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AysviL [449]
3 years ago
11

The distance between the crests of waves

Physics
2 answers:
stepladder [879]3 years ago
8 0
It depends, You have to have the length and the width of the crest wave.
Ivanshal [37]3 years ago
7 0
You need the length and width of the crest waves. -_-
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Does gravity affect your weight???
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Formula for the distance (d) is given by d = rate*time. For example if you are traveling at 60 mph for 3 hours the distance trav
babunello [35]

Explanation:

Distance covered by the particle is given by:

Distance (d) = rate (v) × time (t)                

Speed of Mary, v₁ = 50 mph

Speed of Jim, v₂ = 60 mph

It is assumed that, Mary and Jim leave at the same time. After one hour, Jim is 10 miles ahead.

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Distance travelled by Mary, d₂ = 50t

The distance between Mary and Jim is greater than or equal to 100 miles.

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7 0
3 years ago
A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli
Art [367]

Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

   centripetal force acting on satellite = \dfrac{mv^2}{r}

     gravitational force = \dfrac{GMm}{r^2}

    equating both the above equation

    \dfrac{mv^2}{r} = \dfrac{GMm}{r^2}

      v = \sqrt{\dfrac{GM}{r}}

      v = \sqrt{\dfrac{6.67 \times 10^{-11}\times 5.972 \times 10^{24}}{2 \times 6.4 \times 10^6}}

          v = 5578.5 m/s

b) T= \dfrac{2\pi\ r}{v}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

          T = 14416.92 s

          T = \dfrac{14416.92}{3600}\ hr

          T = 4 hr

c) gravitational force acting

  F = \dfrac{GMm}{r^2}

  F = \dfrac{6.67 \times 10^{-11}\times 545 \times 5.972 \times 10^{24} }{(6.46 \times 10^6)^2}

     F = 5202 N

4 0
3 years ago
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