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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:

The horizontal component of its velocity remains constant and the vertical component of its acceleration is equal to -g.

Explanation:

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Also, the projectile has only vertical component of acceleration and no horizontal component of acceleration since, its horizontal component of velocity remains constant. Thus, no change in the horizontal component of velocity.

The vertical component of acceleration is equal to -g since, the weight is the only vertical force acting on it.

So, <u>the horizontal component of its velocity remains constant and the vertical component of its acceleration is equal to -g.</u>

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3 years ago
What is a charge-coupled device (CCD), and how is it used in astronomy?
Sedaia [141]

Answer:

Charge-coupled device  (CCD) is a device that receives and transfers an electrical charge to the next region

Explanation:

Charge-coupled device  (CCD) is a device that receives and transfers an electrical charge to the next region where it can be modified like changing it to a electronic value.

In astronomy, high-powered telescopes can be used with CCD device image sensor cameras. The imaging system can concentrate for a number of hours on one place in space once the Earth's rotation synchronizes with the telescope.

6 0
4 years ago
An auto race takes place on a circular track. A car completes one lap in a time of 25.0 s, with an average tangential speed of 5
ludmilkaskok [199]

Answer:

(a) Angular speed will be 0.2512 rad/sec

(b) Radius will be 213.375 m

Explanation:

We have given time to complete 1 lap = 25 sec

We know that 1 lap = 2\pi radian

(a) So angular speed =\frac{2\pi }{25}=\frac{2\times 3.14}{25}=0.2512rad/sec

(B) Tangential velocity is given as v = 53.6 m/sec

We know that tangential velocity is given by

v=\omega r

So radius r=\frac{v}{\omega }=\frac{53.6}{0.2512}=213.375m

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4 years ago
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kaheart [24]

Answer:

zero

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In this system, only conservative forces act. Therefore, the mechanical energy, that is, the sum of the kinetic energy and the potential energy, remains constant. When the mass is at its maximum displacement from equilibrium, its potential energy is maximum, therefore, its kinetic energy is minimal, that is to say, that its instantaneous velocity at that point is zero.

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4 years ago
Pls help‼️‼️I have no idea how to solve this question... Can someone pls help! :(
sdas [7]

Answer: Its dragon ball bro

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PART II

7 0
4 years ago
Read 2 more answers
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