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andrew11 [14]
3 years ago
10

If wavelength and speed of a wave are 4 m and 332 m/s respectively, calculate its frequency​

Physics
1 answer:
Furkat [3]3 years ago
4 0

Explanation:

<em>Given </em>

<em>wavelength </em><em>=</em><em> </em><em>4</em><em> </em><em>m</em>

<em>speed </em><em> </em><em>=</em><em> </em><em>3</em><em>3</em><em>2</em><em> </em><em>m/</em><em>s</em>

<em>frequency </em><em>=</em><em> </em><em>?</em>

<em>We </em><em>know </em><em>we </em><em>have </em><em>the </em><em>formula </em>

<em>wavelength</em><em> </em><em>=</em><em> </em><em>speed </em><em>/</em><em> </em><em>frequency </em>

<em>4</em><em> </em><em>=</em><em> </em><em>3</em><em>3</em><em>2</em><em> </em><em>/</em><em> </em><em>frequency </em>

<em>frequency </em><em>=</em><em> </em><em>3</em><em>3</em><em>2</em><em>/</em><em>4</em>

<em>Therefore </em><em> </em><em>frequency </em><em>is </em><em>8</em><em>3</em><em> </em><em>Hertz </em><em>.</em>

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3 years ago
A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a
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Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach  gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g  (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω  / t

(ω = √g/R)

∝ = \frac{1}{t } \sqrt{\frac{g}{R} } }

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = \frac{2FR}{MR^2 + 2mr^2}

Now we have ,

∝ = \frac{1}{t} \sqrt{\frac{g}{R} }  , ∝ = \frac{2FR}{MR^2+2mr^2}

equating both values

We have,

F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N

Each thruster has to applied a force of 294.5N in tangential direction

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3 years ago
A Ferris wheel has diameter of 10 m. It rotates at a uniform rate and makes one revolution in 8.0 seconds. A person weighing 670
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Explanation:

from the question, we have

diameter = 10 m

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weight (Fw) = 670 N

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