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9966 [12]
3 years ago
7

A simple pendulum takes 2.20 s to make one compete swing. If we now triple the length, how long will it take for one complete sw

ing?
Physics
1 answer:
sergiy2304 [10]3 years ago
5 0

Answer:

Time taken for 1 swing = 3.81 second

Explanation:

Given:

Time taken for 1 swing = 2.20 Sec

Find:

Time taken for 1 swing , when triple the length(T2)

Computation:

Time taken for 1 swing = 2π[√l/g]

2.20 = 2π[√l/g].......Eq1

Time taken for 1 swing , when triple the length (3L)

Time taken for 1 swing = 2π[√3l/g].......Eq2

Squaring and dividing the eq(1) by (2)

4.84 / T2² = 1 / 3

T2 = 3.81 second

Time taken for 1 swing = 3.81 second

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You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.
Alika [10]

Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

          z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 +9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

        d = 2021.6 km

7 0
3 years ago
A pair of narrow slits, separated by 1.8 mm, is illuminated by a monochromatic light source. Light waves arrive at the two slits
Nastasia [14]

Answer:

750 nm

Explanation:

d  = separation of the slits = 1.8 mm = 0.0018 m

λ = wavelength of monochromatic light

D  = screen distance = 4.8 m

y = position of first bright fringe = \frac{1cm}{5 fringe} = \frac{0.01}{5} = 0.002 m

n  = order = 1

Position of first bright fringe is given as

y = \frac{nD\lambda }{d}

0.002 = \frac{(1)(4.8)\lambda }{0.0018}

λ = 7.5 x 10⁻⁷ m

λ = 750 nm

3 0
3 years ago
Which of the following tools is used to observe objects that are very far away? A. a anemometer B. a hand lens C. a microscope D
inn [45]

Answer:

D. a telescope

Explanation:

3 0
3 years ago
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Irina-Kira [14]

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5 0
2 years ago
A sled of mass 2.12 kg has an initial speed of 5.49 m/s across a horizontal surface. The coefficient of kinetic friction between
Darya [45]

Answer:

The speed of the sled is 3.56 m/s

Explanation:

Given that,

Mass = 2.12 kg

Initial speed = 5.49 m/s

Coefficient of kinetic friction = 0.229

Distance = 3.89 m

We need to calculate the acceleration of sled

Using formula of acceleration

a = \dfrac{F}{m}

Where, F = frictional force

m = mass

Put the value into the formula

a=\dfrac{\mu mg}{m}

a=\mu g

a=0.229\times9.8

a=2.244\ m/s^2

We need to calculate the speed of the sled

Using equation of motion

v^2=u^2-2as

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

v ^2=(5.49)^2-2\times2.244\times3.89

v=\sqrt{(5.49)^2-2\times2.244\times3.89}

v=3.56\ m/s

Hence, The speed of the sled is 3.56 m/s.

8 0
3 years ago
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