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:
Explanation:
a) Energy stored in spring = 1/2 k x² = .5 x k 0.1²
500 = 5 x 10⁻³ k ,
k = (500/5) x 10³ = 10⁵ N/m
b )
k = 4.5 x 10¹ = 45 N/m
Stored energy = 1/2 k x² = .5 x 45 x 8² x 10⁻⁴ =1440 x 10⁻⁴ J
This energy gets dissipated by friction .
work done by friction = μ mg d
d is the distance traveled under friction
so 1440 x 10⁻⁴ = μ x 3 x 9.8 x 2
μ = 245 x 10⁻⁴ or 0.00245 which appears to be very small. .
Answer:
Given:
Fundamental frequency: 470Hz
T1:310k,T2:315k
Calculating velocity
Recall v=(331m/s)✓[T1/273k)
V=331✓(310/273)
V1=331*(1.0656)=352.72m/s
V2=331✓(315/273)=355.5m/s
Fundamental frequency=4L
F2=F1(V2/V1)
F2=470(355.5/352.72)=474.4Hz
Beat=[F2-F1]=474.4-470=4.4Hz
Explanation:
The distance covered by the acorn is 3.136 m.
<u>Explanation:</u>
The time taken for the acorn to hit the ground is 0.8 s. As it is a free fall, the acorn will be completely under the influence of gravity. So the acceleration will be acceleration due to gravity.
Then using the second law of equation,
Since the initial velocity and time is zero, then the time taken to reach the ground is stated as 0.8 s, so
So the distance covered by the acorn is 3.136 m.
