire
Magnetic force is given by
Net current is given by
The current I is .
Answer:
The centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².
Explanation:
The centripetal acceleration is given by:
Where:
: is the tangential speed = 9.50 m/s
r: is the distance = 6.00 m
Hence, the centripetal acceleration is:
Therefore, the centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².
I hope it helps you!