Unconfined water comes straight from the surface and is more

1. In Newton’s ring experiment, the diameter of the 5th ring is 0.30 cm and diameter of 15th the ring is 0.62 cm. Find the diame

IgorC [24]

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

4 0

3 years ago

a car accelerates uniformly from rest to a speed of 51.9mi/h in 9.37s. find the constant acceleration of the car. answer in unit

Darina [25.2K]

Here is my step-by-step-work. Let me know if you have any questions! :)

3 0

3 years ago

What is the name of the scale used to measure hurricanes

shusha [124]

Yea it’s called the Saffir-Simpson Hurricane scale, made in 1960s and further developed in 1970s

8 0

3 years ago

A standing wave is created on a string of length 1.2 m. If the speed of the wave on the string is 60.0 m/s, what is the fundamen

coldgirl [10]

Answer:

Fundamental frequency in the string will be 25 Hz

Explanation:

We have given length of the string L = 1.2 m

Speed of the wave on the string v = 60 m/sec

We have to find the fundamental frequency

Fundamental frequency in the string is equal to $f=\frac{v}{2L}$ , here v is velocity on the string and L is the length of the string

So frequency will be equal to $f=\frac{v}{2L}=\frac{60}{2\times 1.2}=25Hz$

So fundamental frequency will be 25 Hz

6 0

3 years ago

Read 2 more answers

In a charging process, 4 × 1013 electrons are removed from one small metal sphere and placed on a second identical sphere. Initi

Alina [70]

Answer:

The distance between the two spheres is 914.41 X 10³ m

Explanation:

Given;

4 X 10¹³ electrons, and its equivalent in coulomb's is calculated as follows;

1 e = 1.602 X 10⁻¹⁹ C

4 X 10¹³ e = 4 X 10¹³ X 1.602 X 10⁻¹⁹ C = 6.408 X 10⁻⁶ C

V = Ed

where;

V is the electrical potential energy between two spheres, J

E is the electric field potential between the two spheres N/C

d is the distance between two charged bodies, m

$V = \frac{K*q}{d^2}*d = \frac{K*q}{d}$

$d = \frac{K*q}{V}$

where;

K is coulomb's constant = 8.99 X 10⁹ Nm²/C²

d = (8.99 X 10⁹ X 6.408 X 10⁻⁶)/0.063

d = 914.41 X 10³ m

Therefore, the distance between the two spheres is 914.41 X 10³ m

3 0

4 years ago