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2 years ago

A dog is 60m away while moving at constant velocity of 10m/s towards you. Where is the dog after 4 seconds?

Physics

2 answers:

Kitty [74]2 years ago

6 0

The dog is 20m away.
if the dog moves 10m/s, multiply 10 by 4, subtract 40 from 60. 20

evablogger [386]2 years ago

4 0

20m away

the dog was 60m away from. you subtract 40m since it is 10m/s x 4 seconds

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Answer:

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3 years ago

On the interstate, the speed limit is 60 mi/h (about 100 km/h) a skilled driver can safely decelerate at about 6.1m/s^2. How lon

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Answer:

4.56 seconds

63.25 m

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 60 km/h = 100 km/h = 100000/3600 = 27.78 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -6.1 m/s²

Time taken by the car to stop

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-27.78}{-6.1}\\\Rightarrow t=4.56\ s$

Time taken by the car to stop is 4.56 seconds

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3 years ago

As you know, a common example of a harmonic oscillator is a mass attached to a spring. In this problem, we will consider a horiz

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E= U + K = $\frac{1}{2}$ kx² + $\frac{1}{2}$ mv²

where

'k' represents the spring constant

'x' is the compression/stretching of the spring with respect to its equilibrium position

'm' is the mass of the block attached to the spring

and 'v' is the speed of the block

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x= Asin(ωt+∅)

where

amplitude is 'A'

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency of the motion

t is the time

$\phi$ is the phase (we can take $\phi$ =0 )

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According to the law of conservation of the mechanical energy, this energy must be equal to the energy of the system at its maximum displacement (1), so we can write

$\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2$

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$v_{max}=\sqrt{\frac{kA^2}{m}}=\sqrt{\frac{k}{m}}A=\omega A$

<h2> $v_{max}=\omega A$ </h2>

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