Sorry to say but I know that t(e introduction is first and the coda is last
Answer:
Diameter of Newton’s 5th ring = 0.30 cm
Diameter of Newton’s 15th ring = 0.62 cm
Diameter of Newton’s 25th ring = ?
From Newton’s rings experiment we infer that
D2n+m − D2n = 4λmR
For the 5th and 15th rings we have
D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)
For 15th and 25th rings
D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)
We equate the two derivatives
Equation (2) = Equation (1)
D225 − D215 = D215 − D25
D225 = 2D215 – D25
Substituting the values into the equation
D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2
D25 = 0.8239 cm
Answer:
B. 6 cm
Explanation:
First, we calculate the spring constant of a single spring:

where,
k = spring constant of single spring = ?
F = Force Applied = 10 N
Δx = extension = 4 cm = 0.04 m
Therefore,

Now, the equivalent resistance of two springs connected in parallel, as shown in the diagram, will be:

For a load of 30 N, applying Hooke's Law:

Hence, the correct option is:
<u>B. 6 cm</u>
Answer:
B electrons protons and neutrons
hope i helped...
Explanation: