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yaroslaw [1]
3 years ago
5

How long should you make a simple pendulum so its period is 200. ms??

Physics
1 answer:
Lelu [443]3 years ago
7 0
So we want to know how long should we make the simple pendulum so it's period is T=200 ms = 0.2 s. Since the formula for simple pendulum is T=2*pi*sqrt(L/g) where T is the period, L is the length and g=9.81 m/s^2. Now we need to invert the formula to get the length: T/2pi=sqrt(L/g), we square both sides of the equation: (T/2pi)^2=L/g and multiply both sides with g:
g*(T/2pi)^2=L. Now we input the numbers and get that the length L= 0.00995m.
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3 years ago
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

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3 years ago
The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit
vovikov84 [41]

Gravitational force is given by, F= G\frac{mM}{R^{2}}

Where, m and M are the masses of the objects, R is the distance between them and G gravitational constant.

Gravitational force of the star on planet 1, F_{1}= G\frac{m_{1}M}{R^{2}}

Gravitational force of the star on planet 2, F_{2}= G\frac{3m_{1}M}{(3R)^{2}}

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\frac{F_{1}}{F_{2}}=  \frac{3}{1}

Therefore, the gravitational force of the star on the planet 1 is three times that on planet 2.

6 0
3 years ago
Read 2 more answers
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Igoryamba

While falling, both the sheet of paper and the paper ball experience air resistance. But the surface area of the sheet is much more than that of the spherical ball. And air resistance varies directly with surface area. Hence the sheet experiences more air resistance than the ball and it falls more slowly than the paper ball.

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Please help me solve this and give an explanation​
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