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antoniya [11.8K]
3 years ago
7

A dog runs after a car. The car is travelling at an average speed of 5 m/s.

Physics
2 answers:
n200080 [17]3 years ago
3 0

Answer:

<h2>The answer is 4 s</h2>

Explanation:

The time taken can be found by using the formula

t =  \frac{d}{v}  \\

where

d is the distance

v is the velocity

From the question we have

t =  \frac{20}{5}  \\

We have the final answer as

<h3>4 s</h3>

Hope this helps you

Lina20 [59]3 years ago
3 0

Answer:

4 secs

Explanation:

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kaheart [24]

To solve this exercise it is necessary to use the concepts related to Difference in Phase.

The Difference in phase is given by

\Phi = \frac{2\pi \delta}{\lambda}

Where

\delta = Horizontal distance between two points

\lambda = Wavelength

From our values we have,

\lambda = 500nm = 5*10^{-6}m

\theta = 1\°

The horizontal distance between this two points would be given for

\delta = dsin\theta

Therefore using the equation we have

\Phi = \frac{2\pi \delta}{\lambda}

\Phi = \frac{2\pi(dsin\theta)}{\lambda}

\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}

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6 0
3 years ago
Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the
Doss [256]

Answer:

v_{su} = 19.44 m/s

Explanation:

m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg

T=9.29x10^8\\r_{o}=1.43x10^{12}

If the sun considered as x=0 on the axis to put the center of the mass as a:

m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}

solve to r1

r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}

r_1=1.428x10^9m

Now convert to coordinates centered on the center of mass.  call the new coordinates x' and y' (we won't need y').  Now since in the sun centered coordinates the angular momentum was  

L = \frac{m_{sa}*2*pi*r_1^2}{T}

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum.  So that means

L_{sun}=\frac{m_{sa}*2*\pi *( 2r_{o}*r_1 -r_1^2)}{T}

Since

L_{su}= m_{su}*v_{su}*r_1

then

v_{su}=\frac{m_{sa}*2*pi*(2r_{o}*r_{1}-r_{1}^2)}{T*m_{sa}*r_1}

v_{su} = 19.44 m/s

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3 years ago
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bija089 [108]

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You rapidly swirl a mixture of substances with different densities
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6 0
3 years ago
At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 40.0 N at the s
Mumz [18]

Answer

acceleration due to gravity on Jupiter's moon,g' = 1.81 m/s²

weight of water melon on earth, W = 40 N

acceleration due to gravity on earth, g = 9.8 m/s²

a) Mass on the earth surface

    M = \dfrac{W}{g}

    M = \dfrac{40}{9.8}

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b) Mass on the surface of Lo

 Mass of an object remain same.

  Hence, mass of object at the surface of Lo = 4.08 Kg.

c) Weight at the surface of Lo

   W' = m g'

   W' =4.08 x 1.81

   W' = 7.38 N

8 0
3 years ago
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