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3 years ago

A dog walks 12 meters to the west and then 16 meters back to the east

Physics

2 answers:

kupik [55]3 years ago

5 0

4 meters east dude woot

Anna71 [15]3 years ago

4 0

Answer:

Explanation:

4 meters east

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An experimenter finds that standing waves on a string fixed at both ends occur at 24 Hz and 32 Hz , but at no frequencies in bet

Vera_Pavlovna [14]

Answer:

8 Hz

Explanation:

Given that

Standing wave at one end is 24 Hz

Standing wave at the other end is 32 Hz.

Then the frequency of the standing wave mode of a string having a length, l, is usually given as

f(m) = m(v/2L), where in this case, m could be 1. 2. 3. 4 etc

Also, another formula is given as

f(m) = m.f(1), where f(1) is the fundamental frequency..

Thus, we could say that

f(m+1) - f(m) = (m + 1).f(1) - m.f(1) = f(1)

And as such,

f(1) = 32 - 24

f(1) = 8 Hz

Then, the fundamental frequency needed is 8 Hz

4 0

3 years ago

Why is the law of gravity a scientific law

Yuri [45]

An object in motion will stay in motion unless acted upon another force.

Newton used this to prove that gravity existed. Without an unseen force, we could throw a ball and it would go on forever correct? Unless there was something to pull it down, in this case, gravity.

6 0

3 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$