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Complete Question
Three equal point charges are held in place as shown in the figure below
If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.
A) F1=2F2
B) F1=3F2
C) F1=4F2
D) F1=9F2
Answer:
D) F1=9F2
Explanation:
We are told in the question that there are three equal point charges.
q, Q1, Q2 ,
q = Q1 = Q2
From the diagram we see the distance between the points d
q to Q1 = d
Q1 to Q2 = nd
Assuming n = 2
= 2 × d = 2d
Sum of the two distances = d + 2d = 3d
F1 is the force on q due to Q1 and
F2 is the force on q due to Q2,
Since we have 3 equal point charges and a total sum of distance which is 3d
Hence,
F1 = 9F2
Answer:
33.68 N
Explanation:
Data
W= 32J
d- 0.95m
F= ?
W=Fd
They are asking for the magnitude which is the force, so you need to solve for force.
F=W/d
= 32J/ 0.95m
= 33.68 N
Answer:
a) When its length is 23 cm, the elastic potential energy of the spring is
0.18 J
b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Explanation:
Hi there!
a) The elastic potential energy (EPE) is calculated using the following equation:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretched lenght.
Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).
First, let´s convert the spring constant units into N/m:
4 N/cm · 100 cm/m = 400 N/m
EPE = 1/2 · 400 N/m · (0.03 m)²
EPE = 0.18 J
When its length is 23 cm, the elastic potential energy of the spring is 0.18 J
b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:
EPE = 1/2 · 400 N/m · (0.06 m)²
EPE = 0.72 J
When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Answer:
5.22 x 10^5 V
Explanation:
guessed on castle learning and got it right
