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lara [203]
3 years ago
10

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

ter of the 25th ring.
Physics
1 answer:
IgorC [24]3 years ago
4 0

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

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Answer:

It is possible by increasing the speed of the tennis ball by a factor of (Mass of the tennis ball)/(Mass of the basketball)

Explanation:

The momentum of a body = The bod's mass × The body's velocity

Therefore, the momentum of a given mass of an object, such as a tennis ball can be increased by increasing the velocity or speed of the object. Whereby the speed of the ball, v₁, is increased such that the momentum of the basketball and the tennis ball will be the same, is given by the following equation

Mass of the basketball × v₂ = Mass of the tennis ball × v₁

Therefore, v₁/v₂ = (Mass of the tennis ball)/(Mass of the basketball)

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3 years ago
Plz answer this! I am stumped
Gwar [14]
MgCl2
Mg = magnesium
Cl = chlorine

Magnesium + chlorine = magnesium chloride.

This is because compounds are always written with the METAL FIRST and the NON METAL SECOND. the non metal ends in - ide when it reacts with a metal.

So ur answer would be magnesium chloride. :)
6 0
3 years ago
Two samples of dirt are collected from a suspect's tread in his shoe and a crime scene. She notes very similar characteristics.
elena-14-01-66 [18.8K]

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8 0
3 years ago
A jet can travel the 6000 km distance between washington,
Andrei [34K]
<span> d = r*t is the basic distance equation
 d = 6000 km
 t with the tail wind = 6 hr
 r with the tail wind = speed of the plane + wind speed = s + w
 t with the head wind = 7.5 hr
 r with the head wind = speed of the plane - wind speed = s-w
 (s+w)*6 = 6000
 (s-w)*7.5 = 6000
 s + w = 1000
 s - w = 800
</span><span> 2s = 1800
 s = 900 km/h
 s + w = 1000
 w = 100
 Check the anwer by calculating the return trip.
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3 years ago
A mass is oscillating with amplitude A at the end of a spring.
Dmitry_Shevchenko [17]

A) x=\pm \frac{A}{2\sqrt{2}}

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

E=\frac{1}{2}kA^2 (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2 (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

U=\frac{1}{3}K

Using (2) we can rewrite this as

U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}

And using (1), we find

U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2

Substituting U=\frac{1}{2}kx^2 into the last equation, we find the value of x:

\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}

B) x=\pm \frac{3}{\sqrt{10}}A

In this case, the kinetic energy is 1/10 of the total energy:

K=\frac{1}{10}E

Since we have

K=E-U

we can write

E-U=\frac{1}{10}E\\U=\frac{9}{10}E

And so we find:

\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A

3 0
3 years ago
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