What metals does a magnetic attract

For no apparent reason, a poodle is running at a constant speed of 5.00 m/s in a circle with radius 2.9 m . Let v⃗ 1 be the velo

Nostrana [21]

At a constant speed of 5.00 m/s, the speed at which the poodle completes a full revolution is

$\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\left(\dfrac{1\,\mathrm{rev}}{2\pi(2.9\,\mathrm m)}\right)\approx0.2744\,\dfrac{\mathrm{rev}}{\mathrm s}$

so that its period is $T=3.644\,\frac{\mathrm s}{\mathrm{rev}}$ (where 1 revolution corresponds exactly to 360 degrees). We use this to determine how much of the circular path the poodle traverses in each given time interval with duration $\Delta t$ . Denote by $\theta$ the angle between the velocity vectors (same as the angle subtended by the arc the poodle traverses), then

$\Delta t=0.4\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.4\,\mathrm s}\theta\implies\theta\approx39.56^\circ$

$\Delta t=0.2\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.2\,\mathrm s}\theta\implies\theta\approx19.78^\circ$

$\Delta t=7\times10^{-2}\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{7\times10^{-2}\,\mathrm s}\theta\implies\theta\approx6.923^\circ$

We can then compute the magnitude of the velocity vector differences $\Delta\vec v$ for each time interval by using the law of cosines:

$|\Delta\vec v|^2=|\vec v_1|^2+|\vec v_2|^2-2|\vec v_1||\vec v_2|\cos\theta$

$\implies|\Delta\vec v|=\begin{cases}3.384\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.4\,\mathrm s\\1.718\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.2\,\mathrm s\\0.6038\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

and in turn we find the magnitude of the average acceleration vectors to be

$\implies|\vec a|=\begin{cases}8.460\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.4\,\mathrm s\\8.588\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.2\,\mathrm s\\8.625\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

So that takes care of parts A, C, and E. Unfortunately, without knowing the poodle's starting position, it's impossible to tell precisely in what directions each average acceleration vector points.

5 0

3 years ago

It takes 120 minutes for a man to ride his bicycle up the road to Alpe d'Huez in France. The vertical height of the climb is 1,1

Greeley [361]

First solve the potential energy of the biker. using the fomula:
PE = mgh
where m is the mass of the object
g is the acceleration due to gravity ( 9.81 m/s2)
h is the height

PE = 96 kg ( 1120 m ) ( 9.81 m/s2)
PE = 1054771.2 J
then power = Work / time
P = 1054771.2 J / ( 120 min ) ( 60 s / 1 min)
P = 146.5 W

8 0

4 years ago

Superman throws a boulder of weight 3800 n at an adversary on the surface of the earth, where the magnitude of the acceleration

evablogger [386]

The weight of the boulder is 3800 N, therefore its mass i s
3800/9.8 = 387.755 kg

If the horizontal acceleration is 12.8 m/s², then the horizontal force applied is
F = (387.755 kg)*(12.8 m/s²) = 4.963 x 10³ n = 4.963 kN

Answer: 4.963 kN

5 0

4 years ago

Which characteristic must a food have to receive a “natural” label from the FDA ?

adelina 88 [10]

No artificial ingredients
i only know that because my mom was vegan for two years

4 0

4 years ago

Make a 29 day timeline. Along the timeline draw and label the phases of the Moon starting with a full moon, crescent, first quar

Yanka [14]

Ummm.... this is a type of question that u have to do by yourself for class. Good luck ;)!!!!! Plz mark as brainliest

5 0

3 years ago